Differentiation (Part-38) | S N De

$~8.~~$ A curve passes through the point $~(2,0)~$ and satisfies the equation $~y~dy +4~dx=0;$ show that the curve is a parabola and find the length of its latus rectum.

Sol. We have,

$~~~y~dx+4~dx=0 \\ \text{or,}~~\displaystyle\int{y~dx}+\int{4~dx}=0 \\ \text{or,}~~\frac{y^2}{2}+4x=c\rightarrow(1)\\~~c\rightarrow\text{constant of integration.}$

Since the curve passes through the point $~(2,0)~$,we get from $~(1),$

$~~0+4\times 2=c \Rightarrow c=8.$

So, from $(1),~$ we get the equation of the curve as

$~~\frac{y^2}{2}+4x=8~~[\because ~c=8] \\ \text{or,}~~y^2+8x=16 \\ \text{or,}~~y^2=-8(x-2)\rightarrow(2)$

Hence, from $~(2)~$, we can say that the length of the latus rectum is $~8~$ unit.

$~9.~~$ A motor boat is moving in a straight line. Its velocity is $\,V\,$ when the motor is shut off. If the retardation at any subsequent time is equal to the magnitude of its velocity at that time, find its velocity and the distance traversed in time $\,t\,$ after the motor is shut off.

Sol. By the given condition,

$~~-\frac{dv}{dt}=v \\ \therefore \displaystyle\int_v^{v_0}{\frac{dv}{v}}=-\int_0^t{dt} \\ \text{or,}~~\left[~\log|v|~\right]_v^{v_0}=-[t]_0^t \\ \text{or,}~~-t=\log v_0-\log v \\ \text{or,}~~ \log\left(\frac{v_0}{v}\right)=-t \\ \text{or,}~~\frac{v_0}{v}=e^{-t} \\ \text{or,}~~ v_0=ve^{-t}\rightarrow(1)$

Again, $~~-v\frac{dv}{dx}=v \\ \therefore \displaystyle\int_v^{v_0}{dv}=-\int_0^x{dx} \\ \text{or,}~~[v]_v^{v_0}=-[x]_0^x \\ \text{or,}~~ v_0-v=-x \\ \text{or,}~~ x=v-v_0 \\ \text{or,}~~ x=v-ve^{-t} \\ \text{or,}~~ x=v(1-e^{-t})\rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2)\,$ we can find its velocity and the distance traversed in time $\,t\,$ after the motor is shut off.

$~10.~~$ A particle starts from the origin and moves along a straight line. If the velocity of the particle at time $\,t\,$ seconds be $\,10t\,$ cm/s, find the distance traversed during first $\,4\,$ seconds after the start.

Sol. By the given condition,

$~~v=10t.$

$~~\therefore~~$ If the distance traversed is $~x$, then

$~~\frac{dx}{dt}=v \\ \text{or,}~~\displaystyle\int{dx}=\int{10t~dt} \\ \text{or,}~~ x=10 \times \frac{t^2}{2}+c \\ \text{or,}~~ x=5t^2+c \rightarrow(1)$

If the particle starts from origin, then $~x=0,~t-0$ and so from $~(1)~$ we get,

$~~0=5 \times 0^2+c\Rightarrow c=0.\\ \therefore x=5t^2~~[\text{By (10}]$

Now, when $~t=5~\text{sec},~~x=5 \times 16=80~\text{(cm.)}~~\text{(ans)}$

$\,11.~~$ The velocity of a particle moving in a straight path is given by $\,v=16t-4t^2\,\,$ where and are measured in ft/s and second units respectively. If the initial displacement be $\,10\,$ ft, find  the displacement of the particle in $\,t\,$ seconds.

Sol. Veclocity of the particle is given by : $~~v=16t-4t^2.$

$~\therefore~$ If the displacement of the particle is $~~x,$

then $~~\frac{dx}{dt}=v \\ \therefore \frac{dx}{dt}=16t-4t^2 \\ \text{or,}~~\displaystyle\int{dx}=\int{16t~dt}-\int{4t^2~dt}\\ \text{or,}~~x=16 \times \frac{t^2}{2}-4\times \frac{t^3}{3}+c\rightarrow(1) \\ ~~~c \rightarrow \text{constant of integration.}$

Since the initial displacement be $\,10\,$ ft, 

$~~\therefore~~\text{for}~~t=0,~~x=10.$

So, by $~(1),~$ we get,

$~~10=0-0+c \\ \text{or,}~~c=0.$

Hence, putting the value of $\,c\,$ in $~(1),~$ we get,

$~~x=8t^2-\frac 43t^3+10.$

So, the displacement of the particle in $\,t\,$ seconds is given by :

 $~x=8t^2-\frac 43t^3+10~~\text{ft.}$

$\,12.~~$ The acceleration of a particle moving along a straight line is $\,(6t^2-3t)\,\text{cm/}~s^2$ , at time $\,t\,$ seconds. If the velocity of the particle at time $\,2\,$ seconds be $\,10\,\text{cm/s}~,$  find its velocity at time $\,t\,$ seconds.

Sol. The acceleration of the particle $~f=6t^2-3t.$

If velocity of the particle be $~v,$ then

$~~\frac{dv}{dt}=f \\ \therefore~\frac{dv}{dt}=6t^2-3t \\ \text{or,}~~\displaystyle\int{dv}=\int{6t^2~dt}-\int{3t~dt} \\ \text{or,}~~v=\frac{6t^3}{3}-\frac{3t^2}{2}+c\rightarrow(1)\\ ~~~c \rightarrow \text{constant of integration.}\\ \therefore 10=\frac{6 \times 2^3}{3}-\frac{3 \times 2^2}{2}+c~~[*] \\ \text{or,}~~10=16-6+c \\ \text{or,}~~ 10=10+c \\ \text{or,}~~ c=0\rightarrow(2)$

Hence, from $~(1),~(2)~$ we get,

$~~v=2t^3-\frac 32t^2~~\text{(ans.)}$

Note[*] : Since the velocity of the particle at time $\,2\,$ seconds is given to be $\,10\,\text{cm/s}.$

$~13.~~$ If the acceleration of a particle at time $\,t\,$ seconds be $\,1.2~t^2\,$ ft/$s^2$ and velocity of the particle at time $\,2\,$ seconds be $\,2.2\,$ ft/s, then find its velocity at time $\,5\,$ seconds.

Sol.  The acceleration od the particle $~(f)=1.2~t^2.$

If the velocity of the particle be $\,v,$

$~~\frac{dv}{dt}=1.2~t^2 \\ \text{or,}~~\displaystyle\int{dv}=1.2 \int{t^2~dt} \\ \text{or,}~~v=\frac{1.2}{3}~t^3+c\rightarrow(1)\\~~c\rightarrow\text{constant of integration.}$

Now, for $~t=2 ~\text{sec},~v=2.2~\text{ft/s.}$

So, from $~(1),~$ we get,

$~~2.2=\frac{1.2}{3}\times 8+c \\ \text{or,}~~6.6=1.2 \times 8+3c \\ \text{or,}~~ 6.6-9.6=3c \\ \text{or,}~~ -3=3c \\ \text{or,}~~ c=\frac{-3}{3}=-1.\\ \therefore~v=\frac{1.2}{3} t^3-1~~[\text{By (1)}].$

Now, for $~t=5~\text{sec},~~v=\frac{1.2}{3} \times 125-1=49.$

So, the velocity of the particle $=49~\text{ft/sec.}~~\text{(ans.)}$

$~14.~~$  A particle starts from the origin with a velocity of $\,10\,$ cm/s and moves along a straight line. If its acceleration be $\,(2t^2-3t)\,$ cm/$s^2$ at the end of $\,t\,$ seconds, then find its velocity and the distance from the origin at the end of $\,6\,$ seconds.

Sol. The acceleration of the particle $(f)=2t^2-3t.$

If the velocity of the particle  $\,v,\,$ then $~\frac{dv}{dt}=f=2t^2-3t.$

$~\therefore~\displaystyle\int{dv}=\int{(2t^2-3t)~dt} \\ \text{or,}~~ \displaystyle\int{dv}=2\int{t^2~dt}-3\int{t~dt} \\ \text{or,}~~ v=\frac 23t^3-\frac 32t^2+c\rightarrow(1)\\~~c \rightarrow\text{constant of integration.}$

Since the particle starts from the origin $~(t=0)~$ with a velocity $\,(v)\,$ of $\,10\,$ cm/s, we get from $\,(1),$ 

$~~~10=0-0+c \Rightarrow c=0.$

Hence, putting the value of $\,c,\,$ we get from $\,(1),$

$~~v=\frac{2t^3}{3}-\frac{3t^2}{2}+10 \rightarrow(2)$

So, its velocity $\,(v)\,$ from the origin at the end of $\,6\,$ seconds is :

$~~~~v=\frac{2 \times 6^3}{3}-\frac{3 \times 6^2}{2}+10~~[\text{By (2)}] \\ \text{or,}~~v=144-54+10 \\ \text{or,}~~v=100.$

Hence, the required velocity is : $~~100~$ cm/sec.

Again, if $\,x\,$ be the displacement of the particle from the origin at a time $\,t\,$, then 

$~~v=\frac{dx}{dt} \\ \therefore \frac{dx}{dt}=\frac{2t^3}{3}-\frac{3t^2}{2}+10 \\ \therefore \displaystyle\int{dx}=\frac 23 \int{t^3~dt}-\frac 32 \int{t^2~dt}\\~~~~+10\displaystyle\int{dt} \\ \text{or,}~~x=\frac 23 \cdot \frac{t^4}{4}-\frac 32 \cdot \frac{t^3}{3}+10t+c_1 \\~~[c_1 \rightarrow\text{constant of integration.}]\\ \text{or,}~~x=\frac{t^4}{6}-\frac{t^3}{2}+10t+c_1\rightarrow(3)$

Now, at the origin , $~x=0,t=0.$

So, putting $~x=0,~t=0\,$ we get from $\,(3)\,$ ,

$~~0=0-0+0+c_1 \Rightarrow c_1=0.$

Now,  we have to find out the distance from the origin at the end of $\,6\,$ seconds and so  replacing $~c_1=0, ~t=6$ in $\,(3)\,$ we get,

$~~~~x= \frac 16 \times 6^4-\frac 12 \times 6^3+10 \times 6+0\\ \therefore ~x=166.$

So, the distance of the particle from the origin at the end of $\,6\,$ seconds is $~~166 ~\text{cm.}$

$~15.~~$ A particle starts from the origin with a velocity of $\,2\,$ cm/s and moves along a straight line. If the acceleration of the particle at a distance $\,x\,$ cm from the origin be $\,(2x-3/2)\,$ cm/$s^2$, find the velocity of the particle at a distance $\,7\,$ cm from the origin.

Sol. The acceleration  of the particle $~(f)=2x-\frac 32\,$ and let $\,v\,$ denotes the velocity of particle.

$~~ \therefore ~v\frac{dv}{dx}=2x-\frac 32~~[\because f=v\frac{dv}{dx}] \\ \text{or,}~~ \displaystyle\int{v~dv}=2\int{x~dx}-\frac 32 \int{dx} \\ \text{or,}~~ \frac{v^2}{2}=2\cdot\frac{x^2}{2}-\frac 32 x+c' \\~~c' \rightarrow\text{constant of integration.} \\ \text{or,}~~v^2=2x^2-3x+c \rightarrow(1)\\~~[~c=2c']$

Now, at the origin $~~x=0,~v=2.$

So, from $\,(1)\,$ we get, 

$~2^2=0-0+c \Rightarrow c=4$

Now, for $~x=7,~~v^2=2 \times 7^2-3 \times 7+4 \\ ~~~~~~~~\therefore v^2=98-21+4\\ ~~~~~~~~\text{or,}~~ v^2=81 \\ ~~~~~~~~\text{or,}~~ v=\sqrt{81}=9 $

So, the velocity of the particle at a distance $\,7\,$ cm from the origin is $~~9~\text{cm/sec.}$