$\text { The radius, } r \text {, of a circle is decreasing at a rate of } 5 \text { centimeters per minute. }$ $\text { Find the rate of change of area, } A \text {, when the radius is } 6 .$
A) $\frac { d A } { d t } = - 360 \pi \mathrm { sq } \mathrm { cm } / \mathrm { min }$
B) $\frac { d A } { d t } = 360 \pi \mathrm { sq } \mathrm { cm } / \mathrm { min }$
C) $\frac { d A } { d t } = - 60 \pi \mathrm { sq } \mathrm { cm } / \mathrm { min }$
D) $\frac { d A } { d t } = 60 \pi \mathrm { sq } \mathrm { cm } / \mathrm { min }$
E) $\frac { d A } { d t } = - 30 \pi \mathrm { sq } \mathrm { cm } / \mathrm { min }$

Question

$\text { The radius, } r \text {, of a circle is decreasing at a rate of } 5 \text { centimeters per minute. }$ $\text { Find the rate of change of area, } A \text {, when the radius is } 6 .$ 
A) $\frac { d A } { d t } = - 360 \pi \mathrm { sq } \mathrm { cm } / \mathrm { min }$
B) $\frac { d A } { d t } = 360 \pi \mathrm { sq } \mathrm { cm } / \mathrm { min }$
C) $\frac { d A } { d t } = - 60 \pi \mathrm { sq } \mathrm { cm } / \mathrm { min }$
D) $\frac { d A } { d t } = 60 \pi \mathrm { sq } \mathrm { cm } / \mathrm { min }$
E) $\frac { d A } { d t } = - 30 \pi \mathrm { sq } \mathrm { cm } / \mathrm { min }$

Quizplus · Accepted Answer

The area $ A $ of a circle is given by $ A = \pi r^2 $. Differentiating with respect to time $ t $, we get $ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} $. Given $ \frac{dr}{dt} = -5 $ cm/min and $ r = 6 $ cm, substituting these values gives $ \frac{dA}{dt} = 2\pi(6)(-5) = -60\pi $ sq cm/min.

The radius, r, of a circle is decreasing at a rate of 5 centimeters per minute. \text { The radius, } r \text {, of a circle is decreasing at a rate of } 5 \text { centimeters per minute. } The radius, r, of a circle is decreasing at a rate of 5 centimeters per minute.

$\text { The radius, } r \text {, of a circle is decreasing at a rate of } 5 \text { centimeters per minute. }$