Water leaking onto a floor forms a circular pool. The area of the pool increases at a rate of 25π cm²/min. How fast is the radius of the pool increasing when the radius is 6 cm?

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Water leaking onto a floor forms a circular pool. The area of the pool increases at a rate of 25π cm²/min. How fast is the radius of the pool increasing when the radius is 6 cm?

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Answer 1 · 2016-10-23T08:09:16

$d \frac{r}{d t} = \frac{25}{12}$ $dr/dt = (25)/(12)$ cm/min

Explanation:

so from the question, we know that $d \frac{A}{d t} = 25 π$ $dA/dt=25pi$ , this means that the area of the circular puddle is increasing constantly at this rate.

so in order to find how fast the radius is increasing, we must first determine a relationship between the two values. So, for a circle that is the Area formula $A r e a = π \cdot r^{2}$ $Area = pi*r^2$ .

The next part involves related rates and the chain rule.
We know that $d \frac{r}{d t} = d \frac{A}{d t} \cdot d \frac{r}{d} A$ $dr/dt =dA/dt * dr/dA$
(the rate of radius change expressed as two other rates).
So,
$A = π \cdot r^{2}$ $A=pi*r^2$

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

$d \frac{r}{d} A = \frac{1}{2 π \cdot r}$ $dr/dA = 1/(2pi*r)$

using chain rule now.
$d \frac{r}{d t} = d \frac{A}{d t} \cdot d \frac{r}{d} A$ $dr/dt = dA/dt *dr/dA$

$d \frac{r}{d t} = 25 π \cdot \frac{1}{2 π \cdot r}$ $dr/dt =25pi * 1/(2pi*r)$

Sub in 6cm for radius
$d \frac{r}{d t} = \frac{25 π}{2 \cdot 6 π}$ $dr/dt = (25pi)/(2*6pi)$
$d \frac{r}{d t} = \frac{25}{12}$ $dr/dt = (25)/(12)$ cm/min

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Water leaking onto a floor forms a circular pool. The area of the pool increases at a rate of 25π cm²/min. How fast is the radius of the pool increasing when the radius is 6 cm?

1 Answer

Explanation:

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