Question

A stone is dropped into a lake, creating a circular ripple that travels outward at a speed 25cm/sec. Find the rate at which the area within the circle is increasing after 4sec.

Answer 1

$$
A=\pi r^2\\
{dA\over dt}=2\pi r{dr\over dt}=2\pi rv\\
r=\int v~dt=vt+K\\
r(0)=0\implies K=0\\
r(t)=vt\\
\implies {dA\over dt}=2\pi r{dr\over dt}=2\pi v^2t\\
$$
\(v=0.25\) and \(t=4\)

Does this make sense?

Answer 2

sorry, no. I'm doing basic related rates...

Answer 3

Do you know calculus?

Answer 4

Well, i just started taking the class. I'm on the application of derivatives.

Answer 5

Ok, so first we define Area
Next we define its derivative with respect to time
$$
A=\pi r^2\\
{dA\over dt}=2\pi r{dr\over dt}\\
$$
The only thing that changes is \(r\), so that is the only thing about Area that can change. Does this make sense?

Answer 6

Yes

Answer 7

So since \(r\) is the only thing that can change and we also know that \(r\) is a function of time; when we take the derivative of A with respect to time, we also take the derivative of \(r\) with respect to time, because that is the thing about Area that is changing.

The derivative of \(r^2\) is \(2r\). But \(r\) is also a function of time, so we must multiply \(2r\) by \({dr\over dt}\) to account for this fact.
$$
{dA\over dt}=2\pi r{dr\over dt}\\
$$

We also know that \({dr\over dt}\) is just velocity (distance traveled over time taken). So \({dr\over dt}=v\).

Ok so far?

Answer 8

i think so... okay.

Answer 9

Note that we used a rule called "Implicit Differentiation" when we took the derivative of A:
$$
{dA\over dt}=2\pi r{dr\over dt}\\
$$
http://en.wikipedia.org/wiki/Implicit_function#Implicit_differentiation

It might be better to write \(A\) and \(r\) as a function of time \(A=A(t),r=r(t)\) to remind us that we will need to use this rule:
$$
{dA(t)\over dt}=2\pi r(t){dr(t)\over dt}\\
$$

Answer 10

Does velocity make sense to you? Do you see that \({dr\over dt}\) is just velocity?

Answer 11

I think so, because the radius represents the distance, so there is distance over time, which the velocity.

Answer 12

Next we need to find a relationship between \(r\) and time. To do this, we know that velocity times time is distance (using the same idea that velocity is distance over time). Therefore,
$$
r(t)=vt
$$
This is how big the circle is after some time \(t\). We now have everything we need.
$$
{dA\over dt}=2\pi r{dr\over dt}=2\pi \times vt\times v\\
$$

Answer 13

This represents how fast the Area \(A\) is changing \(dA\) over any change in time \(dt\).

Answer 14

It takes time to process all this material. Read over it and ask any questions you want later. I'll check over the next few days. Ok?

Answer 15

okay, thank you!!

Answer 16

You're welcome. Keep hacking at this, took me some time as well, but you will get it!!

Answer 17

dA/dt=2pi(100)(25)??? @ybarrap

Answer 18

0.25, not 25, answer will be in meters squared per second

Answer 19

But all the values are in centimeters. @ybarrap