- #1

- 6

- 0

## Homework Statement

(Translated, sorry my English) Find how fast does the radius of a circle get smaller when its area is 75*pi cm^2 and it (the area) gets smaller at a rate of 2*pi cm^2/s?

## Homework Equations

Not really any given equations, but I used these:

A = r^2 * pi

s = v*t + s0 (If v is a constant)

t=A/Va (where t is the time when A=0)

## The Attempt at a Solution

From now on I'll call area, A = 75*pi cm^2 and the rate v_a = 2*pi cm^2/s.

I gave myself that the speed (v) was constant, that shouldn't really change the answer.

So s = dv/dt = v*t + s0

By using A = r^2 * pi we know that:

r^2 = A/pi

r = sqrt(A/pi) and since at t = 0 then A = 75*pi so

r = sqrt(75) at t = 0, lets call r from now on s.

Let's find the time until A = 0, and therefore s=0

t = A/v_a = 75*pi/(2*pi) = 37,5 sec

So in 37,5 sec the circle area will be zero with is radius. Therefore we can see that:

s = v*t + s0

0 = v*37,5 + sqrt(75)

-sqrt(75) = v*37,5

v = -sqrt(75)/37,5

v = -2*sqrt(75)/75

v = -2/sqrt(75)

So my final answer was v = -2/sqrt(75) (or v = -2/(5*sqrt(3))

But according to my teacher the correct answer is v = -1/sqrt(75).

--

Here's how my teacher did it:

We know that F(t) = pi * r(t)^2 and

(1) F'(t0) = 2*pi*r(t0)*r'(t0) = -2*pi

(2) F(t0) = pi*r(t)^2 = 75*pi

for a specific t0.

According to (2) we know that r(t0) = sqrt(75) = 5*sqrt(3) and according to (1) we get r'(t0) = -1/r(t0) = -1/sqrt(75)

--

He called my solution "a waste of time" although he said it should work too, but couldn't explain to me why I didn't get the same answer.

Where did I go wrong (or even better, where did he go wrong?)