Question

The radius of a circular puddle is growing at a rate of 5 cm/s.
(a) How fast is its area growing at the instant when the radius is 10 cm?
(b) How fast is the area growing at the instant when it equals 25 cm2?

Answer 1

\[\large r'=5 \;cm/s\]Area of a circle is,\[\large A=\pi r^2\]Part a) is asking,\[\large \text{When r=10 cm find A'}\]

Answer 2

Understand where the r' came from? :O
That's the RATE OF CHANGE of r, that they gave us at the start.

Answer 3

Are you more comfortable doing these problem written in Leibniz notation? Whatever works for you :D\[\large \frac{dr}{dt}=5 \; cm/s\]

Answer 4

no i understand the other way. so i would just have to calculate to find A. and how would i find the answer to B?

Answer 5

Yah for A we just take the derivative of the area function, take the derivative with respect to TIME, so don't forget that an r' will pop out when you apply the chain rule on the r term. Understand that part or should we go through it? :D

Answer 6

i got it

Answer 7

actually can you go through on how to find the derivative after calculating A= pie(10)^2

Answer 8

We want to make sure we have BOTH of our VARIABLES in the problem when we differentiate. Don't plug anything in before differentiating! :O
\[\large A=\pi r^2\]Taking the derivative WRT time gives us,\[\large A=\pi(2r)r'\]

Answer 9

A' should be at the start of that bottom line :3 my bad.

Answer 10

\[\large A'=\pi (2r)r'\]

Answer 11

ah ok

Answer 12

now i would plug in 10 or 5 for r?

Answer 13

You plug in these values, and solve for A'.
\[\large r=10, \quad r'=5\]

Answer 14

100(pie)

Answer 15

which would be 314 after calculations

Answer 16

Mmmmmmm yah looks good.

Answer 17

im assuming its the same for b?

Answer 18

actually b wants to find out the growing rate when it EQUALS 25cm

Answer 19

Part B is a little tricky, right now, we have our Area function written in terms of r.
r is the INDEPENDENT variable. We can plug in any value for the Radius, and from there we can calculate an Area.

In part B, we are given a specific Area, so we're treating the area as the independent variable. So let's rewrite our function in terms of A.

Answer 20

We want it to look like, r=something.

Answer 21

Err actually... let's just find a radius that corresponds to A=25 cm. Then do it the normal way, that will probably be easier :)

Answer 22

Plug in A=25, then solve for r.

Answer 23

\[\large A=\pi r^2 \quad \rightarrow \quad 25=\pi r^2\]
\[\large \frac{25}{\pi}=r^2 \quad \rightarrow \quad r=\sqrt{\frac{25}{\pi}} \quad \rightarrow \quad r=\frac{5}{\sqrt {\pi}}\]Something like that

Answer 24

So now you have the R that corresponds to A=25.
And now you can proceed the same why you did in part A, just with a new R value this time.

Answer 25

ok how would i calculate 2(5/squareroot(pie)?

Answer 26

What do you mean? Like how on a calculator? :O

Answer 27

Oh, just multiply the 2 times the 5.

Answer 28

\[\large A'=\pi(2r)r', \qquad r=\frac{5}{\sqrt{\pi}},\qquad r'=5\]\[\large A'=\pi\left(2\frac{5}{\sqrt{\pi}}\right)5 \quad \rightarrow \quad A'=\frac{50 \pi}{\sqrt{\pi}}\]
You should get something like this when you plug everything in.. I think :O Where you getting stuck?

Answer 29

thats what i got, but not sure if thats the final answer.