Question

The length of a rectangle is given by 6t + 5 and its height is √t, where t is time in seconds and the dimensions are in centimeters. Find the rate of change of the area with respect to time.
A'(t) =???​

Answer 1

MARK AS BRAINLIST

Step-by-step explanation:

L = 6t+5

H = √t

A = L*H

= (6t+5) *√t = 6t√t + 5√t

A'(t) = 6t * 1/2*√t + √t * 6 + 5/2*√t

= 6t/2√t + 6√t + 5/2√t

Answer 2

Expression for the rate of change of the area with respect to time will be,

A' = $9t^{\frac{1}{2}}+\frac{5}{2}t^{-\frac{1}{2}}$ cm per second

It's given in the question,

• Length of a rectangle is given by the expression

      (6t + 5) cm

• Width or height of the rectangle is given by the expression,

       $\sqrt{t}$ cm

Area of the rectangle is given by,

Area = Length × Height

A = (6t + 5)(√t)

A = $6t^{\frac{3}{2}}+5t^\frac{1}{2}$

Now find the derivative of area (A) with respect to time (t).

$\frac{dA}{dt}=\frac{d}{dt}(6t^{\frac{3}{2}}+5t^\frac{1}{2})$

A' = $(6\times \frac{3}{2})t^\frac{1}{2}+\frac{5}{2} t^{-\frac{1}{2}}$

A' = $9t^{\frac{1}{2}}+\frac{5}{2}t^{-\frac{1}{2}}$

     Therefore, expression for the rate of change of the area will be,

A' = $9t^{\frac{1}{2}}+\frac{5}{2}t^{-\frac{1}{2}}$

Learn more,

https://brainly.com/question/12596049?