# Rate of change problem.

• Mar 30th 2010, 09:21 PM
Kakariki
Rate of change problem.
Hey, I am having a really hard time solving this problem. I need all the help you can give me.

Question:
When a small pebble is dropped in a pool of still water, it produces a circular wave that travels outward at a constant speed of 15 cm/s. At what rate is the area inside the wave increasing
a) when the radius is 5 cm?
b) when the area is 400pi cm^2
c) when 4s have elapsed

Solution
Basically I have no idea how to solve this problem. I know that the area of a circle can be found by $\displaystyle A = pi r^2$. However I do not know how to factor in the fact that the wave is traveling at 15m/s.

Any help is greatly appreciated!

I went to move onto the next problem, but it is similar. I need help understanding how to answer these. Here is next question:
Question 2:
The angle of elevation of the sun is decreasing at 1/4 rad/h. How fast is the shadow cast by a building of height 50m lengthening, when the angle of elevation of the sun is pi/4?

Solution
I am having the same problem. I do not know how to go about solving this.
• Mar 30th 2010, 09:49 PM
TKHunny
Consider everything a function of time (t). For your circle:

$\displaystyle A = \pi\cdot r^{2}$ suggests $\displaystyle A(t) = \pi\cdot r(t)^{2}$

We find the derivative with respect to t = time.

On the left-had side, we have simply $\displaystyle dA(t)/dt$.

On the right-hand side, we have $\displaystyle \pi\cdot (2r(t)\cdot (dr(t)/dt))$

Putting it all together (and dispensing with the cumbersome '(t)' notation), we get

$\displaystyle dA/dt = \pi\cdot (2r\cdot (dr/dt))$

Knowing this, one is armed with enough to solve many problem types.

Actually, we may also want to dispose of the cumbersome 'dt' notation and just go with the differential usage.

$\displaystyle dA = \pi\cdot (2r\cdot dr)$

In words, this says, "The change in the area is $\displaystyle 2\pi r$ multiplied by the change in the radius.

We are given dr = 15 cm/s.

Let's see what you get.
• Mar 30th 2010, 10:12 PM
Kakariki
Quote:

Originally Posted by TKHunny
Consider everything a function of time (t). For your circle:

$\displaystyle A = \pi\cdot r^{2}$ suggests $\displaystyle A(t) = \pi\cdot r(t)^{2}$

We find the derivative with respect to t = time.

On the left-had side, we have simply $\displaystyle dA(t)/dt$.

On the right-hand side, we have $\displaystyle \pi\cdot (2r(t)\cdot (dr(t)/dt))$

Putting it all together (and dispensing with the cumbersome '(t)' notation), we get

$\displaystyle dA/dt = \pi\cdot (2r\cdot (dr/dt))$

Knowing this, one is armed with enough to solve many problem types.

Actually, we may also want to dispose of the cumbersome 'dt' notation and just go with the differential usage.

$\displaystyle dA = \pi\cdot (2r\cdot dr)$

In words, this says, "The change in the area is $\displaystyle 2\pi r$ multiplied by the change in the radius.

We are given dr = 15 cm/s.

Let's see what you get.

Thank you for responding! However, I am confused by your explanation.
I do not understand how you can have everything as a function of time. The function:$\displaystyle A(t) = \pi\cdot r(t)^{2}$ just doesn't make sense to me. I realize that the radius is equal to 15*t, so the time when the radius is 5 is 5/15. And putting that through your formula gives me a different answer than putting it through the normal area of a circle formula.

Hopefully this makes sense.
• Mar 31st 2010, 01:28 AM
HallsofIvy
Then until you really understand "function notation" and "composition of functions" you shouldn't be attempting problems like this. That should have been mastered before starting Calculus. I suggest you ask your teacher to help you review this.
• Mar 31st 2010, 08:49 AM
ione
Quote:

Originally Posted by Kakariki
I do not understand how you can have everything as a function of time. The function:$\displaystyle A(t) = \pi\cdot r(t)^{2}$ just doesn't make sense to me.

Over time the circle is getting bigger and the area is getting bigger. So the area is a function of time.

$\displaystyle A(t) = \pi\cdot r(t)^{2}$ means that the area at time t is pi times the square of the radius at time t.

Quote:

Originally Posted by Kakariki
I realize that the radius is equal to 15*t, so the time when the radius is 5 is 5/15. And putting that through your formula gives me a different answer than putting it through the normal area of a circle formula.

So r(5/15) = 5

$\displaystyle A(\frac{5}{15})=\pi (r(\frac{5}{15}))^2=\pi(5)^2=25\pi$ which is the same answer you get using the normal area of a circle formula.
• Mar 31st 2010, 01:41 PM
TKHunny
Quote:

Originally Posted by Kakariki
...normal area of a circle formula.