Exercise 7.1-Additional Problems - Chapter 7 - Applications of Differential Calculus - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions
Question 1.
A water tank has the shape of an inverted circular cone with base radius 2 metres and height 4 metres. If water is being pumped into the tank at a rate of $2 \mathrm{mt}^3 / \mathrm{min}$, find the rate at which the water level is rising when the water is $3 \mathrm{~m}$ deep.
Solution:
We first sketch the cone and label it as in diagram. Let $\mathrm{V}, \mathrm{r}$ and $\mathrm{h}$ be respectively the volume of the water, the radius of the cone and the height at time $t$, where $t$ is measured in minutes.
We are given that $\frac{d \mathrm{~V}}{d t}=2 \mathrm{~m}^3 / \mathrm{min}$ and we are asked to find $\frac{d h}{d t}$ when $\mathrm{h}$ is $3 \mathrm{~m}$.

The quantities $\mathrm{V}$ and $\mathrm{h}$ are related by the equation $\mathrm{V}=\frac{1}{3} \pi r^2 h$. But it is very useful to express $\mathrm{V}$ as function of $\mathrm{h}$ alone.
In order to eliminate $r$ we use similar triangles in diagram to write $\frac{r}{h}=\frac{2}{4} \Rightarrow r=\frac{h}{2}$ and the expression for $\mathrm{V}$ becomes $\mathrm{V}=\frac{1}{3} \pi\left(\frac{h}{2}\right)^2 h=\frac{\pi}{12} h^3$.
Now we can differentiate each side with respect to $t$ and we have
$
\frac{d \mathrm{~V}}{d t}=\frac{\pi}{4} h^2 \frac{d h}{d t} \Rightarrow \frac{d h}{d t}=\frac{4}{\pi h^2} \frac{d \mathrm{~V}}{d t}
$
Substituting $h=3 \mathrm{~m}$ and $\frac{d \mathrm{~V}}{d t}=2 \mathrm{~m}^3 / \mathrm{min}$.
We get, $\frac{d h}{d t}=\frac{4}{\pi(3)^2} 2=\frac{8}{9 \pi} \mathrm{m} / \mathrm{min}$

Question 2.
A car A is travelling from west at $50 \mathrm{~km} / \mathrm{hr}$ and car B is travelling towards north at $60 \mathrm{~km} / \mathrm{hr}$. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car $\mathrm{A}$ is $0.3$ kilometers and car $\mathrm{B}$ is $0.4$ kilometers from the intersection?
Solution:
We draw diagram where $\mathrm{C}$ is the intersection of the two roads. At a given time $t$, let $\mathrm{x}$ be the distance from car $\mathrm{A}$ to $\mathrm{C}$, let $\mathrm{y}$ be the distance from car $\mathrm{B}$ to $\mathrm{C}$ and let $\mathrm{z}$ be the distance between the cars $\mathrm{A}$ and $\mathrm{B}$ where $\mathrm{x}$, $\mathrm{y}$ and $\mathrm{z}$ are measured in kilometers.
We are given that $\frac{d x}{d t}=-50 \mathrm{~km} / \mathrm{hr}$ and $\frac{d y}{d t}-=-60 \mathrm{~km} / \mathrm{hr}$.

Note that $\mathrm{x}$ andy are decreasing and hence the negative sign.
We are asked to find $\frac{d z}{d t}$
The equation that relate $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ is given by the pythagoras theorem $\mathrm{z}^2=\mathrm{x}^2+\mathrm{y}^2$ Differentiating each side with respect to $t$,
we have $2 z \frac{d z}{d t}=2 x \frac{d x}{d t}+2 y \frac{d y}{d t} \Rightarrow \frac{d z}{d t}=\frac{1}{z}\left(x \frac{d x}{d t}+y \frac{d y}{d t}\right)$
When $x=0.3$ and $y=0.4 \mathrm{~km}$, we get $z=0.5 \mathrm{~km}$ and we get
$
\frac{d z}{d t}=\frac{1}{0.5}[0.3(-50)+40(-60)]=-78 \mathrm{~km} / \mathrm{hr} \text {. }
$
i.e., The cars approaching each other at a rate of $78 \mathrm{~km} / \mathrm{hr}$

Question 3.
The distance $x$ metres travelled by a vehicle in time $t$ seconds after the brakes are applied is given by $x=$ $20 \mathrm{t}-\frac{5}{3} \mathrm{t}^2$. Determine
(i) the speed of the vehicle (in $\mathrm{km} / \mathrm{hr}$ ) at the instant the brakes are applied and
(ii) the distance the car travelled before it stops.
Solution:
$x=20 t-\frac{5}{3} t^2$
Velocity $=v=\frac{d x}{d t}=20-\frac{5}{3}(2 t)=20-\frac{10 t}{3}$
(i) Velocity $($ at $t=0)=20-\frac{10}{3}(0)=20 \mathrm{~km} / \mathrm{hr}$
(ii) $v=0 \Rightarrow 20-\frac{10 t}{3}=0$
(i.e) $10 t=60$ or $t=6$

$\begin{aligned}
& \text { Now } x(\text { at } v=0)=x(\text { at } t=6)=20(6)-\frac{5}{3}(6)^2 \\
& =120-\frac{5}{3}(36)=120-60=60 \mathrm{~m} \\
&
\end{aligned}$

Question 4.
At noon, ship A is $100 \mathrm{~km}$ west of ship B. Ship A is sailing east at $35 \mathrm{~km} / \mathrm{hr}$ and ship B is sailing north at $25 \mathrm{~km} / \mathrm{hr}$. How fast is the distance between the ships changing at $4.00 \mathrm{pm}$ ?
Solution:
Distance travelled by A in 4 hrs
$(12$ noon to $4.00 \mathrm{pm})=4 \times 35=140 \mathrm{~km}$

$
\therefore \mathrm{BA}^{\prime}=40 \mathrm{~km}(140-100)
$
Distance travelled by B in $4 \mathrm{hrs}=4 \times 25=100 \mathrm{~km}$
Now in right angled $\Delta \mathrm{A}^{\prime} \mathrm{BB}^{\prime}, \mathrm{x}^2+\mathrm{y}^2=\mathrm{s}^2$
Here, $\mathrm{x}=\mathrm{BA}^{\prime}=40 \mathrm{~km}$
$
\begin{aligned}
& \mathrm{y}=\mathrm{BB}^{\prime}=100 \mathrm{~km} \\
& \Rightarrow \mathrm{s}^2=40^2+100^2 \\
& =1600+10000 \\
& =11600
\end{aligned}
$
$
\begin{aligned}
\therefore s & =\sqrt{11600} \\
& =10 \sqrt{116}=10 \sqrt{4 \times 29}=20 \sqrt{29} \mathrm{~km} \\
x^2+y^2 & =s^2
\end{aligned}
$
So, $2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=2 s \frac{d s}{d t}$
$
(\div \text { by } 2), x \frac{d x}{d t}+y \frac{d y}{d t}=s \frac{d s}{d t}
$
Here, $x=40, y=100, \frac{d x}{d t}=35, \frac{d y}{d t}=25$ and $s=20 \sqrt{29}$

$
\begin{aligned}
\Rightarrow \frac{d s}{d t} & =\frac{x \frac{d x}{d t}+y \frac{d y}{d t}}{s}=\frac{(40)(35)+(100)(25)}{20 \sqrt{29}} \\
& =\frac{1400+2500}{20 \sqrt{29}}=\frac{3900}{20 \sqrt{29}}=\frac{390}{2 \sqrt{29}}=\frac{195}{\sqrt{29}} \mathrm{~km} / \mathrm{hr}
\end{aligned}
$
(i.e) The distance between the ships is increasing at the rate of $\frac{195}{\sqrt{29}} \mathrm{~km} / \mathrm{hr}$.

Question 5.
Gravel is being dumped from a conveyor belt at a rate of $30 \mathrm{ft}^3 / \mathrm{min}$ and its coarsened such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is $10 \mathrm{ft}$ high?
Solution:
Given:
Diameter $=$ Height
$=\mathrm{h} \mathrm{ft}$ (say)
$\begin{aligned} \therefore \text { radius } & =r=\frac{h}{2} \\ \text { Volume }=\mathrm{V} & =\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi\left(\frac{h}{2}\right)^2 h=\frac{1}{3} \pi\left(\frac{h^2}{4}\right)(h)=\frac{1}{12} \pi h^3 \\ \text { Volume }=\mathrm{V} & =\frac{1}{12} \pi h^3 \\ \frac{d \mathrm{~V}}{d t} & =\frac{1}{12} \pi\left[3 h^2 \frac{d h}{d t}\right] ;(i . e), \quad \frac{d \mathrm{~V}}{d t}=\frac{\pi}{4} h^2 \frac{d h}{d t}\end{aligned}$
Here, $\frac{d \mathrm{~V}}{d t}=30 \mathrm{ft}^3$ and $h=10 \mathrm{ft}$
$
\begin{aligned}
& \Rightarrow 30=\frac{\pi}{4}(10)^2 \frac{d h}{d t} \\
& \Rightarrow \frac{d h}{d t}=\frac{30 \times 4}{100 \pi}=\frac{120}{100 \pi}=\frac{6}{5 \pi} \mathrm{ft} / \mathrm{min}
\end{aligned}
$
So, the height is increasing at the rate of $\frac{6}{5 \pi} \mathrm{ft} / \mathrm{min}$