Exercise 7.10-Additional Problems - Chapter 7 - Applications of Differential Calculus - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions Solved
Question 1.
The gradient of the curve $y=-2 x^3+3 x+5$ at $x=2$ is
(a) $-20$
(b) 27
(c) $-16$
(d) $-21$
Solution:
(d) $-21$
Hint:
Given, $y=-2 x^3+3 x+5$
Gradient, $\frac{d y}{d x}=-6 x^2$
$
\left(\frac{d y}{d x}\right)_{(x=2)}=-6(2)^2+3=-24+3=-21
$
Question 2.
The rate of change of area $\mathrm{A}$ of a circle of radius $r$ is
(a) $2 \pi r$
(b) $2 \pi r \frac{d r}{d t}$
(c) $\pi r^2 \frac{d r}{d t}$
(d) $\pi \frac{d r}{d t}$
Solution:
(b) $2 \pi r \frac{d r}{d t}$
Hint: $\quad$ Given $A=\pi r^2$
Differentiating with respect to $t$ on both sides,
$
\frac{d \mathrm{~A}}{d t}=2 \pi r \frac{d r}{d t}
$

Question 3.
A spherical snowball is melting in such a way that its volume is decreasing at a rate $0 \mathrm{~cm}^3 / \mathrm{min}$. The rate at which the diameter is decreasing when the diameter is $10 \mathrm{cms}$ is
(a) $\frac{-1}{50 \pi} \mathrm{cm} / \mathrm{min}$
(b) $\frac{1}{50 \pi} \mathrm{cm} / \mathrm{min}$
(c) $\frac{-11}{75 \pi} \mathrm{cm} / \mathrm{min}$
(d) $\frac{-2}{75 \pi} \mathrm{cm} / \mathrm{min}$
Solution:
(a) $\frac{-1}{50 \pi} \mathrm{cm} / \mathrm{min}$
Hint: Let $x$ be the diameter of the ball $\Rightarrow r=\frac{x}{2}=\pi r^2$
Volume of the sphereical ball $v=\frac{4}{3} \pi r^3$
$
\begin{aligned}
v & =\frac{4}{3} \pi\left(\frac{x}{2}\right)^3 \\
v & =\frac{4}{3} \pi \frac{x^3}{8}=\frac{1}{6} \pi x^3
\end{aligned}
$
Differentiating with respect to " $t "$ on both sides,
when $\frac{d v}{d t}=$ (As volume decreases), $x=10$
(1) $\Rightarrow-1=\frac{\pi}{2}(10)^2 \frac{d x}{d t}$
$
\frac{-2}{100 \pi}=\frac{d x}{d t} \Rightarrow \frac{d x}{d t}=\frac{-1}{50 \pi}
$

Question 4.
The parametric equations of the curve $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$ are
(a) $x=a \sin ^3 \theta ; y=a \cos ^3 \theta$
(b) $x=a \cos ^3 \theta ; y=a \sin ^3 \theta$
(c) $x=a^3 \sin \theta ; y=a^3 \cos \theta$
(d) $x=a^3 \cos \theta ; y=a^3 \sin \theta$
Solution:
(b) $x=a \cos ^3 \theta ; y=a \sin ^3 \theta$
Hint:
The parametric equations of the curve $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$ are $x=a \cos ^3 \theta ; y=a \sin ^3 \theta$

Question 5.
If the normal to the curve $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$ makes an angle $\theta$ with the $\mathrm{x}$ - axis then the slope of the
normal is
(a) $-\cot \theta$
(b) $\tan \theta$
(c) $-\tan \theta$
(d) $\cot \theta$
Solution:
(b) $\tan \theta$
Hint:
Since the normal to the curve makes an angle $\theta$ with the $\mathrm{x}$ axis.
Slope of the normal $=\tan \theta$

Question 6.
What is the surface area of a sphere when the volume is increasing at the same rate as its radius?
(a) 1
(b) $\frac{1}{2 \pi}$
(c) $4 \pi$
(d) $\frac{4 \pi}{3}$
Solution:
(a) 1
Hint:
Let the volume of the sphere $\mathrm{V}=\frac{4}{3} \pi r^3$
Differentiate with respect to $t$ on both sides
$
\begin{aligned}
& \frac{d \mathrm{~V}}{d t}=\frac{4}{3} \pi\left(3 r^2 \frac{d r}{d t}\right) \\
& \frac{d \mathrm{~V}}{d t}=4 \pi r^2 \frac{d r}{d t} \\
& \frac{\frac{d \mathrm{~V}}{d t}}{\frac{d r}{d t}}=4 \pi r^2 \\
& \text { Given, } \frac{d r}{d t}=\frac{d \mathrm{~V}}{d t} \\
& 4 \pi r^2=1 \\
&
\end{aligned}
$

Question 7.
For what values of $x$ is the rate of increase of $x^3-2 x^2+3 x+8$ is twice the rate of increase of $x$.
(a) $\left(-\frac{1}{3},-3\right)$
(b) $\left(\frac{1}{3}, 3\right)$
(c) $\left(-\frac{1}{3}, 3\right)$
(d) $\left(\frac{1}{3}, 1\right)$
Solution:
(d) $\left(\frac{1}{3}, 1\right)$
Hint:
$
\begin{aligned}
& \text { Let } y=x^3-2 x^2+3 x+8 \\
& \text { Given, } \frac{d y}{d x}=2 \frac{d x}{d t} \\
& \text { From }(1) \Rightarrow \frac{d y}{d x}=3 x^2-4 x+3=2 \\
& \Rightarrow 3 x^2-4 x+1=0 \\
& \Rightarrow(3 x-1)(x-1)=0 \\
& x=1 ; \text { or } x=\frac{1}{3}
\end{aligned}
$

Question 8.
If the volume of an expanding cube is increasing at the rate of $4 \mathrm{~cm}^3 / \mathrm{sec}$ then the rate of change of surface area when the volume of the cube is 8 cubic $\mathrm{cm}$ is .......
(a) $8 \mathrm{~cm}^2 / \mathrm{sec}$
(b) $16 \mathrm{~cm}^2 / \mathrm{sec}$
(c) $2 \mathrm{~cm}^2 / \mathrm{sec}$
(d) $4 \mathrm{~cm}^2 / \mathrm{sec}$
Solution:
(a) $8 \mathrm{~cm}^2 / \mathrm{sec}$
Hint:
Let the volume of the cube $\mathrm{V}=a^3 \Rightarrow \frac{d \mathrm{~V}}{d t}=3 a^2 \frac{d a}{d t}$
Given, $\frac{d \mathrm{~V}}{d t}=4 ; a=2$
$
\begin{aligned}
(1) \Rightarrow 4 & =3(4) \frac{d a}{d t} \\
\frac{d a}{d t} & =\frac{1}{3}
\end{aligned}
$
Let surface area of the cube $S=6 \mathrm{a}^2$

Question 9.
If a normal makes an angle $\theta$ with positive $\mathrm{x}$-axis then the slope of the curve at the point where the normal is drawn is ........
(a) $-\cot \theta$
(b) $\tan \theta$
(c) $-\tan \theta$
(d) $\cot \theta$
Solution:
(a) $-\cot \theta$
Hint:
Given, the slope of normal $=-\frac{1}{m}=\tan \theta$
Slope of the tangent $=m=-\frac{1}{\tan \theta}=-\cot \theta$

Question 10.
If the velocity of a particle moving along a straight line is directly proportional to the square of its distance from a fixed point on the line. Then its acceleration is proportional to .......
(a) $\mathrm{s}$
(b) $s^2$
(c) $s^3$
(d) $s^4$
Solution:
(c) $s^3$
Hint:
Let $s$ be the distance of the particle from fixed point on the line at time at $t$ and $\mathrm{V}$ be the velocity. Then $\mathrm{V}$ $\propto s^2$
Velocity $\mathrm{V}=k \mathrm{~s}^2$
$
\begin{aligned}
\text { Acceleration }=\frac{d \mathrm{~V}}{d t} & =k(2 s) \cdot t \\
& =k 2 s \mathrm{~V} \\
\text { Acceleration } & =2 s k\left(k s^2\right)=k 22 s^3 \\
\therefore \text { Acceleration } \propto s^3 &
\end{aligned}
$

Question 11.
The Rolle's constant for the function $y=x^2$ on $[-2,2]$ is
(a) $\frac{2 \sqrt{3}}{3}$
(b) 0
(c) 2
$(d)-2$
Solution:
(b) 0
Hint:
Let $f(x)=y=x^2$
$
\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}=0
$
By Rolle's theorem there exists a ' $c$ ' such that
$
f^{\prime}(c)=0 \Rightarrow 2 c=0 \Rightarrow c=0 \in(-2,2)
$
Question 12.
The value ' $\mathrm{c}$ ' of Lagranges Mean Value Theorem for $\mathrm{f}(\mathrm{x})=\sqrt{x}$ when $\mathrm{a}=1$ and $\mathrm{b}=4$ is $\ldots \ldots \ldots$
(a) $\frac{9}{4}$
(b) $\frac{3}{2}$
(c) $\frac{1}{2}$
(d) $\frac{1}{4}$

Solution:
(a) $\frac{9}{4}$
Hint:
$
\text {} \begin{aligned}
f(x) & =\sqrt{x} \\
f(1) & =\sqrt{1}=1=f(a) \\
f(4) & =\sqrt{4}=4=f(b) \\
f^{\prime}(x) & =\frac{1}{2 \sqrt{x}}
\end{aligned}
$
By Lagranges Mean Value Theorem there exists $f^{\prime}(c)$ such that
$
\begin{aligned}
& f^{\prime}(c)=\frac{1}{2 \sqrt{c}} \Rightarrow \frac{1}{2 \sqrt{c}}=\frac{f(b)-f(a)}{b-a} \\
& f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{1}{2 \sqrt{c}}=6=\frac{2-1}{4-1}=\frac{1}{3} \\
& \Rightarrow \frac{3}{2}=\sqrt{c} \Rightarrow c=\frac{9}{4}
\end{aligned}
$

Question 13.
In a given semi circle of diameter $4 \mathrm{~cm}$ a rectangle is to be inscribed. The maximum area of the rectangle is ........
(a) 2
(b) 4
(c) 8
(d) 16
Solution:
(b) 4
Hint:
Let the side of the rectangle be ' $a$ ' $\mathrm{cm}$
Maximum rectangle inscribed in a circle is a square.
The diagonal of the rectangle $=4 \mathrm{~cm}$

$
\begin{aligned}
& \sqrt{a^2+a^2}=4 \\
& \sqrt{2 a^2}=4 \\
& 2 a^2=16 \\
& a^2=8 \mathrm{~cm}
\end{aligned}
$
The maximum area $=8$
Half of the area of the square $=4 \mathrm{~cm}$

Question 14.
The least possible perimeter of a rectangle of area $100 \mathrm{~m}^2$ is
(a) 10
(b) 20
(c) 40
(d) 60
Solution:
(c) 40
Hint:
Let the length of the rectangle $=\mathrm{x} \mathrm{m}$
Let the length of the breadth $=y \mathrm{y} \mathrm{m}$
Area of the rectangle $A=100$ (given)
$
y=\frac{100}{x}
$
Perimeter of the rectangle $\mathrm{P}=2(x+y)$
$
\begin{aligned}
\mathrm{P} & =2\left(x+\frac{100}{x}\right) \\
\mathrm{P} & =2 x+\frac{200}{x} \\
\frac{d \mathrm{P}}{d x} & =2-\frac{200}{x^2}=0 \\
\Rightarrow 2 & =\frac{200}{x^2} \Rightarrow 2 x^2=200 \Rightarrow x^2=100 \Rightarrow x=10 \\
\therefore y & =\frac{100}{x}=\frac{100}{10}=10 \\
\frac{d^2 \mathrm{P}}{d x^2} & =0-200\left(\frac{-2}{x^3}\right)=\frac{400}{x^3}>0
\end{aligned}
$
The least perimeter $=2(\mathrm{x}+\mathrm{y})=2(10+10)=40 \mathrm{~cm}$

Question 15.
Which of the following curves is concave down?
(a) $y=-x^2$
(b) $y=x^2$
(c) $y=e^x$
(d) $y=x^2+2 x-3$
Solution:
(a) $y=-x^2$
Hint: Given $y=-x^2 \quad y=x^2 \quad y=e^x \quad y=x^2+2 x-3$ $\frac{d y}{d x}=-2 x<0 \quad \frac{d y}{d x}=2 x>0 \quad \frac{d y}{d x}=e^x>0 \quad \frac{d y}{d x}=2 x+2>0$ $y=-x^2$ is concave downward

Question 16.
The point of inflection of the curve $y=x^4$ is at
(a) $x=0$
(b) $x=3$
(c) $x=12$
(c) no where
Solution:
(a) $x=0$
Hint:
$
\text { Given } \begin{aligned}
y & =x^4 \\
\frac{d y}{d x} & =4 x^3 \\
\frac{d^2 y}{d x^2} & =12 x^2=0 \Rightarrow x^2=0 \Rightarrow x=0
\end{aligned}
$