Exercise 7.1 - Chapter 7 Applications of Differential Calculus 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Applications of Differential Calculus
$\operatorname{Ex} 7.1$
Question 1.
A point moves along a straight line in such a way that after $t$ seconds its distance from the origin is $s=2 t^2$ $+3 \mathrm{t}$ metres.
(i) Find the average velocity of the points between $t=3$ and $t=6$ seconds.
(ii) Find the instantaneous velocities at $\mathrm{t}=3$ and $\mathrm{t}=6$ seconds.
Solution:
(i) Given $s=2 t^2+3 t$
(i.e.) $f(t)=2 t^2+3 t$
Now $f(3)=18+9=27=f(a)$
$f(6)=72+18=90=f(b)$
Now $\frac{f(b)-f(a)}{b-a}=\frac{90-27}{6-3}=\frac{63}{3}$
$=21 \mathrm{~m} / \mathrm{s}$
(ii) $f(t)=2 t^2+3 t$
$f^{\prime}(t)=4 t+3$
$\mathrm{f}^{\prime}(3)=4(3)+3=15$
$f^{\prime}(6)=4(3)+3=15$
Question 2.
A camera is accidentally knocked off an edge of a cliff $400 \mathrm{ft}$ high. The camera falls a distance of $\mathrm{s}=16 \mathrm{t}^2$ in $t$ seconds.
(i) How long does the camera fall before it hits the ground?
(ii) What is the average velocity with which the camera falls during the last 2 seconds?
(iii) What is the instantaneous velocity of the camera when it hits the ground?
Solution:
(i) $s=16 t^2$
$16 \mathrm{t}^2=400$
$
\begin{aligned}
& t^2=\frac{400}{16}=25 \\
& \mathrm{t}=5 \mathrm{sec}
\end{aligned}
$
(ii) Last 2 seconds means $t=3$ to $t=5$
$\mathrm{f}(\mathrm{t})=16 \mathrm{t}^2$
$f(3)=16(9)=144=f(a)$

$
\begin{aligned}
& \mathrm{f}(5)=16(25)=400=\mathrm{f}(\mathrm{b}) \\
& \text { So } \frac{f(b)-f(a)}{b-a}=\frac{400-144}{5-3}=\frac{256}{2} \\
& =128 \mathrm{ft} / \mathrm{sec}
\end{aligned}
$
(iii) $f(t)={ }^2$
$f^{\prime}(\mathrm{t})=32 \mathrm{t}$
$f^{\prime}(\mathrm{t})$ at $\mathrm{t}=5=32(5)$
$=160 \mathrm{ft} / \mathrm{sec}$
Question 3.
A particle moves along a line according to the law $s(t)=2 t^3-9 t^2+12 t-4$, where $t \geq 0$.
(i) At what times the particle changes direction?
(ii) Find the total distance travelled by the particle in the first 4 seconds.
(iii) Find the particle's acceleration each time the velocity is zero.
Solution:
$
\begin{aligned}
& \text { (i) } s=f(t)=2 t^3-9 t^2+12 t-4 \\
& V=f^{\prime}(t)=6 t^2-18 t+12 \\
& V=0 \Rightarrow 6\left(t^2-3 t-2\right)=0 \\
& (t-1)(t-2)=0 \\
& t=1,2
\end{aligned}
$
When $\mathrm{t}<1$, (say $\mathrm{t}=0.5$ )
$
\mathrm{V}=6(0.25-1.5+2)=+\mathrm{ve}
$
When $12$, (say $t=3$ )
$
\mathrm{V}=6(9-6+2)=+\mathrm{ve}
$
So the particle changes its direction when $t$ lies between 1 and 2 secs.
(ii) The distance travelled in the first 4 seconds is
$
|s(0)-s(1)|+|s(1)-s(2)|+|s(2)-s(3)|+|s(3)-s(4)|
$
Here, $s(t)=2 t^3-9 t^2+12 t-4$
$
\begin{aligned}
& s(0)=-4 \\
& s(1)=1 \\
& s(2)=0 \\
& s(3)=5 \\
& \text { and } s(4)=28
\end{aligned}
$
$\therefore$ Distance travelled in first 4 seconds
$
\begin{aligned}
& =|-4-1|+|1-0|+|0-5|+|5-28| \\
& =5+1+5+23=34 \mathrm{~m}
\end{aligned}
$

(iii)
$
\begin{aligned}
\mathrm{V}= & \frac{d s}{d t}=f^{\prime}(t)=6 t^2-18 t+12 \\
\mathrm{~V}= & 06\left(t^2-3 t+2\right)=0 \\
& \Rightarrow(t-1)(t-2)=0 \Rightarrow t=1 \text { or } 2
\end{aligned}
$
Now, $\quad a=\frac{d \mathrm{~V}}{d t}=12 t-18$
$\mathrm{a}(\mathrm{at} \mathrm{V}=0)$ is ' $\mathrm{a}$ ' at $\mathrm{t}=1$ and 2
Now a $($ at $t=1)=12-18=-6 \mathrm{~m} / \mathrm{sec}^2$
a $($ at $\mathrm{t}=2)=24-18=+6 \mathrm{~m} / \mathrm{sec}^2$

Question 4.

If the volume of a cube of side length
$\mathrm{x}$ is $\mathrm{V}=\mathrm{x}$
${ }^3$. Find the rate of change of the volume with respect to $x$
when $\mathrm{x}=5$ units.
Solution:
$
\begin{aligned}
\mathrm{V} & =x^3 \\
\frac{d \mathrm{~V}}{d x} & =3 x^2 \\
\text { at } x & =5, \frac{d \mathrm{~V}}{d x}=3\left(5^2\right)=3(25)=75 \text { units }
\end{aligned}
$

Question 5.
If the mass $\mathrm{m}(\mathrm{x})$ (in kilograms) of a thin rod of length $\mathrm{x}$ (in metres) is given by, $\mathrm{m}(\mathrm{x})=\sqrt{3 x}$ then what is the rate of change of mass with respect to the length when it is $x=3$ and $x=21$ metres.
Solution:
$
\begin{aligned}
m & =\sqrt{3 x}=\sqrt{3} \sqrt{x} \\
\frac{d m}{d x} & =\sqrt{3}\left[\frac{1}{2 \sqrt{x}}\right] \\
\frac{d m}{d x}(\text { at } x=3) & =\frac{\sqrt{3}}{2 \sqrt{3}}=\frac{1}{2} \mathrm{~kg} / \mathrm{m} \\
\frac{d m}{d x}(\text { at } x=27) & =\frac{\sqrt{3}}{2 \sqrt{27}}=\frac{\sqrt{3}}{2(3) \sqrt{3}}=\frac{1}{6} \mathrm{~kg} / \mathrm{m}
\end{aligned}
$

Question 6.
A stone is dropped into a pond causing ripples in the form of concentric circles. The radius $\mathrm{r}$ of the outer ripple is increasing at a constant rate at $2 \mathrm{~cm}$ per second. When the radius is $5 \mathrm{~cm}$ find the rate of changing of the total area of the disturbed water?
Solution:
$
\begin{aligned}
& \mathrm{A}=\pi r^2 \\
& \frac{d \mathrm{~A}}{d t}=\pi(2 r) \frac{d r}{d t} \\
& \therefore \frac{d \mathrm{~A}}{d t}=\pi(2)(5)(2)=20 \pi \text { sq.cm } / \mathrm{sec} \\
&
\end{aligned}
$
Question 7.
A beacon makes one revolution every 10 seconds. It is located on a ship which is anchored $5 \mathrm{~km}$ from a straight shore line. How fast is the beam moving along the shore line when it makes an angle of $45^{\circ}$ with the shore?
Solution:
Time for 1 rotation $=10 \mathrm{sec}$
So angular velocity $=\frac{d \theta}{d t}=\frac{2 \pi}{10}=\frac{\pi}{5}$
From the diagram $\tan 45^{\circ}=\frac{x}{5}=1 \Rightarrow x=5$

$
\text { Again } \begin{aligned}
\tan \theta & =\frac{x}{5} \Rightarrow x=5 \tan \theta \\
\Rightarrow \frac{d x}{d t} & =5 \sec ^2 \theta \frac{d \theta}{d t}=5\left[\sec ^2\left(45^{\circ}\right)\right] \frac{\pi}{5} \\
& =5(\sqrt{2})^2\left(\frac{\pi}{5}\right)=(\sqrt{2})^2 \pi=2 \pi \mathrm{km} / \mathrm{sec}
\end{aligned}
$

Question 8.
A conical water tank with vertex down of 12 metres height.has a radius of 5 metres at the top. If water flows into the tank at a rate $10 \mathrm{cubic} \mathrm{m} / \mathrm{min}$, how fast is the depth of the water increases when the water is 8 metres deep?
Solution:
Radius of cone $r=5 \mathrm{~m}$. Height of cone $h=12 \mathrm{~m}$,
$
\therefore \frac{r}{h}=\frac{5}{12} \Rightarrow r=\frac{5}{12} h
$
Volume of cone $(\mathrm{V})=\frac{1}{3} \pi r^2 h$

$
\text { (i.e) } \begin{aligned}
\mathrm{V} & =\frac{1}{3} \pi\left(\frac{5}{12} h\right)^2(h)=\frac{25}{432} \pi h^3 \\
\frac{d \mathrm{~V}}{d t} & =\frac{25 \pi}{432}\left[3 h^2 \frac{d h}{d t}\right]
\end{aligned}
$
Here $h=8 \mathrm{~m}$ and $\frac{d \mathrm{~V}}{d t}=10$ cubic $\mathrm{m}$
$
\Rightarrow 10=\frac{25 \pi}{432}(3)\left(8^2\right) \frac{d h}{d t} \Rightarrow 10 \times \frac{432}{25 \pi} \times \frac{1}{3 \times 64}=\frac{d h}{d t}
$
(i.e) $\frac{d h}{d t}=\frac{9}{10 \pi} \mathrm{m} /$ minute

Question 9.
A ladder 17 metre long is leaning against the wall. The base of the ladder is pulled away from the Waif at a rate of $5 \mathrm{~m} / \mathrm{s}$. When the base of the ladder is 8 metres from the wall
(i) How fast is the top of the ladder moving down the wall?
(ii) At what rate, the area of the triangle formed by the ladder, wall, and the floor, is changing?
Solution:

(i) $x^2+y^2=s^2=17^2$
When $x=8$
$
\begin{aligned}
8^2+y^2 & =17^2 \\
y^2 & =17^2-8^2=15^2 \Rightarrow y=15 \\
x^2+y^2 & =17^2 \\
2 x \frac{d x}{d t}+2 y \frac{d y}{d t} & =0 \\
\left(\div \text { by 2) } x \frac{d x}{d t}+y \frac{d y}{d t}\right. & =0
\end{aligned}
$
Here $x=8 \mathrm{~m}, \frac{d x}{d t}=5 \mathrm{~m} / \mathrm{sec}$ and $y=15 \mathrm{~m}$
To find $\frac{d y}{d t}$ :
$
\text { (8) } \begin{aligned}
(5)+(15)\left(\frac{d y}{d t}\right) & =0 \\
\therefore \frac{d y}{d t} & =\frac{-8 \times 5}{15}=\frac{-8}{3} \mathrm{~m} / \mathrm{sec} .
\end{aligned}
$
So the top of the ladder is moving down the wall at $\frac{-8}{3} \mathrm{~m} / \mathrm{sec}$

(ii) $
\begin{aligned}
\text { Area }=\mathrm{A}= & \frac{1}{2} b h \\
& \therefore \frac{d \mathrm{~A}}{d t}=\frac{1}{2}\left[b \frac{d h}{d t}+h \frac{d b}{d t}\right]
\end{aligned}
$
Here $b=$ base $=8 \mathrm{~m}$ and $h=$ height $=15 \mathrm{~m}$
$
\begin{aligned}
\frac{d b}{d t} & =5 \mathrm{~m} / \mathrm{sec} \text { and } \frac{d h}{d t}=\frac{-8}{3} \mathrm{~m} / \mathrm{sec} \\
\therefore \frac{d \mathrm{~A}}{d t} & =\frac{1}{2}\left[(8)\left(\frac{-8}{3}\right)+(15)(5)\right] \\
& =\frac{1}{2}\left[\frac{-64}{3}+75\right]=\frac{161}{6}=26.83 \mathrm{sq} \mathrm{m} / \mathrm{sec}
\end{aligned}
$

Question 10.
A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding car that has turned and moving straight east. When the jeep is $0.6 \mathrm{~km}$ north of the intersection and the car is $0.8$ $\mathrm{km}$ to the east. The police determine with a radar that the distance between them and the car is increasing at $20 \mathrm{~km} / \mathrm{hr}$. If the jeep is moving at $60 \mathrm{~km} / \mathrm{hr}$ at the instant of measurement, what is the speed of the car?
Solution:
$\begin{aligned} & s=\sqrt{x^2+y^2} \\ &=\sqrt{0.8^2+0.6^2} \\ &=\sqrt{1}=1 \\ & \text { Given } x=0.8, y=0.6 \frac{d y}{d t}=-60, \frac{d s}{d t}=20 \\ & \text { the figure } s^2=x^2+y^2\end{aligned}$
From the figure $s^2=x^2+y^2$ differentiate w.r to $t$

$
\begin{aligned}
2 s \frac{d s}{d t} & =2 x \frac{d x}{d t}+2 y \frac{d y}{d t} \\
(\div \text { by } 2) \Rightarrow \quad s \frac{d s}{d t} & =x \frac{d x}{d t}+y \frac{d y}{d t} \\
1(20) & =(0.8) \frac{d x}{d t}+(0.6)(-60) \\
\frac{d x}{d t} & =\frac{20+36}{0.8}=\frac{56}{0.8}=70 \\
\frac{d x}{d t} & =70 \mathrm{~km} / \mathrm{hr}
\end{aligned}
$