Optimization Problems

1.        A manufacturer wants to design an open box having a square base and a surface area of $ \displaystyle 108$ square inches, as shown in figure . What dimensions will produce a box with maximum volume?

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$ \displaystyle \begin{array}{l}\text{ Total}\ \text{surface}\ \text{area =108 i}{{\text{n}}^{\text{2}}}\\\\\text{ (area of base)+(area of four sides) = 108}\\\\\ \ \ \ {{x}^{2}}+4xh=108\end{array}$

$ \displaystyle \ \ \ \ h=\frac{{108-{{x}^{2}}}}{{4x}}$

$ \displaystyle \begin{array}{l}\text{ Let the volume of the box be }V.\\\\\therefore \ \ V={{x}^{2}}h\end{array}$

$ \displaystyle \therefore \ \ V={{x}^{2}}\left( {\frac{{108-{{x}^{2}}}}{{4x}}} \right)=27x-\frac{1}{4}{{x}^{3}}$

$ \displaystyle \therefore \ \ \frac{{dV}}{{dx}}=27-\frac{3}{4}{{x}^{2}}$

$ \displaystyle \ \ \ \ \frac{{dV}}{{dx}}=0\ \text{when}$

$ \displaystyle \ \ \ \ 27-\frac{3}{4}{{x}^{2}}=0$

$ \displaystyle \therefore \ \ {{x}^{2}}=36\Rightarrow x=6\,\ \ \left[ {\because x>0} \right]$

$ \displaystyle \ \ \ \ \frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}=-\frac{3}{2}x$

$ \displaystyle \therefore \ \ {{\left. {\frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}} \right|}_{{x=6}}}=-\frac{3}{2}(6)=-9<0$

$ \displaystyle \therefore \ \ V\ \text{is maximum when }x=6\ \text{cm}\text{.}$

$ \displaystyle \therefore \ \ h=\frac{{108-{{6}^{2}}}}{{24}}=3\ \text{cm}$

$ \displaystyle \begin{array}{l}\therefore \ \ \text{The box has maximum volume when }\\\ \ \ \ x=6\ \text{cm and }h=3\ \text{cm}.\end{array}$

2.         Which points on the curve $ \displaystyle y = 4 − x^2$ are closest to the point $ \displaystyle (0, 2)?$

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$ \displaystyle \begin{array}{l}\text{ Let (}x,y)\ \text{be the required point on the curve}\text{.}\\\\\text{ Curve: }y=4-{{x}^{2}}\\\\\ \ \ \ \text{Let the distance between (0,2) and (}x,y)\ \text{be}\ s.\\\\\therefore \ \ \ s=\sqrt{{{{{(x-0)}}^{2}}+{{{(y-2)}}^{2}}}}\ \left[ {\text{using distance formula}} \right]\\\\\therefore \ \ \ s=\sqrt{{{{x}^{2}}+{{{(4-{{x}^{2}}-2)}}^{2}}}}\\\\\therefore \ \ \ s=\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}\end{array}$

$ \displaystyle \therefore \ \ \ \frac{{ds}}{{dx}}=\frac{{4{{x}^{3}}-6x}}{{2\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}$

$ \displaystyle \therefore \ \ \ \frac{{ds}}{{dx}}=\frac{{x(2{{x}^{2}}-3)}}{{\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}$

$ \displaystyle \ \ \ \ \ \frac{{ds}}{{dx}}=0\ \text{when}$

$ \displaystyle \ \ \ \ \ \frac{{x(2{{x}^{2}}-3)}}{{\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}=0$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Since }\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}\ne 0,\\\\\ \ \ \ \ x(2{{x}^{2}}-3)=0\end{array}$

$ \displaystyle \therefore \ \ \ x=0\ (\text{or) }x=-\sqrt{{\frac{3}{2}}}\ (\text{or)}\ x=\sqrt{{\frac{3}{2}}}\ $

$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}=\frac{{2{{x}^{6}}-9{{x}^{4}}+24{{x}^{2}}-12}}{{{{{\left( {{{x}^{4}}-3{{x}^{2}}+4} \right)}}^{{\frac{3}{2}}}}}}$

$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=0}}}=-\frac{3}{2}<0$

$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=-\sqrt{{\frac{3}{2}}}}}}=\frac{{12}}{{\sqrt{7}}}>0$

$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=\sqrt{{\frac{3}{2}}}}}}=\frac{{12}}{{\sqrt{7}}}>0$

$ \displaystyle \therefore \ \ s\text{ is maximum volume when }$

$ \displaystyle \ \ \ \ x=-\sqrt{{\frac{3}{2}}}\text{ and }x=-\sqrt{{\frac{3}{2}}}.$

$ \displaystyle \therefore \ \ \text{When }x=-\sqrt{{\frac{3}{2}}},y=\frac{5}{2}.$

$ \displaystyle \ \ \ \ \text{When }x=\sqrt{{\frac{3}{2}}},y=\frac{5}{2}$

$ \displaystyle \therefore \ \ \text{The closest ponits on the curve are}$

$ \displaystyle \ \ \ \ \left( {-\sqrt{{\frac{3}{2}}},\frac{5}{2}} \right)\text{and }\left( {\sqrt{{\frac{3}{2}}},\frac{5}{2}} \right).$

3.        A rectangular page is to contain $ \displaystyle 24$ square inches of print. The margins at the top and bottom of the page are to be $ \displaystyle 1 \frac{1}{2}$ inches, and the margins on the left and right are to be $ \displaystyle 1$ inch (see Figure). What should the dimensions of the page be so that the least amount of paper is used?

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$ \displaystyle \begin{array}{l} \ \ \ \text{Let the length and width of the printed }\\\ \ \ \text{area be }x\ \text{and }y\text{ respectively}\text{.}\\\\\text{ By the problem, }xy=24\end{array}$

$ \displaystyle \therefore \ \ y=\frac{{24}}{x}$

$ \displaystyle \ \ \ \ \text{Let the area of the paper needed be }A.$

$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {A=(x+3)(y+2)} \end{array}$

$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {A=(x+3)(\frac{{24}}{x}+2)} \end{array}=30+2x+\frac{{72}}{x}$

$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=2-\frac{{72}}{{{{x}^{2}}}}$

$ \displaystyle \ \ \ \ \ \frac{{dA}}{{dx}}=0\ \text{when }2-\frac{{72}}{{{{x}^{2}}}}=0.$

$ \displaystyle \begin{array}{l}\therefore \ \ \ {{x}^{2}}=36\\\\\ \ \ \ \ x=6\ \ \ (\because x>0)\end{array}$

$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}=\frac{{144}}{{{{x}^{3}}}}$

$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}} \right|}_{{x=6}}}=\frac{{144}}{{{{6}^{3}}}}>0$

$ \displaystyle \therefore \ \ A\text{ is maximum volume when }x=6$

$ \displaystyle \therefore \ \ \text{ }y=\frac{{24}}{6}=4\ \text{cm}.$

$ \displaystyle \therefore \ \ \text{The dimenssions of the paper}=4\ \text{cm}\times 6\text{cm}.$

4.         Two posts, one $ \displaystyle 12$ feet high and the other $ \displaystyle 28$ feet high, stand $ \displaystyle 30$ feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post. Where should the stake be placed to use the least amount of wire?

 

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$ \displaystyle \begin{array}{l} \ \ \ \text{Let the length of wire required to attach 12-foot post be }y\ \ \ \\\ \ \ \ \text{and that required to attach 28-foot post be }z\text{.}\\\\\ \ \ \text{Let the distance of the stack from 12-foot post be }x.\\\\\therefore \ \ {{x}^{2}}+{{12}^{2}}={{y}^{2}}\Rightarrow y=\sqrt{{144+{{x}^{2}}}}\\\\\ \ \ \ {{(30-x)}^{2}}+{{28}^{2}}={{z}^{2}}\Rightarrow z=\sqrt{{{{x}^{2}}-60x+1684}}.\\\\\ \ \ \ \text{Let the total length of wire required be }W.\end{array}$

$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {W=x+y} \end{array}=\sqrt{{144+{{x}^{2}}}}+\sqrt{{{{x}^{2}}-60x+1684}}.$

$ \displaystyle \therefore \ \ \ \frac{{dW}}{{dx}}=\frac{x}{{\sqrt{{144+{{x}^{2}}}}}}+\frac{{x-30}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}$

$ \displaystyle \ \ \ \ \ \frac{{dW}}{{dx}}=0\ \text{when }\frac{x}{{\sqrt{{144+{{x}^{2}}}}}}+\frac{{x-30}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}=0.$

$ \displaystyle \therefore \ \ \ \frac{x}{{\sqrt{{144+{{x}^{2}}}}}}=\frac{{30-x}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ x\sqrt{{{{x}^{2}}-60x+1684}}=(30-x)\sqrt{{{{x}^{2}}+144}}\\\\\ \ \ \ \ {{x}^{2}}({{x}^{2}}-60x+1684)={{(30-x)}^{2}}(x2+144)\\\\\ \ \ \ \ {{x}^{4}}-60{{x}^{3}}+1684{{x}^{2}}={{x}^{4}}-60{{x}^{3}}+1044{{x}^{2}}-8640x+129,600\\\\\ \ \ \ \ 640{{x}^{2}}+8640x-129,600=0\\\\\ \ \ \ \ 2{{x}^{2}}+27x-405=0\\\\\ \ \ \ \ (x-9)(2x+45)=0\\\\\ \ \ \ \ \text{Since }x>0,\ x=9\\\\\ \ \ \ \ W=\sqrt{{144+{{x}^{2}}}}+\sqrt{{{{x}^{2}}-60x+1684}}.\\\\\therefore \ \ \ \text{When }x=0,\\\\\ \ \ \ \ W=\sqrt{{144}}+\sqrt{{1684}}\\\\\ \ \ \ \ \ \ \ \ \ \ =12+41.04=53.04\\\\\ \ \ \ \ \text{When }x=9,\\\\\ \ \ \ \ W=\sqrt{{144+81}}+\sqrt{{81-540+1684}}\\\\\ \ \ \ \ \ \ \ \ =15+35=50\\\\\ \ \ \ \ \text{When }x=30,\\\\\ \ \ \ \ W=\sqrt{{144+900}}+\sqrt{{900-1800+1684}}\\\\\ \ \ \ \ \ \ \ \ \ \ =32.31+28=60.31\\\\\therefore \ \ \ W\text{ is maximum when }x=9.\\\\\therefore \ \ \ \text{The}\ \text{wires}\ \text{should}\ \text{be}\ \text{staked}\ \text{at}\ \text{9}\ \text{feet}\ \text{from}\ \text{the}\ \text{12-foot}\ \text{pole}.\end{array}$

5.       The sum of the perimeters of a circle and square is $ \displaystyle k,$ where $ \displaystyle k$ is some constant. Prove that the sum of their areas is least, when the side of the square is double the radius of the circle.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Let the radius of the circle be }r\ \text{and the length }\\\ \ \ \ \ \text{of each side of the square be }x.\\\\\ \ \ \ \ \text{Sum of perimeters}=k\ \ \left[ {\text{given}} \right]\\\\\therefore \ \ \ \ 4x+\text{ }2\pi r=k\end{array}$

$ \displaystyle \therefore \ \ \ r=~\frac{{k-4x}}{{2\pi }}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Let the total are be }A.\\\\\therefore \ \ \ A={{x}^{2}}+\pi {{r}^{2}}\end{array}$

$ \displaystyle \therefore \ \ \ A={{x}^{2}}+\pi {{\left( {\frac{{k-4x}}{{2\pi }}} \right)}^{2}}$

$ \displaystyle \therefore \ \ \ A={{x}^{2}}+\frac{{{{{(k-4x)}}^{2}}}}{{4\pi }}$

$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=2x+\frac{{2(-4)(k-4x)}}{{4\pi }}$

$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=2\left( {x+\frac{{4x-k}}{\pi }} \right)$

$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=\frac{2}{\pi }\left[ {(\pi +4)x-k} \right]$

$ \displaystyle \ \ \ \ \ \frac{{dA}}{{dx}}=0\ \text{when}\ \frac{2}{\pi }\left[ {(\pi +4)x-k} \right]=0$

$ \displaystyle \therefore \ \ \ x=\frac{k}{{\pi +4}}$

$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}=\frac{{2(\pi +4)}}{\pi }>0$

$ \displaystyle \therefore \ \ \ A\ \text{is minimum when }x=\frac{k}{{\pi +4}}.$

$ \displaystyle \therefore \ \ \ r=\frac{1}{{2\pi }}[k-\frac{{4k}}{{\pi +4}}]$

$ \displaystyle \ \ \ \ \ \ \ \ =\frac{{\pi k+4k-4k}}{{2\pi (\pi +4)}}$

$ \displaystyle \ \ \ \ \ \ \ \ =~\frac{1}{2}\left( {\frac{k}{{\pi +4}}} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ =~\frac{1}{2}x$

$ \displaystyle \begin{array}{l}\therefore \ \ x=2r\\\\\ \ \ \text{Hence the sum of their areas is least},\\\ \ \ \text{when the side of the square is double }\\\ \ \ \text{the radius of the circle}.\end{array}$

6.        Find two positive numbers that the sum of the first number cubed and the second number is $ \displaystyle 500$ and the product is a maximum.

7.        An open box of maximum volume is to be made from a square piece of material, $ \displaystyle 24$ inches on a side, by cutting equal squares from the corners and turning up the sides (see figure). How much material should be cut?

 

8.        A Norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window (see figure). Find the dimensions of a Norman window of maximum area when the total perimeter is $ \displaystyle 16$ feet.

9.        Find the area of the largest isosceles triangle that can be inscribed in a circle of radius $ \displaystyle 6$ (see figure).
(a) Solve by writing the area as a function of $ \displaystyle h$.

(b) Solve by writing the area as a function of $ \displaystyle α$.

(c) Identify the type of triangle of maximum area.

10.     Twenty feet of wire is to be used to form two figures: equilateral triangle and square, how much wire should be used for each figure so that the total enclosed area is maximum?

11.     A right triangle is formed in the first quadrant by the $ \displaystyle x$- and $ \displaystyle y$-axes and a line through the point $ \displaystyle (1, 2)$ (see figure).

(a) Write the length $ \displaystyle L$ of the hypotenuse as a function of $ \displaystyle x.$

(b) Find the vertices of the triangle such that its area is a minimum.

တြက္ၾကည့္ပါ။ အေျဖမ်ားကို ေနာက္ရက္တြင္ တင္ေပး ပါမည္။

Definition of a Limit

$ \displaystyle f(x)=\frac{{{{x}^{3}}-1}}{{x-1}}$ ဆိုတဲ့ function တစ္ခုကို စဥ္စားၾကည့္ရေအာင္ $ \displaystyle f(x)$ ဟာ ထည့္လိုက္တဲ့ $ \displaystyle x$ တန္ဖိုးတိုင္းအတြက္ output (image) တစ္ခု ထုတ္ေပးမွာေပါ့။ 

အေပၚက slider ကို ဆြဲၾကည့္ပါ။ 

ဒါေပမယ့္ $ \displaystyle x=1$ ျဖစ္သြားရင္ေတာ့ $ \displaystyle f(x)$ ဟာ ဘာလုပ္ရမွန္းမသိေတာ့ပါဘူး။ ဘာလို႔လဲ ဆိုေတာ့ $ \displaystyle f(1)=\frac{{1-1}}{{1-1}}=\frac{0}{0}$ ဆိုတာကို $ \displaystyle f(x)$ က နားမလည္ေတာ့ဘူးေလ။ 

ဘယ္ကိန္းကို မဆို ာ့ $ \displaystyle 0$ နဲ႔ စားျခင္းက အဓိပါယ္မရွိလို႔ပါပဲ။ ဒါဆိုရင္ $ \displaystyle x$ က $ \displaystyle 1$ ျဖစ္လို႔ မရဘူးေပါ့။ $ \displaystyle x=1$ မျဖစ္ရဘူးဆိုရင္ ဘယ္ေလာက္အထိ ျဖစ္ခြင့္ ရွိပါသလဲ။ 

$ \displaystyle x=1$ မျဖစ္ဖို႔ပဲ လိုတာ။ ။ $ \displaystyle 1$ မဟုတ္တဲ့ ဘယ္ကိန္းစစ္မဆို ျဖစ္လို႔ ရပါတယ္။ 

ကၽြန္ေတ့ာ္ ေက်ာင္းသားေတြကို သင္လိုက္သလို ေျပာရရင္ ။ $ \displaystyle 1$ ကို ထိလို႔ မရဘူး ထိရင္ ေရွာ့ပဲ။ မထိနဲ႔လို႔ ေျပာတာ.. မကပ္နဲ႔ လို႔ မေျပာတဲ့ အတြက္ ။ $ \displaystyle 1$ ရဲ့ အနားကို ကပ္ႏိုင္ သေလာက္ထိ ကပ္ၾကည့္မယ္..။ 

တကယ္ေတာ့ ကပ္ႏိုင္သေလာက္ဆိုတာ ဘယ္ေလာက္လဲ တိတိက်က် တန္ဖိုးေတာ့ မရွိဘူးေပါ့။ ေအာက္က ဇယားကိုၾကည့္ပါ။

$ \displaystyle x$ က $ \displaystyle 1$ အနားကို နီးကပ္လာေလေလ ... $ \displaystyle f(x)$ က $ \displaystyle 3$ အနားကို ေရာက္လာေလေလေပါ့။ 

$ \displaystyle x=1$ ဆိုၿပီး တိုက္ရိုက္ထည့္ လိုက္ရင္ေတာ့ ေရွာ့ပဲ...။ စိတ္ရွည္ရွည္နဲ႔ ကပ္ၾကည့္ လိုက္ရင္ေတာ့ ရလဒ္တစ္ခု ထြက္လာတာေပါ့။ 

ဒါကို Calculus မွာ $ \displaystyle f(x)$ ရဲ့ limit လို႔ သတ္မွတ္ၿပီး သေကၤတ အားျဖင့္ $ \displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,f(x)=3$ (The limit of f(x) as x approaches 1 is 3) လို႔ သတ္မွတ္ပါတယ္။ 

လက္ေတြ႔ပုစာၦေျဖရွင္းရာမွာေတာ့ ပုစာၦတိုင္းကို အခုလို ခ်ဥ္းကပ္ေျဖရွင္းဖို႔ မျဖစ္ႏိုင္ေတာ့တဲ့အတြက္ တိုက္ရိုက္အစားသြင္းၾကည့္တဲ့ အခါ $ \displaystyle f(1)=\frac{0}{0}$ (indeterminate form လို႔ ေခၚပါတယ္) ျဖစ္ရင္ discontinuity ကေန continuous condition (အစားသြင္းလို႔ ရမယ့္ အေျခအေနကို) ေျပာင္းလဲ ေျဖရွင္းၿပီးမွ တြက္ၾကရတာေပါ့။ 

အခုလို တြက္ပါမယ္။

$ \displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-1}}{{x-1}}=\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(x-1)({{x}^{2}}+x+1)}}{{x-1}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,({{x}^{2}}+x+1)\ \ \ \ \ \ \left[ {\text{continuous}\ \text{condition}} \right]$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1+1+1\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 3\end{array}$

Definition of a limit

If $ \displaystyle f(x)$ becomes arbitrarily close to a single number $ \displaystyle L$ as $ \displaystyle x$ approaches $ \displaystyle c$ from either side, then the limit of $ \displaystyle f(x),$ as $ \displaystyle x$ approaches $ \displaystyle c$, is $ \displaystyle L$. This limit is written as

$ \displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,f(x)=L$

Limit ပုစာၦမ်ား တြက္ရမွာ သိထားရမယ့္ properties ေတြကို ေတာ့ 👉 ဒီေနရာမွာ 👈 ဖတ္ၾကည့္ပါ။

Calculus - Limits

Limit ပုစ္ဆာများ တွက်ရာတွင် သိရှိနားလည်ထားရမည့် ဂုဏ်သတ္တိများ ကို လေ့လာသိရှိပြီးမှ ပုစ္ဆာများကို တွက်လျှင် ပို၍ နားလည်လွယ်ကူစေပါသည်။

1.      $ \displaystyle \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-1}}{{x-1}}$

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$ \displaystyle \ \ \ \ \ \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}-1}}{{x-1}}=\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(x-1)({{x}^{2}}+x+1)}}{{x-1}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\left[ {{{x}^{2}}+x+1} \right]$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\left( {{{x}^{2}}} \right)+\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( x \right)+\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( 1 \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1+1+1$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 3$

2.      $ \displaystyle \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{x-1}}{{\sqrt[3]{x}-1}}$

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$ \displaystyle \ \ \ \ \ \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{x-1}}{{\sqrt[3]{x}-1}}\ \ =\underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{{\left( {\sqrt[3]{x}} \right)}}^{3}}-{{1}^{3}}}}{{\sqrt[3]{x}-1}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{(\sqrt[3]{x}-1)\left[ {{{{\left( {\sqrt[3]{x}} \right)}}^{2}}+\sqrt[3]{x}+1} \right]}}{{\sqrt[3]{x}-1}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,\left[ {{{{\left( {\sqrt[3]{x}} \right)}}^{2}}+\sqrt[3]{x}+1} \right]$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to 1}}{\mathop{{\lim }}}\,{{\left( {\sqrt[3]{x}} \right)}^{2}}+\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( {\sqrt[3]{x}} \right)+\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( 1 \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1+1+1$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 3$

3.      $ \displaystyle \underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}-9}}{{{{x}^{2}}+\sqrt{3}x-6}}\ $

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$ \displaystyle \underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}-9}}{{{{x}^{2}}+\sqrt{3}x-6}}\ \ =\underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{({{x}^{2}}-3)({{x}^{2}}+3)}}{{(x-\sqrt{3})(x+2\sqrt{3})}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{(x-\sqrt{3})(x+\sqrt{3})({{x}^{2}}+3)}}{{(x-\sqrt{3})(x+2\sqrt{3})}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \underset{{x\to \sqrt{3}}}{\mathop{{\lim }}}\,\frac{{(x+\sqrt{3})({{x}^{2}}+3)}}{{(x+2\sqrt{3})}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{{(\sqrt{3}+\sqrt{3})(3+3)}}{{(\sqrt{3}+2\sqrt{3})}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{{12\sqrt{3}}}{{3\sqrt{3}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 4$

4.      $ \displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{{{x}^{2}}-x-6}}\ $

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$ \displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{{{x}^{2}}-x-6}}\ =\underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{(x+2)({{x}^{2}}+5)}}{{(x+2)(x-3)}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to -2}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+5}}{{x-3}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{4+5}}{{-2-3}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{9}{5}$

5.      $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{3{{x}^{3}}-x-6}}\ $

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$ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}+2{{x}^{2}}+5x+10}}{{3{{x}^{3}}-x-6}}\ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{3}}\left( {1+\frac{2}{x}+\frac{5}{{{{x}^{2}}}}+\frac{{10}}{{{{x}^{3}}}}} \right)}}{{{{x}^{3}}\left( {3-\frac{1}{{{{x}^{2}}}}-\frac{6}{{{{x}^{3}}}}} \right)}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{1+\frac{2}{x}+\frac{5}{{{{x}^{2}}}}+\frac{{10}}{{{{x}^{3}}}}}}{{3-\frac{1}{{{{x}^{2}}}}-\frac{6}{{{{x}^{3}}}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{1+0+0+0}}{{3-0-0}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{3}$

6.      $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{{-3}}}+5{{x}^{{-2}}}+8}}{{5{{x}^{{-3}}}-{{x}^{{-1}}}-2}}\ $

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$ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{{-3}}}+5{{x}^{{-2}}}+8}}{{5{{x}^{{-3}}}-{{x}^{{-1}}}-2}}\ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{{{{x}^{3}}}}+\frac{5}{{{{x}^{2}}}}+8}}{{\frac{5}{{{{x}^{3}}}}-\frac{1}{x}-2}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{{{{x}^{3}}}}+\frac{5}{{{{x}^{2}}}}+8}}{{\frac{5}{{{{x}^{3}}}}-\frac{1}{x}-2}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{0+0+8}}{{0-0-2}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-4$

7.       $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2{{x}^{4}}+3{{x}^{3}}-7{{x}^{2}}-12x-4}}{{{{x}^{3}}-8}}\ $

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$ \displaystyle \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2{{x}^{4}}+3{{x}^{3}}-7{{x}^{2}}-12x-4}}{{{{x}^{3}}-8}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}\left( {2+\frac{3}{x}-\frac{7}{{{{x}^{2}}}}-\frac{{12}}{{{{x}^{3}}}}-\frac{4}{{{{x}^{4}}}}} \right)}}{{{{x}^{3}}\left( {1-\frac{8}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{x\left( {2+\frac{3}{x}-\frac{7}{{{{x}^{2}}}}-\frac{{12}}{{{{x}^{3}}}}-\frac{4}{{{{x}^{4}}}}} \right)}}{{\left( {1-\frac{8}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,x\cdot \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2+\frac{3}{x}-\frac{7}{{{{x}^{2}}}}-\frac{{12}}{{{{x}^{3}}}}-\frac{4}{{{{x}^{4}}}}}}{{1-\frac{8}{{{{x}^{3}}}}}}\ $

$ \displaystyle =\infty $

8.       $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+3x+2}}{{{{x}^{3}}+32{{x}^{2}}-5x-20}}\ $

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$ \displaystyle \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+3x+2}}{{{{x}^{3}}+32{{x}^{2}}-5x-20}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}\left( {1+\frac{3}{x}+\frac{2}{{{{x}^{2}}}}} \right)}}{{{{x}^{3}}\left( {1+\frac{{32}}{x}-\frac{5}{{{{x}^{2}}}}-\frac{{20}}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\left( {1+\frac{3}{x}+\frac{2}{{{{x}^{2}}}}} \right)}}{{x\left( {1+\frac{{32}}{x}-\frac{5}{{{{x}^{2}}}}-\frac{{20}}{{{{x}^{3}}}}} \right)}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{1}{x}\cdot \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{1+\frac{3}{x}+\frac{2}{{{{x}^{2}}}}}}{{1+\frac{{32}}{x}-\frac{5}{{{{x}^{2}}}}-\frac{{20}}{{{{x}^{3}}}}}}\ $

$ \displaystyle =0$

9.       $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left( {\sqrt{{{{x}^{2}}+2x+1}}-\sqrt{{{{x}^{2}}+1}}} \right)\ $

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$ \displaystyle \ \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left( {\sqrt{{{{x}^{2}}+2x+1}}-\sqrt{{{{x}^{2}}+1}}} \right)\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\left[ {\left( {\sqrt{{{{x}^{2}}+2x+1}}-\sqrt{{{{x}^{2}}+1}}} \right)\times \frac{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}} \right]\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+2x+1-({{x}^{2}}+1)}}{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}\ $

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{\sqrt{{{{x}^{2}}+2x+1}}+\sqrt{{{{x}^{2}}+1}}}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{\sqrt{{{{x}^{2}}\left( {1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}} \right)}}+\sqrt{{{{x}^{2}}\left( {1+\frac{1}{{{{x}^{2}}}}} \right)}}}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{x\sqrt{{1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}}}+x\sqrt{{1+\frac{1}{{{{x}^{2}}}}}}}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{2x}}{{x\left( {\sqrt{{1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}}}+\sqrt{{1+\frac{1}{{{{x}^{2}}}}}}} \right)}}$

$ \displaystyle =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{2}{{\sqrt{{1+\frac{2}{x}+\frac{1}{{{{x}^{2}}}}}}+\sqrt{{1+\frac{1}{{{{x}^{2}}}}}}}}$

$ \displaystyle =\frac{2}{{\sqrt{{1+0+0}}+\sqrt{{1+0}}}}$

$ \displaystyle =\frac{2}{{\sqrt{1}+\sqrt{1}}}$

$ \displaystyle =1$

10.      $ \displaystyle \ \ \ \ \ \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+3+...+n}}{{{{n}^{2}}}}$

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Since $ \displaystyle 1 + 2 + 3 + … + n$ is an arithmetic series with the first term $ \displaystyle 1$ and the common difference $ \displaystyle 1$ having $ \displaystyle n$ terms.

$ \displaystyle \therefore 1+2+3+\ldots +n=\frac{n}{2}(1+n)$

$ \displaystyle \ \ \ \ \ \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+3+...+n}}{{{{n}^{2}}}}$

$ \displaystyle =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{{n+{{n}^{2}}}}{2}}}{{{{n}^{2}}}}$$ \displaystyle =\frac{1}{2}\cdot \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{n+{{n}^{2}}}}{{{{n}^{2}}}}$

$ \displaystyle =\frac{1}{2}\cdot \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{n+{{n}^{2}}}}{{{{n}^{2}}}}$

$ \displaystyle =\frac{1}{2}\underset{{n\to \infty }}{\mathop{{\cdot \lim }}}\,\frac{{{{n}^{2}}\left( {\frac{1}{n}+1} \right)}}{{{{n}^{2}}}}$

$ \displaystyle =\frac{1}{2}\underset{{n\to \infty }}{\mathop{{\cdot \lim }}}\,\frac{{\frac{1}{n}+1}}{1}$

$ \displaystyle =\frac{1}{2}\left( 1 \right)$

$ \displaystyle =\frac{1}{2}$

11.      $ \displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+4+...+{{2}^{{n-1}}}}}{{{{2}^{n}}+1}}$

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Since $ \displaystyle 1 + 2 + 4 + … + 2^{n – 1}$ is a geometric series with the first term $ \displaystyle 1$ and the common ratio $ \displaystyle 2$ having $ \displaystyle n$ terms.

$ \displaystyle \therefore 1+2+4+\ldots +{{2}^{{n-1}}}=\frac{{1({{2}^{n}}-1)}}{{2-1}}={{2}^{n}}-1$

$ \displaystyle \ \ \ \ \ \ \ \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1+2+4+...+{{2}^{{n-1}}}}}{{{{2}^{n}}+1}}\ \ \ \ $

$ \displaystyle \ \ \ =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{{{2}^{n}}-1}}{{{{2}^{n}}+1}}$

$ \displaystyle \ \ \ =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{{{2}^{n}}(1-\frac{1}{{{{2}^{n}}}})}}{{{{2}^{n}}(1+\frac{1}{{{{2}^{n}}}})}}$

$ \displaystyle \ \ \ =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{1-\frac{1}{{{{2}^{n}}}}}}{{1+\frac{1}{{{{2}^{n}}}}}}$

$ \displaystyle \ \ \ =\frac{{1-0}}{{1+0}}$

$ \displaystyle \ \ \ =1$

12.      $ \displaystyle \ \underset{{t\to \infty }}{\mathop{{\lim }}}\,\frac{{\sqrt{t}+{{t}^{2}}}}{{2t-{{t}^{2}}}}$

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Let $ \displaystyle x=\sqrt{t}$ then $ \displaystyle t = x^2$

As $ \displaystyle t\to \infty, t\to \infty $

$ \displaystyle \therefore \ \ \ \ \ \ \underset{{t\to \infty }}{\mathop{{\lim }}}\,\frac{{\sqrt{t}+{{t}^{2}}}}{{2t-{{t}^{2}}}}$

$ \displaystyle \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{x+{{x}^{4}}}}{{2x-{{x}^{4}}}}$

$ \displaystyle \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{{{x}^{4}}(\frac{1}{{{{x}^{3}}}}+1)}}{{{{x}^{4}}(\frac{2}{{{{x}^{3}}}}-1)}}$

$ \displaystyle \ \ \ =\underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{{\frac{1}{{{{x}^{3}}}}+1}}{{\frac{2}{{{{x}^{3}}}}-1}}$

$ \displaystyle \ \ \ =\frac{{0+1}}{{0-1}}$

$ \displaystyle \ \ \ =-1$

13.      $ \displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{{{4}^{x}}-{{4}^{{-x}}}}}{{{{4}^{x}}+{{4}^{{-x}}}}}$

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$ \displaystyle \ \ \ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{{{4}^{x}}-{{4}^{{-x}}}}}{{{{4}^{x}}+{{4}^{{-x}}}}}$

$ \displaystyle =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{{{4}^{x}}-\displaystyle \frac{1}{{{{4}^{x}}}}}}{{{{4}^{x}}+\displaystyle \frac{1}{{{{4}^{x}}}}}}$

$ \displaystyle =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{\left( {{{4}^{x}}-\displaystyle \frac{1}{{{{4}^{x}}}}} \right)\times \displaystyle \frac{1}{{{{4}^{x}}}}}}{{\left( {{{4}^{x}}+\displaystyle \frac{1}{{{{4}^{x}}}}} \right)\times \displaystyle \frac{1}{{{{4}^{x}}}}}}$

$ \displaystyle =\ \ \underset{{x\to \infty }}{\mathop{{\lim }}}\,\displaystyle \frac{{1-\displaystyle \frac{1}{{{{4}^{{2x}}}}}}}{{1+\displaystyle \frac{1}{{{{4}^{{2x}}}}}}}$

$ \displaystyle \begin{array}{l}=\ \ \displaystyle \frac{{1-0}}{{1+0}}\\\\=1\end{array}$

Calculus : Approximation

1.       Given that $ \displaystyle y = 3x^3 - 7x^2 + 8$, find the value of $ \displaystyle \frac{dy}{dx}$ at the point $ \displaystyle (2, 4)$. Hence, find the approximate increase in $ \displaystyle x$ which will cause $ \displaystyle y$ to increase from $ \displaystyle 4$ to $ \displaystyle 4.04$.

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$ \displaystyle \begin{array}{l}\ \ \ y=3{{x}^{3}}-7{{x}^{2}}+8\\\\\ \ \ \text{Let}\ ({{x}_{0}},{{y}_{0}})=(2,4)\ \text{and}\ {{y}_{1}}=4.04\\\\\therefore \ \delta y={{y}_{1}}-{{y}_{0}}\\\\\ \ \ \ \ \ \ \ =4.04-4\\\\\ \ \ \ \ \ \ \ =0.04\end{array}$

$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=9{{x}^{2}}-14x$

$ \displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{0}},{{y}_{0}})}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{(2,4)}}}=9{{(2)}^{2}}-14(2)=8$

$ \displaystyle \ \ \ \text{By linear approximation,}$

$ \displaystyle \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{0}},{{y}_{0}})}}}\cdot \delta x$

$ \displaystyle \begin{array}{l}\therefore \ 0.04\simeq 8\cdot \delta x\\\\\therefore \delta x\simeq 0.005\\\\\therefore \ x\ \text{is increased by 0}\text{.005}\text{.}\end{array}$

2.       Given that $ \displaystyle y = 2x^2 + 3x$, find the approximate percentage change in $ \displaystyle y$ when $ \displaystyle x$ decreases from $ \displaystyle 2$ to $ \displaystyle 1.97$.

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$ \displaystyle \begin{array}{l}\ \ \ \ y=2{{x}^{2}}+3x\\\\\ \ \ \ \text{Let}\ {{x}_{0}}=2\ \text{and}\ {{x}_{1}}=1.97\\\\\therefore \ \ \delta x=1.97-2=-0.03\\\\\ \ \ \ \text{When}\ {{x}_{0}}=2,\ {{y}_{0}}=14\end{array}$

$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=4x+3$

$ \displaystyle \ \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{2}}=4(2)+3=11$

$ \displaystyle \ \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =11(-0.03)\\\\\ \ \ \ \ \ \ \ \ =-0.33\\\\\ \ \ \ \text{Approximate percentage change in }y\end{array}$

$ \displaystyle \ \ \ \ =\frac{{\delta y}}{{{{y}_{0}}}}\times 100 $ %

$ \displaystyle \ \ \ \ =\frac{{-0.33}}{{14}}\times 100$ %

$ \displaystyle \ \ \ \ =-2.36$ %

$ \displaystyle \therefore \ \ \text{ }y\ \text{is decreased by}\ 2.36$ %.

3.       Use approximation to approximate the following values $ \displaystyle \sqrt{80}$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ y=\sqrt{x}\ \text{and}\ {{x}_{0}}=81\\\\\therefore \ \ {{y}_{0}}=9\ \ \\\\\ \ \ \ \text{Let}\ {{x}_{1}}=\ 80.\\\\\therefore \ \ {{y}_{1}}=\sqrt{{80}}\\\\\therefore \ \ \delta x=80-81=-1\end{array}$

$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=\frac{1}{{2\sqrt{x}}}$

$ \displaystyle \ \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{81}}}=\frac{1}{{2\sqrt{{81}}}}=\frac{1}{{18}}$

$ \displaystyle \ \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$

$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{1}{{18}}(-1)$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =-0.056\\\\\therefore \ \ {{y}_{1}}={{y}_{0}}+\delta y=9-0.056=8.944\\\\\therefore \ \ \sqrt{{80}}\simeq 8.944\end{array}$

4.       The side of a square is $ \displaystyle 5\ \text{cm}$ . How much will the area of the square increase when the side expands by $ \displaystyle 0.01\ \text{cm}$?

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length of the side of a square be }x\\\\\ \ \ \ \operatorname{and}\ \text{the area of the square be }A.\\\\\therefore \ \ A={{x}^{2}}\\\\\ \ \ \ \text{Let }{{x}_{0}}=5\ \text{then}\ {{A}_{0}}=\ 25\\\\\ \ \ \ \text{When the sides expand 0}\text{.01 cm, }\\\\\ \ \ \ \delta x=0.01\ \text{cm}\text{.}\end{array}$

$ \displaystyle \ \ \ \ \frac{{dA}}{{dx}}=2x$

$ \displaystyle \ \ \ \ {{\left. {\frac{{dA}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dA}}{{dx}}} \right|}_{5}}=2(5)=10$

$ \displaystyle \ \ \ \ \delta A\simeq {{\left. {\frac{{dA}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =10(0.01)\\\\\ \ \ \ \ \ \ \ \ =0.1\ \text{c}{{\text{m}}^{2}}\\\\\therefore \ \ \text{The area of the square is increased by }0.1\ \text{c}{{\text{m}}^{2}}.\end{array}$

5.       If the sides of a square are increased by $ \displaystyle 20$ %, by what percentage does the area of the square increase?

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$ \displaystyle \ \ \ \ \text{Solution (1) - Using Approximation}$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length each side of a square be}\ x\\\ \ \ \ \text{and}\ \text{the area be }A.\text{ }\end{array}$

$ \displaystyle \therefore \ \ A={{x}^{2}}.$

$ \displaystyle \therefore \ \ \frac{{dA}}{{dx}}=2x$

$ \displaystyle \ \ \ \ \text{When the lengths of the sides are incresed by 20}\ \text{%,}$

$ \displaystyle \ \ \ \ \frac{{\delta x}}{x}=\text{20 } \text{% }=0.2\Rightarrow \delta x=0.2x$

$ \displaystyle \ \ \ \ \delta A\simeq \frac{{dA}}{{dx}}\times \delta x$

$ \displaystyle \begin{array}{l}\ \ \ \ \delta A\simeq 2x\times 0.2x\\\\\therefore \ \ \delta A\simeq 0.4{{x}^{2}}\\\\\ \ \ \ \delta A\simeq 0.4A\end{array}$

$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}\simeq \text{0}\text{.4}=\text{40}\ \text{%}$

$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 40}\ \text{%.}$

$ \displaystyle \ \ \ \ \text{Solution (2) - Without Calculus}$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length each side of a square be}\ x\\\\\ \ \ \ \text{and}\ \text{the area be }A.\\\\\therefore \ \ {{A}_{0}}={{x}^{2}}\\\\\ \ \ \ \text{When the lengths of the sides are incresed by 20}\ \text{%,}\\\\\ \ \ \ {{x}_{1}}=x+0.2x=1.2x\\\\\therefore \ \ {{A}_{1}}={{(1.2x)}^{2}}=1.44{{x}^{2}}\\\\\ \ \ \ \delta A\simeq {{A}_{1}}-{{A}_{0}}=0.44{{x}^{2}}\\\\\therefore \ \ \ \delta A=0.44A\end{array}$

$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}=\text{0}\text{.44}=\text{44}\ \text{% }$

$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 44}\ \text{%.}$

6.       One side of a rectangle is three times the other. If the perimeter increases by $ \displaystyle 2$ % what is the percentage increase in area?

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{As one side of the rectangle is three times the other},\\\ \ \ \ \text{let the length and the breadth be 3}x\ \text{and}\ x.\end{array}$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ \text{the perimetre and area of the rectangle be }\\\ \ \ \ p\ \text{and }A\ \text{respectively}\text{.}\end{array}$

$ \displaystyle \therefore \ \ p=2(3x+x)=8x\ \text{and}\ A=3{{x}^{2}}.$

$ \displaystyle \therefore \ \ x=\frac{p}{8}\Rightarrow A=3{{\left( {\frac{p}{8}} \right)}^{2}}=\frac{{3{{p}^{2}}}}{{64}}$

$ \displaystyle \therefore \ \ \frac{{dA}}{{dp}}=\frac{{3p}}{{32}}$

$ \displaystyle \ \ \ \ \text{When the perimeter is incresed by 2}$ %,

$ \displaystyle \ \ \ \ \frac{{\delta p}}{p}=\text{2 %}=0.02\Rightarrow \delta p=0.02p$

$ \displaystyle \ \ \ \ \delta A\simeq \frac{{dA}}{{dp}}\times \delta p$

$ \displaystyle \ \ \ \ \delta A\simeq \frac{{3p}}{{32}}\times 0.02p$

$ \displaystyle \therefore \ \ \delta A\simeq \frac{{3{{p}^{2}}}}{{32\times 2}}\times 0.02\times 2$

$ \displaystyle \therefore \ \ \delta A\simeq \frac{{3{{p}^{2}}}}{{64}}\times 0.04$

$ \displaystyle \ \ \ \ \delta A\simeq 0.04A$

$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}\simeq \text{0}\text{.04}=\text{4}$ %

$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 4}$ %.

7.      The kinetic energy $ \displaystyle K$ of a body of mass $ \displaystyle m$ moving with speed $ \displaystyle v$ is given by $ \displaystyle K =\frac{1}{2}mv^2$ . If a body’s speed is increased by $ \displaystyle 1.5$ % what is the approximate percentage change in the kinetic energy?

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$ \displaystyle \ \ \ \ \text{Kinetic Energy}=\ K,\text{ mass}=m,\text{ speed}=v$

$ \displaystyle \ \ \ \ K=\frac{1}{2}m{{v}^{2}}$

$ \displaystyle \ \ \ \ \text{If a body's speed is incresed by 1.5}$ %,

$ \displaystyle \ \ \ \ \frac{{\delta v}}{v}=\text{1}\text{.5}\ \text{%}=0.015\Rightarrow \delta v=0.015v$

$ \displaystyle \ \ \ \ \frac{{dK}}{{dv}}=mv$

$ \displaystyle \ \ \ \ \delta K\simeq \frac{{dK}}{{dv}}\times \delta v$

$ \displaystyle \begin{array}{l}\ \ \ \ \delta K\simeq mv\times \delta v\\\\\therefore \ \ \delta K\simeq mv\times 0.015v\end{array}$

$ \displaystyle \therefore \ \ \delta K\simeq \frac{1}{2}m{{v}^{2}}\times 0.03$

$ \displaystyle \therefore \ \ \delta K\simeq K\times 0.03$

$ \displaystyle \therefore \ \ \frac{{\delta K}}{K}\simeq 0.03=3$ %

$ \displaystyle \therefore \ \ \text{The kinetic energy is increased by}$ 3%.

8.       The two equal sides of an isosceles triangle with fixed base $ \displaystyle b$ are decreasing at the rate of $ \displaystyle 3\ \text{cm}$ per second. How fast is the area decreasing when the two equal sides are equal to the base ?

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length of two equal sides be }x\ \text{and }\\\ \ \ \ \text{the the length of}\ \text{altitude of the triangle on base}\ b\text{ be }h.\text{ }\end{array}$.

$ \displaystyle \therefore \ \ \text{By Pythagoras' } \text{Theorem,}$

$ \displaystyle \ \ \ \ h=\sqrt{{{{x}^{2}}-{{{\left( {\frac{b}{2}} \right)}}^{2}}}}=\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}$

$ \displaystyle \ \ \ \ \text{By the problem, }\frac{{dx}}{{dt}}=3\ \text{cm/s.}$

$ \displaystyle \ \ \ \ \text{Let the area of the triangle be }A.$

$ \displaystyle \therefore \ \ A=\frac{1}{2}bh=\frac{1}{2}b\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}$

$ \displaystyle \therefore \ \ \frac{{dA}}{{dx}}=\frac{1}{2}b\times \frac{{2x}}{{2\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}}}=\frac{{bx}}{{2\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}}}$

$ \displaystyle \ \ \ \ \text{When }x=b,\frac{{dA}}{{dx}}=\frac{{{{b}^{2}}}}{{2\sqrt{{{{b}^{2}}-\frac{{{{b}^{2}}}}{4}}}}}=\frac{{{{b}^{2}}}}{{2b\sqrt{3}}}=\frac{b}{{\sqrt{3}}}$

$ \displaystyle \ \ \ \ \text{By Chain Rule,}\frac{{dA}}{{dt}}=\frac{b}{{\sqrt{3}}}\times 3=\sqrt{3}b\ \text{c}{{\text{m}}^{2}}\text{/s.}$

$ \displaystyle \therefore \ \ \text{The area of the triangle is decreasing at a rate of }\sqrt{3}b\ \text{c}{{\text{m}}^{2}} \text{/s.}$

9.       The volume of the cube is increasing at a rate of $ \displaystyle 9\ \text{cm}^3/ \text{s}$. How fast is the surface area increasing when the edge is $ \displaystyle 10\ \text{cm}$ long ?

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length of each sides be }x,\ \\\ \ \ \ \text{the volume be }V\text{and the surface area be }A.\end{array}$

$ \displaystyle \therefore \ \ V={{x}^{3}}\ \text{and}\ A=6{{x}^{2}}$

$ \displaystyle \therefore \ \ \frac{{dV}}{{dx}}=3{{x}^{2}}\ \text{and}\ \frac{{dA}}{{dx}}=12x.$

$ \displaystyle \ \ \ \ \text{By the problem, }\frac{{dV}}{{dt}}=9\ \text{c}{{\text{m}}^{3}}\text{/s}\text{.}$

$ \displaystyle \ \ \ \ \text{Since }\frac{{dV}}{{dt}}=\frac{{dV}}{{dx}}\times \frac{{dx}}{{dt}},\ \ \ \ \ \left[ {\text{Chain Rule}} \right]$

$ \displaystyle \ \ \ \ 3{{x}^{2}}\times \frac{{dx}}{{dt}}=9\Rightarrow \frac{{dx}}{{dt}}=\frac{3}{{{{x}^{2}}}}$

$ \displaystyle \ \ \ \ \text{Similarly }\frac{{dA}}{{dt}}=\frac{{dA}}{{dx}}\times \frac{{dx}}{{dt}}$

$ \displaystyle \therefore \ \ \frac{{dA}}{{dt}}=12x\left( {\frac{3}{{{{x}^{2}}}}} \right)=\frac{{36}}{x}$

$ \displaystyle \ \ \ \ \text{When }x=10\ \text{cm},\ \frac{{dA}}{{dt}}=\frac{{36}}{{10}}=3.6\ \text{c}{{\text{m}}^{2}}\text{/s.}$

$ \displaystyle \therefore \ \ \text{The surface area of the cube is increasing at a rate of }3.6\ \text{c}{{\text{m}}^{2}}\text{/s.}$

No.8 ႏွင့္ No.9 သည္ Composite Function မ်ား ျဖစ္ေနေသာေၾကာင့္ Chain Rule ကို သံုးရပါသည္။

Calculus : Proof of Differential Equations (Problems and Solutions)

1.       If $ \displaystyle y = \ln (\sin^3 2x)$, then prove that If $ \displaystyle 3(\frac{{d}^{2}y}{d{x}^{2}}) + (\frac{dy}{dx})^2+36=0$.

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$ \displaystyle \begin{array}{l}\ \ \ y=\ln ({{\sin }^{3}}2x)\ \ \\\ \\\ \ \ \ \ =\ln {{(\sin 2x)}^{3}}\\\\\ \ \ \ \ =3\ln (\sin 2x)\end{array}$

$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=\frac{3}{{\sin 2x}}\cdot \frac{d}{{dx}}(\sin 2x)$

$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{3}{{\sin 2x}}\cdot \cos 2x\cdot \frac{d}{{dx}}(2x)$

$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{{6\cos 2x}}{{\sin 2x}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ =6\cot 2x$

$ \displaystyle \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6(-{{\operatorname{cosec}}^{2}}2x)\frac{d}{{dx}}(2x)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ =-12{{\operatorname{cosec}}^{2}}2x$

$ \displaystyle \therefore 3(\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}})+{{(\frac{{dy}}{{dx}})}^{2}}+36$

$ \displaystyle \begin{array}{l}=3(-12{{\operatorname{cosec}}^{2}}2x)+{{(6\cot 2x)}^{2}}+36\\\\=36(-{{\operatorname{cosec}}^{2}}2x+{{\cot }^{2}}2x+1)\\\\=36(-{{\operatorname{cosec}}^{2}}2x+{{\operatorname{cosec}}^{2}}2x)\ \ \ \ \left[ {1+{{{\cot }}^{2}}2x={{{\operatorname{cosec}}}^{2}}2x} \right]\\\\=0\end{array}$

2.       If $ \displaystyle y = (3 + 4x) e^{-2x}$ , show that $ \displaystyle \frac{{d}^{2}y}{d{x}^{2}} + 4(\frac{dy}{dx})+4y=0$.

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$ \displaystyle \ \ \ y=(3+4x){{e}^{{-2x}}}$

$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=(3+4x)\cdot \frac{d}{{dx}}({{e}^{{-2x}}})+{{e}^{{-2x}}}\frac{d}{{dx}}(3+4x)$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ =(3+4x){{e}^{{-2x}}}\frac{d}{{dx}}(-2x)+4{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ =-2(3+4x){{e}^{{-2x}}}+4{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ =(-6-8x+4){{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ =(-8x-2){{e}^{{-2x}}}\end{array}$

$ \displaystyle \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=(-8x-2)\cdot \frac{d}{{dx}}({{e}^{{-2x}}})+{{e}^{{-2x}}}\frac{d}{{dx}}(-8x-2)$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ =(-8x-2){{e}^{{-2x}}}\frac{d}{{dx}}(-2x)-8{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ \ \ =-2(-8x-2){{e}^{{-2x}}}-8{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ \ \ =(16x+4-8){{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ \ \ =(16x-4){{e}^{{-2x}}}\end{array}$

$ \displaystyle \therefore \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}+4\left( {\frac{{dy}}{{dx}}} \right)+4y$

$ \displaystyle \begin{array}{l}=(16x-4){{e}^{{-2x}}}+4(-8x-2){{e}^{{-2x}}}+4(3+4x){{e}^{{-2x}}}\\\\=(16x-4-32x-8+12+16x){{e}^{{-2x}}}\\\\=0\end{array}$

3.       If $ \displaystyle y = a e ^{\sin x}$, where $ \displaystyle a$ is a constant, prove that $ \displaystyle \frac{{d}^{2}y}{d{x}^{2}} = (\cos x-\tan x)\frac{dy}{dx}$.

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$ \displaystyle \ \ \ y=a{{e}^{{\sin x}}}$

$ \displaystyle \therefore \ \frac{{dy}}{{dx}}=a{{e}^{{\sin x}}}\frac{d}{{dx}}(\sin x)$

$ \displaystyle \ \ \ \ \ \ \ \ =a{{e}^{{\sin x}}}\cdot \cos x$

$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}\frac{d}{{dx}}(\cos x)+a\cos x\frac{d}{{dx}}({{e}^{{\sin x}}})$

$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}(-\sin x)+a{{e}^{{\sin x}}}\cos x\frac{d}{{dx}}(\sin x)$

$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}(-\sin x)+a{{e}^{{\sin x}}}{{\cos }^{2}}x$

$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}({{\cos }^{2}}x-\sin x)$

$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}\cdot \cos x\left( {\frac{{{{{\cos }}^{2}}x}}{{\cos x}}-\frac{{\sin x}}{{\cos x}}} \right)$

$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}\cdot \cos x\left( {\cos x-\tan x} \right)$

$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\left( {\cos x-\tan x} \right)\frac{{dy}}{{dx}}$

4.       If $ \displaystyle y=e^x(\sin 2x+ \cos 2x)$, prove that $ \displaystyle y''-2y'+5y=0$.

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$ \displaystyle \ \ \ \ y={{e}^{x}}(\sin 2x+\cos 2x)$

$ \displaystyle \therefore \ \ {y}'={{e}^{x}}\frac{d}{{dx}}(\sin 2x+\cos 2x)+(\sin 2x+\cos 2x)\frac{d}{{dx}}({{e}^{x}})$

$ \displaystyle \begin{array}{l}\therefore \ \ {y}'={{e}^{x}}(2\cos 2x-2\sin 2x)+{{e}^{x}}(\sin 2x+\cos 2x)\\\\\therefore \ \ {y}'={{e}^{x}}(2\cos 2x-2\sin 2x+\sin 2x+\cos 2x)\\\\\therefore \ \ {y}'={{e}^{x}}(3\cos 2x-\sin 2x)\\\\\therefore \ \ {y}''={{e}^{x}}\frac{d}{{dx}}(3\cos 2x-\sin 2x)+(3\cos 2x-\sin 2x)\frac{d}{{dx}}({{e}^{x}})\\\\\therefore \ \ {y}''={{e}^{x}}(-6\sin 2x-2\cos 2x)+{{e}^{x}}(3\cos 2x-\sin 2x)\\\\\therefore \ \ {y}''={{e}^{x}}(-6\sin 2x-2\cos 2x+3\cos 2x-\sin 2x)\\\\\therefore \ \ {y}''={{e}^{x}}(-7\sin 2x+\cos 2x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ------(1)\\\\\therefore \ -2{y}'={{e}^{x}}(2\sin 2x-6\cos 2x)\ \ \ \ \ \ \ \ \ \ \ \ \ ------(2)\\\\\therefore \ \ 5y={{e}^{x}}(5\sin 2x+5\cos 2x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ------(3)\\\\\text{By Equation (1) + Equation (2) + Equation (3)},\\\\\ \ \ \ {y}''-2{y}'+5y=0\end{array}$

5.       Find the value of $ \displaystyle x$ between $ \displaystyle 0$ and $ \displaystyle \frac{\pi }{2}$ for which the curve $ \displaystyle y = e^x \cos x$ has a stationary point. Determine whether it is a maximum or a minimum point.

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$ \displaystyle \ \ \ \ y={{e}^{x}}\cos x,\ 0<x<\frac{\pi }{2}$

$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=-{{e}^{x}}\sin x+{{e}^{x}}\cos x$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ ={{e}^{x}}(\cos x-\sin x)\\\\\ \ \ \ \frac{{dy}}{{dx}}=0\ \text{when}\ {{e}^{x}}(\cos x-\sin x)=0\\\\\ \ \ \ \text{Since}\ {{e}^{x}}\ne 0\ \text{for every }x\in R,\ \\\\\ \ \ \ \cos x-\sin x=0.\\\\\therefore \ \ \cos x=\sin x\\\\\therefore \ \ \tan x=1\end{array}$

$ \displaystyle \therefore \ \ x=\frac{\pi }{4}$ [Calculus တြင္ ေထာင့္မ်ားကို တိုင္းတာရာတြင္ linear scale ျဖစ္ေသာ radian စနစ္ကိုသာ သံုးရမည္။ degree ျဖင့္ အေျဖ မေပးရပါ]

$ \displaystyle \ \ \ \ \text{When }x=\frac{\pi }{4},y={{e}^{{\frac{\pi }{4}}}}\cos \frac{\pi }{4}=\frac{{\sqrt{2}}}{2}{{e}^{{\frac{\pi }{4}}}}$

$ \displaystyle \therefore \ \ \text{The stationary point is }\left( {\frac{\pi }{4},\frac{{\sqrt{2}}}{2}{{e}^{{\frac{\pi }{4}}}}} \right).$
$ \displaystyle \left( {\frac{\pi }{4},\frac{{\sqrt{2}}}{2}{{e}^{{\frac{\pi }{4}}}}} \right)$ သည္ $ \displaystyle \left( {0.79,1.55} \right)$ ခန္႔ရွိသည္။

$ \displaystyle \ \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}={{e}^{x}}(-\sin x-\cos x)+{{e}^{x}}(\cos x-\sin x)$

$ \displaystyle \therefore \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}={{e}^{x}}(-\sin x-\cos x+\cos x-\sin x)$

$ \displaystyle \therefore \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-2{{e}^{x}}\sin x$

$ \displaystyle \therefore \ \ {{\left. {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right|}_{{x=\frac{\pi }{4}}}}=-2{{e}^{{\frac{\pi }{4}}}}\sin \frac{\pi }{4}=-\sqrt{2}{{e}^{{\frac{\pi }{4}}}}<0$

$ \displaystyle \therefore \ \ \text{The stationary point is a maximum turning point}$.

Calculus : Differentiation (Chain Rule, Product Rule and Quotient Rule)

Differentiate the following with respect to $ \displaystyle x$.

(i)  $ \displaystyle (2−x^2) \ln(3x+1)$

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(i) $ \displaystyle \ \ \ \ \frac{d}{{dx}}\left[ {(2-{{x}^{2}})\ln (3x+1)} \right]$

$ \displaystyle =(2-{{x}^{2}})\frac{d}{{dx}}\ln (3x+1)+\ln (3x+1)\frac{d}{{dx}}(2-{{x}^{2}})$

$ \displaystyle =\frac{{2-{{x}^{2}}}}{{3x+1}}\frac{d}{{dx}}\left( {3x+1} \right)-2x\ln (3x+1)$

$ \displaystyle =\frac{{3\left( {2-{{x}^{2}}} \right)}}{{3x+1}}-2x\ln (3x+1)$

(ii)  $ \displaystyle {\frac{{4-\tan 2x}}{{5x}}}$

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(ii) $ \displaystyle \ \ \ \ \frac{d}{{dx}}\left[ {\frac{{4-\tan 2x}}{{5x}}} \right]$

$ \displaystyle =\frac{{5x\frac{d}{{dx}}\left( {4-\tan 2x} \right)-\left( {4-\tan 2x} \right)\frac{d}{{dx}}(5x)}}{{25{{x}^{2}}}}$

$ \displaystyle =\frac{{5x(-{{{\sec }}^{2}}2x)\frac{d}{{dx}}(2x)-5\left( {4-\tan 2x} \right)}}{{25{{x}^{2}}}}$

$ \displaystyle =\frac{{-10x{{{\sec }}^{2}}2x+5\tan 2x-20}}{{25{{x}^{2}}}}$

(iii)  Given that a curve has equation $ \displaystyle y=\frac{1}{x}+2\sqrt{x}$ where $\ x>0$, Find $ \displaystyle \frac{{dy}}{{dx}}$ and $ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$.
Hence or otherwise find the coordinates and nature of the stationary point of the curve.

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(iii)   $ \displaystyle y=\frac{1}{x}+2\sqrt{x},x>0$

$ \displaystyle \frac{{dy}}{{dx}}=-\frac{1}{{{{x}^{2}}}}+\frac{1}{{\sqrt{x}}}$

$ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\frac{2}{{{{x}^{3}}}}-\frac{1}{{2x\sqrt{x}}}$

At the stationary point, $ \displaystyle \frac{{dy}}{{dx}}=0$.

$ \displaystyle \therefore -\frac{1}{{{{x}^{2}}}}+\frac{1}{{\sqrt{x}}}=0$

$ \displaystyle \therefore \frac{1}{{\sqrt{x}}}=\frac{1}{{{{x}^{2}}}}$

$ \displaystyle \therefore x\sqrt{x}=1$

$ \displaystyle \therefore x=1$

When $ \displaystyle x=1$, $ \displaystyle y=\frac{1}{1}+2\sqrt{1}=3$.

Therefore, the stationary point is $ \displaystyle (1,3)$.

$ \displaystyle {{\left. {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right|}_{{(1,3)}}}=\frac{2}{{{{1}^{2}}}}-\frac{1}{{2(1)\sqrt{1}}}=\frac{3}{2}>0$

Therefore, the stationary point (1,3) is a minimum turning point.

Problem Study : Trigonometric Limit

Evaluate $ \displaystyle \underset{{x\,\to \,\pi }}{\mathop{{\lim }}}\,\frac{{1-\cos 7(x-\pi )}}{{5{{{(x-\pi )}}^{2}}}}$.

$\displaystyle x$ ေနရာမွာ $\displaystyle \pi$ ကို တိုက္႐ိုက္ အစားသြင္းၾကည့္မယ္။

$ \displaystyle \underset{{x\to \pi }}{\mathop{{\lim }}}\,\frac{{1-\cos 7(x-\pi )}}{{5{{{(x-\pi )}}^{2}}}}=\frac{{1-\cos 7(\pi -\pi )}}{{5{{{(x-\pi )}}^{2}}}}=\frac{{1-1}}{0}=\frac{0}{0}$

တိုက္ရိုက္ အစားသြင္းၾကည့္တဲ့ အခါ indeterminate form ျဖစ္သြားပါတယ္။ Limit ကို ဆက္ရွာလို႔ မရေတာ့ပါဘူး။ အမွန္တကယ္က $ \displaystyle {x\to \pi }$ ဆိုတာ $\displaystyle x=\pi$ မဟုတ္ပါဘူး $\displaystyle \pi$ ေလာက္နီးနီးရွိေသာ တန္ဖိုးတစ္ခု ျဖစ္တာေၾကာင့္ limit တန္ဖိုး တစ္ခုရွိပါတယ္။ ဒါေၾကာင့္ indetrminate form ကို ေက်ာ္လႊားဖို႔ trigonometric identity တစ္ခ်ိဳ႕ကို သံုးပါမယ္။

Trigonometric Limit ဆိုင္ရာ မွန္ကန္ခ်က္ တစ္ခုျဖစ္တဲ့ $ \displaystyle \underset{{u\to 0}}{\mathop{{\lim }}}\,\frac{{\sin u}}{u}=1$ ဆိုတာရယ္ Limit ရဲ့ ဂုဏ္သတိၱ မ်ားျဖစ္တဲ့ $ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left[ {Cf(x)} \right]=C\underset{{x\to a}}{\mathop{{\lim }}}\,f(x)$ နဲ႔ $\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left[ {f(x)\cdot g(x)} \right]=\underset{{x\to a}}{\mathop{{\lim }}}\,f(x)\cdot \underset{{x\to a}}{\mathop{{\lim }}}\,g(x)$ ဆိုတာကို သိရွိထားရပါမယ္။

အျမင္ရွင္းသြားေအာင္ $ \displaystyle {x-\pi =t}$ လို႔ထားလိုက္မယ္။ ဒါဆိုရင္ $ \displaystyle x$ က $\displaystyle \pi$ ကို ခ်ဥ္းကပ္သြားတဲ့အခါ $\displaystyle t$ က $\displaystyle 0$ ကို ခ်ဥ္းကပ္ သြားမွာေပါ့။ တြက္ၾကည့္ၾကစို႔။

Solution

Let $ \displaystyle {x-\pi }=t$.

When  $ \displaystyle {x\to \pi }$, then $ \displaystyle {t\to 0 }$.

$ \displaystyle \therefore \underset{{x\to \pi }}{\mathop{{\lim }}}\,\frac{{1-\cos 7(x-\pi )}}{{5{{{(x-\pi )}}^{2}}}}=\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{1-\cos 7t}}{{5{{t}^{2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{1-\cos 2\left( {\frac{{7t}}{2}} \right)}}{{5{{t}^{2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{5}\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{1-\cos 2\left( {\frac{{7t}}{2}} \right)}}{{{{t}^{2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{5}\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{1-\left[ {1-2{{{\sin }}^{2}}\left( {\frac{{7t}}{2}} \right)} \right]}}{{{{t}^{2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2}{5}\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\frac{{49}}{4}\times {{{\sin }}^{2}}\left( {\frac{{7t}}{2}} \right)}}{{\frac{{49}}{4}\times {{t}^{2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2}{5}\times \frac{{49}}{4}\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{{{{\sin }}^{2}}\left( {\frac{{7t}}{2}} \right)}}{{{{{\left( {\frac{{7t}}{2}} \right)}}^{2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2}{5}\times \frac{{49}}{4}\underset{{t\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {\frac{{7t}}{2}} \right)\sin \left( {\frac{{7t}}{2}} \right)}}{{\left( {\frac{{7t}}{2}} \right)\left( {\frac{{7t}}{2}} \right)}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2}{5}\times \frac{{49}}{4}(1)(1)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{49}}{{10}}$

Calculus : Gradient of Tangent

ေပးထားတဲ့ curve (အစိမ္းေရာင္) က $ \displaystyle y=f(x)$ ျဖစ္ပါတယ္။ ခရမ္းေရာင္ $ \displaystyle PQ$ မ်ဥ္းကေတာ့ $ \displaystyle \text{curve}$ ေပၚမွာရွိတဲ့ အမွတ္ႏွစ္ခုကို ျဖတ္သြားလို႔ $ \displaystyle \text{secant}$ လို႔ ေခၚမယ္။ $ \displaystyle P$ အမွတ္မွာ $ \displaystyle \text{curve}$ ကို ထိသြားတဲ့ အနက္ေရာင္မ်ဥ္းကိုေတာ့ $ \displaystyle \text{tangent}$ လို႔ေခၚမယ္။


Given (ေပးခ်က္) : Curve : $ \displaystyle y=f(x)$


Claim (ရွာရန္) : Gradient (Slope) of tangent at P


Explanation (ရွင္းလင္းခ်က္)


ေပးထားတာ curve : $ \displaystyle y=f(x)$  ရွိတာေၾကာင့္ curve equation ထဲကို $ \displaystyle x$ ေတြထည့္ရင္ $ \displaystyle y$ ရမွာေပါ့။ တနည္းေျပာရင္ curve ေပၚမွာရွိတဲ့ အမွတ္ေတြကို ႀကိဳက္သေလာက္ ရွာႏိုင္ပါတယ္။ curve ေပၚမွာရွိတဲ့ အမွတ္ႏွစ္ကို ျဖတ္ဆြဲရင္ secant ေပါ့။ အမွတ္ႏွစ္မွတ္ သိမွေတာ့ gradient (slope) ကိုလည္း ရွာလို႔ရၿပီေပါ့။


$ \displaystyle \begin{array}{l}\ \ \ y=f(x)\\\\\ \ \ {{y}_{1}}=f({{x}_{1}})\Rightarrow y+\delta y=f(x+\delta x)\\\\\therefore \ \text{Gradient of secant}\ PQ\\\\=\frac{{{{y}_{1}}-y}}{{{{x}_{1}}-x}}\\\\=\frac{{y+\delta y-y}}{{x+\delta x-x}}\ \text{or }\frac{{f(x+\delta x)-f(x)}}{{x+\delta x-x}}\\\\=\frac{{\delta y}}{{\delta x}}\ \text{or }\frac{{f(x+\delta x)-f(x)}}{{\delta x}}\end{array}$


ဒါေပမယ့္ အခုလိုခ်င္တာက gradient of tangent ျဖစ္ပါတယ္။ secant မဟုတ္ဘူး။ ျပသနာက tangent က curve ေပၚမွာ အမွတ္တစ္မွတ္ကိုပဲ ထိသြားတာ၊ ႏွစ္မွတ္မရွိဘူး။ တစ္မွတ္ထဲကို ႏွစ္ကိုယ္ခြဲတြက္ေပါ့ လို႔ေျပာလိုက ေျပာႏိုင္ပါေသးတယ္။ ႏွစ္ကိုယ္ခြဲလိုက္မွေတာ့ $ \displaystyle \frac{{{{y}_{1}}-y}}{{{{x}_{1}}-x}}=\frac{{y-y}}{{{{y}_{1}}-x}}=\frac{0}{0}$ ဆိုတာ indeterminate form ျဖစ္သြားၿပီ ဘာမွ ဆက္လုပ္လို႔မရေတာ့ဘူး။


ေရ တစ္စည္ထဲကို ေရတစ္ခြက္ ေပါင္းထည့္လို႔၊ ေရတစ္စည္ထဲက ေရတစ္ခြက္ ခပ္ထုတ္လိုက္လို႔ ေရ တစ္စည္ကို တိုးလားတယ္ ေလ်ာ့သြားတယ္လို႔ ေျပာေလ့မရွိၾကပါဘူး။ ဘာေၾကာင့္လဲ ဆိုေတာ့ $\displaystyle \frac{{\operatorname{ေရတစ္ခြက္}}}{{\operatorname{ေရတစ္စည္}}}\approx0$ ျဖစ္တာေၾကာင့္ပါ။ သခၤ်ာ႐ွဳေထာင့္က ၾကည့္ရင္ေတာ့ တစ္ခြက္တိုးတိုး တစ္စက္ တိုးတိုး အတိုး ရွိတာေပါ့့။


အလားတူပါပဲ တစ္မွတ္ထဲပဲ ရွိတဲ့ tangent ရဲ့ gradient ကို မရွာႏိုင္ေပမယ့္ အမွတ္ $ \displaystyle P$ နားကို အလြန္နီးကပ္ေနတဲ့ အမွတ္တစ္ခုကို ယူလိုက္ရင္ေတာ့ အမွတ္ႏွစ္ခု ျဖစ္သြားလို႔ gradient ရွာႏိုင္ၿပီေပါ့။  tangent ေတာ့မဟုတ္ဘူး tangent နား အလြန္ကပ္ေနတဲ့ secant ရဲ့ gradient ေပါ့။ Calculus မွာေတာ့ $ \displaystyle Q$ က $ \displaystyle P$ အနားကို လံုေလာက္ေအာင္ နီးကပ္သြားရင္ Gradient of tangent = Gradient of Secant လို႔ သတ္မွတ္ပါတယ္။


ပံုမွာ ျမင္ေတြ႔ရတဲ့ အတိုင္းေပါ့။ $ \displaystyle Q$ က Curve တေလွ်ာက္ $ \displaystyle P$ အနားကို ကပ္သြားဖို႔ $ \displaystyle {{{x}_{1}}}$ ရဲ့ တန္ဖိုး ေလ်ာ႔သြားဖို႔လိုပါတယ္။ $ \displaystyle {{x}_{1}}=x+\delta x$ ျဖစ္တာေၾကာင့္ $ \displaystyle {{{x}_{1}}}$ ရဲ့ တန္ဖိုး ေလ်ာ့သြားဖို႔ ဆိုတာက $ \displaystyle {\delta x}$ တန္ဖိုး ေလ်ာ့သြားမွ ျဖစ္မွာေပါ့။ ပံုမွာ $ \displaystyle {\delta x}$ တန္ဖိုးသတ္မွတ္ထားတဲ့ slider ကို ဘယ္ဘက္ကို ေရႊ႕ၾကည့္ပါ။


$ \displaystyle \begin{array}{*{20}{l}} {\text{When }\delta x\to 0,\ } \\ {} \\ {\text{Gradient of secant}\to \text{Gradient of tangent}} \\ {} \\ {\text{Therefore the gradient of secant approaches }} \\ {\text{the gradient of tangent when }\delta x\ \text{approaches 0}\text{.}} \\ {} \\ \begin{array}{l}\text{By limit notation,}\\\text{ }\end{array} \\ \begin{array}{l}\text{Gradient of tangent =}\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}\\\\\text{Gradient of tangent =}\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}\end{array} \end{array}$


Gradient of tangent ကိုေတာ့ သေကၤတ $ \displaystyle \frac{{dy}}{{dx}}$ (သို႔) $ \displaystyle y'$ (သို႔) $ \displaystyle f'(x)$ (သို႔) $\displaystyle \frac{d}{{dx}}\left[ {f(x)} \right]$ ျဖင့္သတ္မွတ္ပါတယ္။ ဒါ့ေၾကာင့္ ...


$ \displaystyle \begin{array}{l}\frac{{dy}}{{dx}}={y}'=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}\\{f}'(x)=\frac{d}{{dx}}\left[ {f(x)} \right]=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}\end{array}$

Differentiation from the First Principle

$ \displaystyle \frac{{dy}}{{dx}}=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}$ or $ \displaystyle {f}'(x)=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}$  ဘယ္လို ျဖစ္သြားလဲ...

Calculus : Tangent Line and Normal Line

Find the equation of the normal to the curve $ \displaystyle y = 1 - x^2$ at the point where the curve crosses the positive $ \displaystyle x$-axis. Find also the coordinates of the point where the normal meets the curve again.

$ \displaystyle y = 1 - x^2$ ဆိုတဲ့ curve က အေပါင္း $ \displaystyle x$ ၀င္႐ိုးကို ျဖတ္သြားတဲ့ ေနရာမွာ ရွိတဲ့ normal line equation ကို ရွာေပးပါ။ ၎ normal line က curve ကို ေနာက္ တစ္ႀကိမ္ ျဖတ္သြားတဲ့ အမွတ္ကိုလည္း ရွာေပးပါ။ 

Curve က $ \displaystyle x$ ၀င္႐ိုးကို $ \displaystyle y=0$ ျဖစ္တဲ့ ေနရာမွာ ျဖတ္တာေပါ့။ ဒါဆိုရင္ ေပးထားတဲ့ curve : $ \displaystyle 1 - x^2$ ကို 0 နဲ႔ ညီေပးလိုက္ရင္ curve ျဖတ္သြားတဲ့ $ \displaystyle x$ ၀င္႐ိုးေပၚက အမွတ္ေတြ ရၿပီေပါ့။ အေပါင္း $ \displaystyle x$ ၀င္႐ိုးလို႔ ေျပာထားတာလို႔ $ \displaystyle x$ $ \displaystyle \text{coordinate}$ မွာ အႏႈတ္ပါလာရင္ အႏႈတ္ တန္ဖိုးကို ပယ္ရပါမယ္။ 

Normal Line ရဲ့ equation ကို ရွာဖို႔ gradient (slope) ကိုလည္း သိရမယ္။ Normal Line ဆိုတာ tangent ေပၚ ေထာင့္မတ္က်လို႔ tangent gradient ရဲ့ အႏႈတ္လွန္ကိန္း (negative reciprocal) ျဖစ္ပါတယ္။ gradient of tangent =$ \displaystyle m$ ျဖစ္တယ္ဆိုရင္ normal ရဲ့ gradient က $ \displaystyle -\frac{1}{m}$ ျဖစ္မွာေပါ့။ ဒါဆိုရင္ tangent ရဲ့ gradient ကို သိရင္ normal line equation ကို $ \displaystyle (y-{{y}_{1}})=-\frac{1}{m}(x-{{x}_{1}})$ ဆိုတာနဲ႔ ရွာလိုက္ရင္ ရၿပီ။ 

Gradient of tangent = $ \displaystyle \frac{dy}{dx}$ ဆိုတာ differentiation from the first principles ကတည္းက သိၿပီးျဖစ္မွာပါ။ tangent ရဲ့ gradient ကို ရဖို႔ ေပးထားတဲ့ curve ကို differentiate လုပ္ရင္ ရပါၿပီ။

Curve ကို differentiate လုပ္လို႕ရတဲ့ tangent ရဲ့ gradient ဆိုတာ general form (curve ေပၚမွာရွိတဲ့ မည္သည့္ေနရာကို မဆို ဆိုလိုတယ္)။ ေမးခြန္းက ေမးထားတာက အေပါင္း $ \displaystyle x$ ၀င္႐ိုးကို ျဖတ္ေသာေနရာလို႔ ဆိုထားလို႔ ရွာထားတဲ့  $ \displaystyle x$ ၀င္႐ိုး ျဖတ္မွတ္ [ $ \displaystyle (x_{1},y_{1})$ လို႔ ဆိုၾကပါဆို႔] အစားသြင္းလိုက္မွ လိုခ်င္တဲ့ tangent line ရဲ့ gradient ကို ရမွာေပါ့။ ဆိုလိုတာက $ \displaystyle m={{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{1}},{{y}_{1}})}}}$ ပါ။

Normal Line equation ရရင္ ေမးခြန္းက ထပ္ဆင့္ေမးထားတာက ၎ Normal Line က Curve ကို ေနာက္တစ္ႀကိမ္ ျဖတ္တဲ့အမွတ္ကို ရွာေပးပါလို႔ ဆိုထားပါတယ္။

Line ႏွစ္ေၾကာင္းျဖတ္မွတ္ဆိုတာ အလယ္တန္း အဆင့္မွာကတည္းက သိခဲ့ၿပီးပါၿပီ။ ၎ Line ႏွစ္ေၾကာင္းကို တၿပိဳင္နက္ ေျပလည္ေစတဲ့ $ \displaystyle x$ - $ \displaystyle \text{coordinate}$ နဲ႔ $ \displaystyle y$ - $ \displaystyle \text{coordinate}$ ကို ရွာခိုင္းတာ ျဖစ္ပါတယ္။

ထိုနည္းတူ ပါပဲ။ Curve နဲ႔ Normal Line တို႔ျဖတ္မွတ္ဆိုတာ ၎ Equation ႏွစ္ေၾကာင္းကို တၿပိဳင္နက္ ေျပလည္ေစတဲ့ $ \displaystyle x$ - $ \displaystyle \text{coordinate}$ နဲ႔ $ \displaystyle y$ - $ \displaystyle \text{coordinate}$ ကို ရွာေပးရမွာ ျဖစ္ပါတယ္။ ဒါဆိုရင္ ေျဖရွင္းရမယ့္ အဆင့္ေတြကို သိေလာက္ၿပီလို႔ ထင္ပါတယ္။ တြက္ၾကည့္ၾကစို႔။

Solution

$ \displaystyle \begin{array}{l}\text{Curve : }y=1-{{x}^{2}}\\\\\text{When the curve cuts the }x-\text{axis, }y=0.\\\\\therefore 1-{{x}^{2}}=0\Rightarrow {{x}^{2}}=1\Rightarrow x=\pm 1\\\\\text{But the required points lies on the }\\\text{positive }x-\text{axis, }\\\\\therefore x=1\\\\\text{Let }({{x}_{1}},{{y}_{1}})=(1,0)\\\\y=1-{{x}^{2}}\Rightarrow \frac{{dy}}{{dx}}=-2x\\\\\therefore m={{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{1}},{{y}_{1}})}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{(1,0)}}}=-2(1)=-2\\\\\text{Hence the equation of normal line at }({{x}_{1}},{{y}_{1}})\ \text{is}\\\\(y-{{y}_{1}})=-\frac{1}{m}(x-{{x}_{1}})\\\\y-0=\frac{1}{2}(x-1)\Rightarrow y=\frac{1}{2}(x-1)\\\\\text{When this normal line meets the curve,}\\\\\frac{1}{2}(x-1)=1-{{x}^{2}}\Rightarrow 2{{x}^{2}}+x-3=0\\\\\therefore (2x+3)(x-1)=0\\\\\therefore x=-\frac{3}{2}\ \text{or}\ x=1\\\\\text{When }x=-\frac{3}{2}\ ,y=\frac{1}{2}(-\frac{3}{2}-1)=-\frac{5}{4}.\end{array}$

$ \displaystyle \text{Therefore, the normal line meets}$
$ \displaystyle \text{the curve again at the point (}-\frac{3}{2},-\frac{5}{4}).$

Calculus : Tangent Line

Find the equation of the tangent to the curve y=2x² which is parallel to the secant of the curve which passes through the points on the curve which have the coordinates x=−1 and x=2.

Solution

$ \displaystyle \begin{array}{l}\text{Curve    : }y=2{{x}^{2}}\\\\\text{When }x=-1,y=2{{(-1)}^{2}}=2\\\\\text{When }x=2,y=2{{(2)}^{2}}=8\\\\\therefore \text{Gradient of secant = }\frac{{8-2}}{{2-(-1)}}=2\\\\\text{The gradient of tangent is }\frac{{dy}}{{dx}}.\\\\\therefore \ \frac{{dy}}{{dx}}=4x\\\\\text{By the problem, tangent }\parallel \ \text{secant}\text{.}\\\\\therefore \ \frac{{dy}}{{dx}}=2\Rightarrow 4x=2\Rightarrow x=\frac{1}{2}\\\\\text{When }x=\frac{1}{2},y=2{{(\frac{1}{2})}^{2}}=\frac{1}{2}\end{array}$

$\displaystyle \begin{array}{l}\therefore \text{The tangent line touches the curve at (}\frac{1}{2},\frac{1}{2}\text{)}\text{.}\\\\\therefore \text{Equation of tangent line to the curve at (}\frac{1}{2},\frac{1}{2}\text{)}\ \text{is}\\\\\ \ y-\ \frac{1}{2}=2(x-\frac{1}{2})\Rightarrow y=2x-\frac{1}{2}.\end{array}$

Properties of Limits

1.     $ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,f(x)=f(a)$ when $ \displaystyle f(x)$ is continuous at $ \displaystyle x=a$.
        Function သည် $ \displaystyle x=a$ တွင် continuous ဖြစ်နေလျှင် [$ \displaystyle f(a)$ တန်ဖိုးရှာနိုင်လျှင် (သို့) $ \displaystyle f(a)$ သည် indeterminate form
        မဖြစ်လျှင်] $ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,f(x)=f(a)$ ဟု ရေးနိုင်ပါသည်။

       Assume that the limits of functions $\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,f(x)$ and $\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,g(x)$ exist and that $ \displaystyle c$ is any constant. Then 

2.  The limit of a constant times a function is the constant times the limit.
 

$ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left( {c\cdot f(x)} \right)=c\cdot \underset{{x\to a}}{\mathop{{\lim }}}\,f(x)$

      
       Function ကို constant ဖြင့် မြှောက်ထားလျှင် function ကိုသာ limit ယူပြီး constant ဖြင့် မြှောက်ပေးရပါသည်။

              Example :

          $ \displaystyle \ \ \ \underset{{x\to 3}}{\mathop{{\lim }}}\,(4x)$

          $ \displaystyle =4\underset{{x\to 3}}{\mathop{{\lim }}}\,(x)$

          $ \displaystyle =4(3)$

          $ \displaystyle =12$

3.  The limit of a constant times a function is the constant times the limit. If $ \displaystyle h(x)=c$ for all $ \displaystyle x$, then

$ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,h(x)=\underset{{x\to a}}{\mathop{{\lim }}}\,c=c$

      Function သည်ကိန်းသေ (constant function) ဖြစ်လျှင် function ကို limit ယူလျှင် မူလတန်ဘိုး constant သာ ပြန်ရသည်။

             Example :
             $ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,(3)=3$
             $ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,(2018)=2018$
             $ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,(\sqrt{5})=\sqrt{5}$

4.     The limit of a sum or difference is the sum or difference of the limits.

$ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,[f\left( x \right)\pm g\left( x \right)]\text{ }=~\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( x \right)\pm \underset{{x\to a}}{\mathop{{\lim }}}\,g\left( x \right)$

        Function များ၏ ပေါင်းလဒ် နှုတ်လဒ်ကို Limit ယူလျှင် Function တစ်ခုချင်းစီကို Limit ယူပြီးမှ ပေါင်း၊ နှုတ် လုပ်ရသည်။

           Example :
           $ \displaystyle \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\left( {3{{x}^{2}}+4x+1} \right)$
           $ \displaystyle =\underset{{x\to 1}}{\mathop{{\lim }}}\,3{{x}^{2}}+\underset{{x\to 1}}{\mathop{{\lim }}}\,4x+\underset{{x\to 1}}{\mathop{{\lim }}}\,1$ 
           $ \displaystyle =3+4+1$
           $ \displaystyle =8$

          လက်တွေ့ပုစ္ဆာများကို ဖြေရှင်းရာတွင် ရိုးရှင်းသော အလယ်အဆင့်များကို ကျော်တွက်လေ့ရှိသည်။ 

5.     The limit of the product of two functions is the product of their limits (if they exist):   
 

$ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,[f\left( x \right)\cdot g\left( x \right)]\text{ }=~\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( x \right)\cdot \underset{{x\to a}}{\mathop{{\lim }}}\,g\left( x \right)$

        Function များ၏မြှောက်လဒ်ကို Limit ယူလျှင် Function တစ်ခုချင်းစီကို Limit ယူပြီးမှ မြှောက် ရသည်။

           Example :
           $ \displaystyle \ \ \ \underset{{x\to 4}}{\mathop{{\lim }}}\,\left( {x\cdot \sqrt{x}} \right)$
           $ \displaystyle =\underset{{x\to 4}}{\mathop{{\lim }}}\,x\cdot \underset{{x\to 4}}{\mathop{{\lim }}}\,\sqrt{x}$ 
           $ \displaystyle =4\sqrt{4}$
           $ \displaystyle =2$

      

          လက်တွေ့ပုစ္ဆာများကို ဖြေရှင်းရာတွင် ရိုးရှင်းသော အလယ်အဆင့်များကို ကျော်တွက်လေ့ရှိသည်။  

6.     The limit of quotient of two functions is the quotient of their limits, provided that 
        the limit in the denominator function is not zero: 

$ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{f(x)}}{{g(x)}}=\frac{{\underset{{x\to a}}{\mathop{{\lim }}}\,f(x)}}{{\underset{{x\to a}}{\mathop{{\lim }}}\,g(x)}}$ if $ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,g(x)\ne 0$

        Function များ၏ စားလဒ်ကို Limit ယူလျှင် Function တစ်ခုချင်းစီကို Limit ယူပြီးမှ စား ရသည်။ 
          ပိုင်းခြေ Function ၏ Limit သည် 0 မဖြစ်ရပါ။

          Example :
          $ \displaystyle \ \ \ \ \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+1}}{{2x-1}}$
          $ \displaystyle =\frac{{\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( {{{x}^{2}}+1} \right)}}{{\underset{{x\to 1}}{\mathop{{\lim }}}\,\left( {2x-1} \right)}}$ 
          $ \displaystyle =\frac{{1+1}}{{2-1}}$
          $ \displaystyle =2$

     

      လက်တွေ့ပုစ္ဆာများကို ဖြေရှင်းရာတွင် ပိုင်းခြေ 0 မဖြစ်လျှင် ရိုးရှင်းသော အလယ်အဆင့်များ ကို ကျော်တွက်လေ့ရှိသည်။ 

7 .

$ \displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,{{\left[ {f\left( x \right)} \right]}^{n}}={{\left[ {\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( x \right)} \right]}^{n}},\underset{{x\to a}}{\mathop{{\lim }}}\,\sqrt[n]{{f(x)}}=\sqrt[n]{{\underset{{x\to a}}{\mathop{{\lim }}}\,f(x)}}$ where the powr $ \displaystyle n$ can be any real number.

       ထပ်ကိန်း (ကိန်းရင်း) Function များကို Limit ယူလျှင် Function ကို Limit ယူပြီးမှ ထပ်ကိန်း (ကိန်းရင်း) ကို ရှာရပါသည်။

          Example (1): 

          $ \displaystyle \ \ \ \underset{{x\to 2}}{\mathop{{\lim }}}\,{{x}^{5}}$

          $ \displaystyle ={{\left( {\underset{{x\to 2}}{\mathop{{\lim }}}\,x} \right)}^{5}}$ 

          $ \displaystyle ={{2}^{5}}$

          $ \displaystyle =32$

       

          Example (2): 

          $ \displaystyle \ \ \ \underset{{x\to 2}}{\mathop{{\lim }}}\,{{\left[ {2x-3} \right]}^{3}}$

          $ \displaystyle ={{\left[ {\underset{{x\to 2}}{\mathop{{\lim }}}\,(2x-3)} \right]}^{3}}$ 

          $ \displaystyle ={{\left[ {2\underset{{x\to 2}}{\mathop{{\lim }}}\,x-\underset{{x\to 2}}{\mathop{{\lim }}}\,3} \right]}^{3}}$

          $ \displaystyle ={{\left[ {2(2)-3} \right]}^{3}}$

          $ \displaystyle =1$

      

          Example (3): 

          $ \displaystyle \ \ \ \underset{{x\to 3}}{\mathop{{\lim }}}\,\sqrt[3]{{3{{x}^{2}}}}$

          $ \displaystyle =\sqrt[3]{{\underset{{x\to 3}}{\mathop{{\lim }}}\,\left( {3{{x}^{2}}} \right)}}$

          $ \displaystyle =\sqrt[3]{{\left( {3\cdot \underset{{x\to 3}}{\mathop{{\lim }}}\,{{{\left( x \right)}}^{2}}} \right)}}$ 

          $ \displaystyle =\sqrt[3]{{3\cdot {{{(3)}}^{2}}}}$

          $ \displaystyle =3$

       လက်တွေ့ပုစ္ဆာများကို ဖြေရှင်းရာတွင် ရိုးရှင်းသော အဆင့်များကို ကျော်တွက်လေ့ရှိသည်။