$\text { The length of a rectangle is } 5 t + 5 \text { and its height is } t ^ { 4 } \text {, where } t \text { is time in seconds and }$ the dimensions are in inches. Find the rate of change of area, A, with respect to time.
A) $\frac { d A } { d t } = t ^ { 4 } ( 20 + 25 t )$ square inches/second
B) $\frac { d A } { d t } = t ^ { 4 } ( 15 + 25 t )$ square inches/second
C) $\frac { d A } { d t } = t ^ { 3 } ( 20 + 20 t )$ square inches/second
D) $\frac { d A } { d t } = t ^ { 3 } ( 20 + 25 t )$ square inches/second
E) $\frac { d A } { d t } = t ^ { 3 } ( 25 t + 5 )$ square inches/second

Question

$\text { The length of a rectangle is } 5 t + 5 \text { and its height is } t ^ { 4 } \text {, where } t \text { is time in seconds and }$ the dimensions are in inches. Find the rate of change of area, A, with respect to time. 
A) $\frac { d A } { d t } = t ^ { 4 } ( 20 + 25 t )$ square inches/second
B) $\frac { d A } { d t } = t ^ { 4 } ( 15 + 25 t )$ square inches/second
C) $\frac { d A } { d t } = t ^ { 3 } ( 20 + 20 t )$ square inches/second
D) $\frac { d A } { d t } = t ^ { 3 } ( 20 + 25 t )$ square inches/second
E) $\frac { d A } { d t } = t ^ { 3 } ( 25 t + 5 )$ square inches/second

Quizplus · Accepted Answer

The area of the rectangle is given by 
$$A = (	ext{length}) 	imes (	ext{height}) = (5t+5)(t^4) = 5t^5 + 5t^4$$
Taking the derivative with respect to time, we get
$$\frac{dA}{dt} = 25t^4 + 20t^3 = t^3(25t+20)$$
Therefore, the correct answer is $\frac{dA}{dt} = t^3(25t+20)$, which is choice D.