Water leaking onto a floor forms a circular pool. The radius of the pool increases at a rate of 4 cm/min. How fast is the area of the pool increasing when the radius is 5 cm?

1 Answer
mason m
Jul 26, 2016

#40pi# #"cm"^2"/ min"#

Explanation:

First, we should begin with an equation we know relating the area of a circle, the pool, and its radius:

#A=pir^2#

However, we want to see how fast the area of the pool is increasing, which sounds a lot like rate... which sounds a lot like a derivative.

If we take the derivative of #A=pir^2# with respect to time, #t#, we see that:

#(dA)/dt=pi*2r*(dr)/dt#

(Don't forget that the chain rule applies on the right hand side, with #r^2#--this is similar to implicit differentiation.)

So, we want to determine #(dA)/dt#. The question told us that #(dr)/dt=4# when it said "the radius of the pool increases at a rate of #4# cm/min," and we also know that we want to find #(dA)/dt# when #r=5#. Plugging these values in, we see that:

#(dA)/dt=pi*2(5)*4=40pi#

To put this into words, we say that:

The area of the pool is increasing at a rate of #bb40pi# cm#""^bb2#/min when the circle's radius is #bb5# cm.

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