circle word problem

**yoman360** · June 29th 2009, 08:42 PM

A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 60 cm/s.

a) Express the radius r of this circle as a function of the time t (in seconds).

b) If A is the area of this circle as a function of the radius, A(r(x)) and interpret it.

Please help! I don't even know how to start this problem.

**pickslides** · June 29th 2009, 09:06 PM

Originally Posted by yoman360

A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 60 cm/s.

a) Express the radius r of this circle as a function of the time t (in seconds).

$\frac{dr}{dt}= 60 \Rightarrow r = \int \left(\frac{dr}{dt}\right)dt= r = \int 60 dt= 60t+c$

as $r = 0$ at $t = 0 \Rightarrow c=0$

therefore $r = 60t$

**pickslides** · June 29th 2009, 09:11 PM

Originally Posted by yoman360

b) If A is the area of this circle as a function of the radius, A(r(x)) and interpret it.

$A= \pi r^2$ and $r = 60t$ therefore $A= \pi (60t)^2 = 3600\pi t^2$

$A = 3600\pi t^2$ is the Area of the circle at time $t$

**mr fantastic** · June 29th 2009, 09:40 PM

Originally Posted by pickslides

$\frac{dr}{dt}= 60 \Rightarrow r = \int \left(\frac{dr}{dt}\right)dt= r = \int 60 dt= 60t+c$

as $r = 0$ at $t = 0 \Rightarrow c=0$

therefore $r = 60t$

If the speed is constant (and I assume it is because the question has been posted in the PRE-calculus subforum) then r = 60t can immediately be written (where unit of r is cm).

**yoman360** · July 10th 2009, 12:05 PM

Originally Posted by pickslides

$\frac{dr}{dt}= 60 \Rightarrow r = \int \left(\frac{dr}{dt}\right)dt= r = \int 60 dt= 60t+c$

as $r = 0$ at $t = 0 \Rightarrow c=0$

therefore $r = 60t$

We don't learn integrals in pre-cal. So I can't use integrals if we never learned about them.

**yoman360** · July 10th 2009, 12:48 PM

Can anyone else do this without using integrals?

**e^(i*pi)** · July 10th 2009, 01:13 PM

Originally Posted by yoman360

Can anyone else do this without using integrals?

Mr F did in post 4

**yoman360** · July 10th 2009, 01:18 PM

Originally Posted by e^(i*pi)

Mr F did in post 4

well how did he get r=60t?

**Prove It** · July 10th 2009, 01:55 PM

Originally Posted by yoman360

well how did he get r=60t?

The radius is extending at the rate of 60cm per second.

After 1 second, the radius is 60cm
After 2 seconds, the radius is 2 x 60cm
After 3 seconds, the radius is 3 x 60cm, etc.

So if t represents the time in seconds, the radius is t x 60cm.

This is where the formula $r = 60t$ comes from.

**yoman360** · July 10th 2009, 03:31 PM

Originally Posted by Prove It

The radius is extending at the rate of 60cm per second.

After 1 second, the radius is 60cm
After 2 seconds, the radius is 2 x 60cm
After 3 seconds, the radius is 3 x 60cm, etc.

So if t represents the time in seconds, the radius is t x 60cm.

This is where the formula $r = 60t$ comes from.

Thanks!

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