well, a = πr^2
da/dt = 2πr dr/dt
...
Water leaking onto a floor creates a circular pool with an area that increases at the rate of 3 square inches per minute. How fast is the radius of the pool increasing when the radius is 10 inches?
3 answers
is 188.5 inch/sec. correct?
You have to be kidding!! The entire area is only growing at 3 in^2/s.
You need to learn to do a sanity check on any answer you get, especially in a new situation.
Using the given equation and numbers, we have
3 = 2π*10 dr/dt
dr/dt = 3/(20π) = 0.0477 in/s
Your answer is 60π. You plugged in 3 for dr/dt, but that was incorrect.
You need to learn to do a sanity check on any answer you get, especially in a new situation.
Using the given equation and numbers, we have
3 = 2π*10 dr/dt
dr/dt = 3/(20π) = 0.0477 in/s
Your answer is 60π. You plugged in 3 for dr/dt, but that was incorrect.