MATH SOLVE

3 months ago

Q:
# 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)=

Accepted Solution

A:

Answer: [tex]A'(t)=3\sqrt{t}+\frac{3}{2\sqrt{t}}[/tex] Step-by-step explanation:Given : Length of rectangle = 2t+3 Height of rectangle = [tex]\sqrt{t}[/tex]To Find: Find the rate of the change of the area with respect to time.
Solution:Area of rectangle = [tex]Length \times Width[/tex] = [tex](2t+3) \times \sqrt{t}[/tex] = [tex]2t^{\frac{3}{2}}+3t^{\frac{1}{2}}[/tex] = [tex]2t^{\frac{3}{2}}+3t^{\frac{1}{2}}[/tex]So, [tex]A(t)=2t^{\frac{3}{2}}+3t^{\frac{1}{2}}[/tex][tex]\frac{d}{dx} (x^n)=nx^{n-1}[/tex] [tex]A'(t)=\frac{3}{2} \times 2t^{\frac{3}{2}-1}+\frac{1}{2} \times 3t^{\frac{1}{2}-1}[/tex] [tex]A'(t)=3t^{\frac{1}{2}}+\frac{3}{2}t^{\frac{-1}{2}}[/tex] [tex]A'(t)=3t^{\frac{1}{2}}+\frac{3}{2}t^{\frac{-1}{2}}[/tex] [tex]A'(t)=3\sqrt{t}+\frac{3}{2\sqrt{t}}[/tex] Hence the rate of the change of the area with respect to time is [tex]A'(t)=3\sqrt{t}+\frac{3}{2\sqrt{t}}[/tex]