math quetion on rate of change

[lil^$tar]

solve dis quetion 4 me...a viscious liquid is poured on a flat surface.it forms a circular patch whose area grows at a steady rate of 5cm^2/s.find in terms of pie,the radius of the patch 20 seconds after pouring has commenced

 

[MAVtKnmJ]

Kindly show your steps and someone will tell you what you are doing wrong.

P.S

This could be solved via the Chain Rule Method

 

[anzaar]

here is the solution
dA/dt= 5
A= πr^2
take its derivative with respect to r
you will get
dA/dt=2πr
use chain rule
to find dr/dt
dr/dt=(dr/dA)x(dA/dt)=(1/2πr)x5= your ans

 

[MAVtKnmJ]

@anzaar

I'd suggest you let the OP make an effort before spoon feeding them

 

[anzaar]

ohk.

 

[sijanthapa]

The answer is:
In 1s the area of the patch is 5 cm^2
Then, in 20 seconds the area of the patch is 20*5 cm^2=100 cm^2

if A=100 then (pi)r^2=100
or,r^2=100/(pi)
or, r=10/root(pi) (at the instant when t=20seconds)
But i think the question has asked to find the rate of increase of radius in 20 seconds..
To find it,
d(pi*r^2)/dr * dr/dt=dA/dt
or 2*(pi)*r *dr/dt=5
or dr/dt=5/(2*pi*100/root(pi))
or, dr/dt=1/(20 *root(pi))

Ans=1/(20*root(pi))

 

[sijanthapa]

i am sorry. Actually i wrote it in hurry so..
To edit it the correct answer is:
dr/dt=5/(2*pi*r)
or, dr/dt=5/(2*pi*10/root pi)
solve it you will get,
dr/dt=1/(4*root pi)

This is the correct answer...