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Chapter 2: Problem 83

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

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

Step by step solution

01

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 \( A = (6t + 5) \sqrt{t} \).
02

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 \( v = \sqrt{t} = t^{1/2} \).
03

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 \( \frac{dA}{dt} = 6 \sqrt{t} + \frac{6t + 5}{2\sqrt{t}} \).
04

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, \( \frac{dA}{dt} = 6 \sqrt{t} + \frac{6t}{2\sqrt{t}} + \frac{5}{2\sqrt{t}} = 6 \sqrt{t} + 3 \sqrt{t} + \frac{5}{2\sqrt{t}} = 9 \sqrt{t} + \frac{5}{2\sqrt{t}} \).

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Most popular questions from this chapter

Sketching a Graph Sketch the graph of a differentiable function \(f\) such that \(f(2)=0, f^{\prime} < 0\) for \(-\infty< x <2,\) and \(f^{\prime}>0\) for \(2< x <\infty\) . Explain how you found your answer.

Linear and Quadratic Approximations The linear and quadratic approximations of a function \(f\) at \(x=a\) are $$\begin{array}{l}{P_{1}(x)=f^{\prime}(a)(x-a)+f(a) \text { and }} \\\ {P_{2}(x)=\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}+f^{\prime}(a)(x-a)+f(a)}\end{array}$$ In Exercises 123 and \(124,\) (a) find the specified linear and quadratic approximations of \(f,(b)\) use a graphing utility to graph \(f\) and the approximations, (c) determine whether \(P_{1}\) or \(P_{2}\) is the better approximation, and (d) state how the accuracy changes as you move farther from \(x=a\) . $$ f(x)=\tan x ; \quad a=\frac{\pi}{4} $$

Moving Point In Exercises \(5-8,\) a point is moving along the graph of the given function at the rate \(d x / d t .\) Find \(d y / d t\) for the given values of \(x .\) $$ \begin{array}{l}{y=\cos x ; \frac{d x}{d t}=4 \text { centimeters per second }} \\ {\begin{array}{llll}{\text { (a) } x=\frac{\pi}{6}} & {\text { (b) } x=\frac{\pi}{4}} & {\text { (c) } x=\frac{\pi}{3}}\end{array}}\end{array} $$

Proof Let \(u\) be a differentiable function of \(x .\) Use the fact that \(|u|=\sqrt{u^{2}}\) to prove that \(\frac{d}{d x}[|u|]=u^{\prime} \frac{u}{|u|}, \quad u \neq 0\)

True or False? In Exercises \(93-96\) , determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If a function has derivatives from both the right and the left at a point, then it is differentiable at that point.
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