Mathematics: 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)=

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The length pf a rectangle is given by 2t +3 and

MIAkwicc39

09.03.2022

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

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 = $Length \times Width$

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

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

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

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

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

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

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

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

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