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?

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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?

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Answer 1 · 2016-07-26T22:33:05

$40 π$ $40pi$ ${cm}^{2} / min$ $"cm"^2"/ min"$

Explanation:

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

$A = π r^{2}$ $A=pir^2$

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 $A = π r^{2}$ $A=pir^2$ with respect to time, $t$ $t$ , we see that:

$\frac{d A}{d t} = π \cdot 2 r \cdot \frac{d r}{d t}$ $(dA)/dt=pi*2r*(dr)/dt$

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

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

$\frac{d A}{d t} = π \cdot 2 (5) \cdot 4 = 40 π$ $(dA)/dt=pi*2(5)*4=40pi$

To put this into words, we say that:

The area of the pool is increasing at a rate of $40 π$ $bb40pi$ cm $^{2}$ $""^bb2$ /min when the circle's radius is $5$ $bb5$ cm.

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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?

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