# 4.10 Antiderivatives  (Page 5/10)

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Solve the initial value problem $\frac{dy}{dx}=3{x}^{-2},y\left(1\right)=2.$

$y=-\frac{3}{x}+5$

Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car. We are interested in how long it takes for the car to stop. Recall that the velocity function $v\left(t\right)$ is the derivative of a position function $s\left(t\right),$ and the acceleration $a\left(t\right)$ is the derivative of the velocity function. In earlier examples in the text, we could calculate the velocity from the position and then compute the acceleration from the velocity. In the next example we work the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function to determine the position function.

## Decelerating car

A car is traveling at the rate of $88$ ft/sec $\left(60$ mph) when the brakes are applied. The car begins decelerating at a constant rate of $15$ ft/sec 2 .

1. How many seconds elapse before the car stops?
2. How far does the car travel during that time?
1. First we introduce variables for this problem. Let $t$ be the time (in seconds) after the brakes are first applied. Let $a\left(t\right)$ be the acceleration of the car (in feet per seconds squared) at time $t.$ Let $v\left(t\right)$ be the velocity of the car (in feet per second) at time $t.$ Let $s\left(t\right)$ be the car’s position (in feet) beyond the point where the brakes are applied at time $t.$
The car is traveling at a rate of $88\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$ Therefore, the initial velocity is $v\left(0\right)=88$ ft/sec. Since the car is decelerating, the acceleration is
$a\left(t\right)=-15{\phantom{\rule{0.2em}{0ex}}\text{ft/s}}^{2}.$

The acceleration is the derivative of the velocity,
${v}^{\prime }\left(t\right)=15.$

Therefore, we have an initial-value problem to solve:
${v}^{\prime }\left(t\right)=-15,v\left(0\right)=88.$

Integrating, we find that
$v\left(t\right)=-15t+C.$

Since $v\left(0\right)=88,C=88.$ Thus, the velocity function is
$v\left(t\right)=-15t+88.$

To find how long it takes for the car to stop, we need to find the time $t$ such that the velocity is zero. Solving $-15t+88=0,$ we obtain $t=\frac{88}{15}$ sec.
2. To find how far the car travels during this time, we need to find the position of the car after $\frac{88}{15}$ sec. We know the velocity $v\left(t\right)$ is the derivative of the position $s\left(t\right).$ Consider the initial position to be $s\left(0\right)=0.$ Therefore, we need to solve the initial-value problem
${s}^{\prime }\left(t\right)=-15t+88,s\left(0\right)=0.$

Integrating, we have
$s\left(t\right)=-\frac{15}{2}{t}^{2}+88t+C.$

Since $s\left(0\right)=0,$ the constant is $C=0.$ Therefore, the position function is
$s\left(t\right)=-\frac{15}{2}{t}^{2}+88t.$

After $t=\frac{88}{15}$ sec, the position is $s\left(\frac{88}{15}\right)\approx 258.133$ ft.

Suppose the car is traveling at the rate of $44$ ft/sec. How long does it take for the car to stop? How far will the car travel?

$2.93\phantom{\rule{0.2em}{0ex}}\text{sec},\phantom{\rule{0.2em}{0ex}}64.5\phantom{\rule{0.2em}{0ex}}\text{ft}$

## Key concepts

• If $F$ is an antiderivative of $f,$ then every antiderivative of $f$ is of the form $F\left(x\right)+C$ for some constant $C.$
• Solving the initial-value problem
$\frac{dy}{dx}=f\left(x\right),y\left({x}_{0}\right)={y}_{0}$

requires us first to find the set of antiderivatives of $f$ and then to look for the particular antiderivative that also satisfies the initial condition.

For the following exercises, show that $F\left(x\right)$ are antiderivatives of $f\left(x\right).$

$F\left(x\right)=5{x}^{3}+2{x}^{2}+3x+1,f\left(x\right)=15{x}^{2}+4x+3$

${F}^{\prime }\left(x\right)=15{x}^{2}+4x+3$

$F\left(x\right)={x}^{2}+4x+1,f\left(x\right)=2x+4$

$F\left(x\right)={x}^{2}{e}^{x},f\left(x\right)={e}^{x}\left({x}^{2}+2x\right)$

${F}^{\prime }\left(x\right)=2x{e}^{x}+{x}^{2}{e}^{x}$

$F\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x,f\left(x\right)=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x$

$F\left(x\right)={e}^{x},f\left(x\right)={e}^{x}$

${F}^{\prime }\left(x\right)={e}^{x}$

For the following exercises, find the antiderivative of the function.

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