Question

The length pf a rectangle is given by 2t +3 and its height is square root t, where t is time in seconds and the dimensions are in centimeters. Find the rate of the change of the area with respect to time.

A'(t)=

Answer 1

Answer:

$A'(t)=3\sqrt{t}+\frac{3}{2\sqrt{t}}$          

Step-by-step explanation:

Given : Length of rectangle = 2t+3

           Height of rectangle = $\sqrt{t}$

To Find: Find the rate of the change of the area with respect to time.

Solution:

Area of rectangle = $Length \times Width$

                              = $(2t+3) \times \sqrt{t}$

                              = $2t^{\frac{3}{2}}+3t^{\frac{1}{2}}$

                              = $2t^{\frac{3}{2}}+3t^{\frac{1}{2}}$

So, $A(t)=2t^{\frac{3}{2}}+3t^{\frac{1}{2}}$

$\frac{d}{dx} (x^n)=nx^{n-1}$

$A'(t)=\frac{3}{2} \times 2t^{\frac{3}{2}-1}+\frac{1}{2} \times 3t^{\frac{1}{2}-1}$

$A'(t)=3t^{\frac{1}{2}}+\frac{3}{2}t^{\frac{-1}{2}}$

$A'(t)=3t^{\frac{1}{2}}+\frac{3}{2}t^{\frac{-1}{2}}$          

$A'(t)=3\sqrt{t}+\frac{3}{2\sqrt{t}}$              

Hence the rate of the change of the area with respect to time is    $A'(t)=3\sqrt{t}+\frac{3}{2\sqrt{t}}$