# Implicit Differentiation word problem

• Apr 4th 2010, 09:05 AM
satx
Implicit Differentiation word problem
"A stone is dropped into a still pond. Concentric circular ripples spread out and the radius of the disturbed area increases at a rate of 16 cm/sec. At what rate does the area of the disturbed region increase when its radius is 4 cm?"

don't really know where to start.

I said A = πr^2 --> r = sqrt(A/π) --> d/dt(r) = sqrt(A/π) . d/dt

Now I'm lost...
• Apr 4th 2010, 10:21 AM
CaptainBlack
Quote:

Originally Posted by satx
"A stone is dropped into a still pond. Concentric circular ripples spread out and the radius of the disturbed area increases at a rate of 16 cm/sec. At what rate does the area of the disturbed region increase when its radius is 4 cm?"

don't really know where to start.

I said A = πr^2 --> r = sqrt(A/π) --> d/dt(r) = sqrt(A/π) . d/dt

Why did you do that? (and what is it you think you have done for that matter?)

$\displaystyle A=\pi r^2$

Now just differentiate this with respect to time.

CB
• Apr 4th 2010, 10:26 AM
satx
So, d/dt (A) = πr^2(d/dt) = π4^2/.25s = 64π cm^2 per second?
• Apr 4th 2010, 10:37 AM
Jhevon
Quote:

Originally Posted by satx
So, d/dt (A) = (d/dt)πr^2 = π4^2/.25s = 64π cm^2 per second?

first note what is in red above

Now, $\displaystyle A = \pi r^2$

$\displaystyle \Rightarrow \frac {dA}{dt} = 2 \pi r \cdot \frac {dr}{dt}$

Now continue
• Apr 4th 2010, 10:45 AM
satx
So (2π4^2)/.25s = 128π sq cm per sec?
• Apr 4th 2010, 10:57 AM
Jhevon
Quote:

Originally Posted by satx
So (2π4^2)/.25s = 128π sq cm per sec?

are you looking at what's posted? where is 4^2 coming from? There is nothing squared in what I posted. And what's that 0.25?

$\displaystyle \frac {dA}{dt} = 2 \pi (4) (16) ~\frac {\text {cm}^2}{\text{sec}}= 128 \pi ~\frac {\text {cm}^2}{\text{sec}}$
• Apr 4th 2010, 11:05 AM
satx
Quote:

Originally Posted by Jhevon
first note what is in red above

Now, $\displaystyle A = \pi r^2$

$\displaystyle \Rightarrow \frac {dA}{dt} = 2 \pi r \cdot \frac {dr}{dt}$

Now continue

Quote:

Originally Posted by Jhevon
are you looking at what's posted? where is 4^2 coming from? There is nothing squared in what I posted. And what's that 0.25?

$\displaystyle \frac {dA}{dt} = 2 \pi (4) (16) ~\frac {\text {cm}^2}{\text{sec}}= 128 \pi ~\frac {\text {cm}^2}{\text{sec}}$

I was multiplying the r by the r in dr/dt. And the .25 sec is the time it takes the radius to reach 4 cm. I told you I didn't really know where to start...

Anyway, so the end result is right then?
• Apr 4th 2010, 11:18 AM
Jhevon
Quote:

Originally Posted by satx
I was multiplying the r by the r in dr/dt. And the .25 sec is the time it takes the radius to reach 4 cm. I told you I didn't really know where to start...

Anyway, so the end result is right then?

dr/dt is notation. The r in that is not available to be multiplied.

Do not complicate things. You do exactly as I did, and just plug in. dr/dt is the rate at which the radius is changing. That was given in the problem, it is 16 cm/sec, so you just plug in dr/dt = 16 the form I gave you. Then, r = 4, and compute.
• Apr 4th 2010, 10:13 PM
CaptainBlack
Quote:

Originally Posted by satx
I was multiplying the r by the r in dr/dt. And the .25 sec is the time it takes the radius to reach 4 cm. I told you I didn't really know where to start...

Anyway, so the end result is right then?

You are not yet ready for implicit differentiation, you need to go back and revise the basics of differentiation.

CB