Question

A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 2.8 ft/s.
(a) How rapidly is the area enclosed by the ripple increasing when the radius is 4 feet?

Answer 1

It says the radius is 4 ft. the rate it is increasing is 2.8 ft/s

Answer 2

so it's asking us how fast is the area being enclosed. so we will need to know the equation for area of a circle

Answer 3

oh. A=pir^2. Am I suppose to take the derivative of the equation so 2pir(dr/dt)?

Answer 4

Yeeeeah lol

Answer 5

Related Rates

Answer 6

I'm bad at this stuff.
repost this and get some help from another person.

Answer 7

it's okay. lol Neither am I. :)

Answer 8

Thanks, though!

Answer 9

A = (pi)(r^2)
dA/dt = 2(pi)rdr/dt

Put r = 4 and dr/dt = 2.8 and find dA/dt.

Answer 10

Thanks! I got that!!

Answer 11

cool. yw.

Answer 12

uh if it asks for the area after 7.2 seconds, where would I put it?

Answer 13

@ranga

Answer 14

*put the time?

Answer 15

dr/dt = 2.8
Integrate both sides and express r as a function of time t first.

Answer 16

dr/dt = 2.8
Integrate both sides:
r = 2.8t + C
when t = 0, r = 0
0 = 0 + C. So C = 0
r = 2.8t
At time t = 7.2 find r. Then Area = pi(r^2)

Answer 17

Ohh. Okay. Yes, I get it! Thanks!!

Answer 18

you are welcome.