Question

A rectangle is growing such that the length of a rectangle is 5t+4 and its height is √t, where t is time in seconds and the dimensions are in inches. Find the rate of change of area, A, with respect to time.

Answer 1

Answer:

$3\sqrt{t} +\frac{2}{\sqrt{t}}$ square inches per seconds.

Step-by-step explanation:

given that a rectangle is growing such that the length of a rectangle is 5t+4 and its height is √t, where t is time in seconds and the dimensions are in inches

Area of rectangle = length * width

A = $\sqrt{t} (5t+4)\\= 5t\sqrt{t} +4\sqrt{t}$

To find rate of change of A with respect to t, we can find derivative of A with respect to t.

WE get

$\frac{dA}{dt} =5(\frac{3}{2} )\sqrt{t} +\frac{4}{2\sqrt{t} } \\= 3\sqrt{t} +\frac{2}{\sqrt{t}}$

Hence rate of change of area with respect to time is

$3\sqrt{t} +\frac{2}{\sqrt{t}}$ square inches per seconds.