Question

Area The length of a rectangle is given by \(6 t+5\) and its height is \(\sqrt{t},\) where \(t\) is time in seconds and the dimensions are in centimeters. Find the rate of change of the area with respect to time.

Short answer

The rate of change of the area with respect to time is \(9 \sqrt{t} + \frac{5}{2\sqrt{t}}\) cm²/s.

Step-by-step solution

Write down the formula for the area of a rectangle

The area \( A \) of a rectangle is given by the product of its length and width. Here, the length \( l \) is represented as \(6t + 5\) and the width \( w \) is represented as \( \sqrt{t} \). Therefore, the area can be represented as \( A = (6t + 5) \sqrt{t} \).

Differentiate the area with respect to time

Differentiating both sides of \( A = (6t + 5) \sqrt{t} \) with respect to time \( t \), we get \( \frac{dA}{dt} = \frac{d}{dt} [ (6t + 5) \sqrt{t} ] \). Now we must apply the product rule of differentiation (if \( z = uv \), then \( dz/dt = u'(t)v + u(t)v'(t) \)), with \( u = 6t + 5 \) and \( v = \sqrt{t} = t^{1/2} \).

Compute the derivatives

The derivative \( u' = 6 \) and \( v' = \frac{1}{2} t^{-1/2} = \frac{1}{2\sqrt{t}} \). Now substitute back into the product rule to obtain \( \frac{dA}{dt} = 6 \sqrt{t} + \frac{6t + 5}{2\sqrt{t}} \).

Simplify the derivative

It's always best to simplify the derivative if it's possible. So, do that by clubbing the similar terms together. Hence, \( \frac{dA}{dt} = 6 \sqrt{t} + \frac{6t}{2\sqrt{t}} + \frac{5}{2\sqrt{t}} = 6 \sqrt{t} + 3 \sqrt{t} + \frac{5}{2\sqrt{t}} = 9 \sqrt{t} + \frac{5}{2\sqrt{t}} \).