OBJECTIVES

When you finish this section, you should be able to:

1. Find antiderivatives (p. 329)
2. Solve a differential equation (p. 331)
3. Solve applied problems modeled by differential equations (p. 333)

DEFINITION Antiderivative

A function $F$ is called an antiderivative of the function $f$ if $F^\prime (x) = f(x)$ for all $x$ in the domain of $f$.

COROLLARY

If $f$ and $g$ are differentiable functions and if $f^\prime (x) =g^\prime (x)$ for all numbers $x$ in an interval $(a,b),$ then there exists a number $C$ for which $f(x) =g(x) +C$ on $(a,b)$.

THEOREM

If a function $F$ is an antiderivative of a function $f$ defined on an interval $I$, then any other antiderivative of $f$ has the form $F(x) +C$, where $C$ is an (arbitrary) constant.

Find all the antiderivatives of:

Solution (a) Since the derivative of a constant function is $0,$ all the antiderivatives of $f$ are of the form $F(x) =C,$ where $C$ is a constant.

(b) Since $\dfrac{d}{d\theta}\cos \theta = -\sin \theta$, all the antiderivatives of $g(\theta) = -\sin \theta$ are of the form $G(\theta ) =\cos \theta +C$.

(c) The derivative of $\dfrac{2}{3}x^{3/2}$ is $\left( \dfrac{2}{3}\right) \ \left( \dfrac{3}{2}x^{\hbox{$\frac{3}{2}-1$}}\right) =x^{1/2}$.

So, all the antiderivatives of $h(x)=x^{1/2}$ are of the form $H(x) ={{\dfrac{2}{3}}x^{3/2}+C}$, where ${C}$ is a constant.

NOW WORK

Problem 17.

IN WORDS

An antiderivative of the sum of two functions equals the sum of the antiderivatives of the functions.

THEOREM Sum Rule

If the functions $F_{1}$ and $F_{2}$ are antiderivatives of the functions $f_{1}$ and $f_{2}$, respectively, then $F_{1}+F_{2}$ is an antiderivative of $f_{1}+f_{2}$.

IN WORDS

An antiderivative of a number times a function equals the number times an antiderivative of the function.

THEOREM Constant Multiple Rule

If $k$ is a real number and if $F$ is an antiderivative of $f$, then $kF$ is an antiderivative of $kf$.

Find all the antiderivatives of $f(x)=e^{x}+\dfrac{6}{x^{2}}-\sin x$.

Solution Since $f$ is the sum of three functions, we use the Sum Rule. That is, we find the antiderivatives of each function individually and then add. $$f(x) =e^{x}+6x^{-2}-\sin x$$

An antiderivative of $e^{x}$ is $e^{x}.$ An antiderivative of $6x^{-2}$ is $$6\cdot \frac{x^{-2+1}}{-2+1}=6\cdot \dfrac{x^{-1}}{-1}=-\dfrac{6}{x}$$

Finally, an antiderivative of $\sin x$ is $-\;\cos x$. Then all the antiderivatives of the function $f$ are given by $$F(x) =e^{x}-\frac{6}{x}+\cos x+C$$

where $C$ is a constant.

NOW WORK

Problem 27.

Solve the differential equation $\dfrac{dy}{dx}=x^{2}+2x+1$ with the boundary condition when $x=3,$ then $y={-}1.$

Solution We begin by finding the general solution of the differential equation, namely $$y=\frac{x^{3}}{3}+x^{2}+x+C$$

To determine the number $C$, we use the boundary condition when $x=3,$ then $y={-}1$. $$\begin{array}{rl@{\qquad}l} -1 &=\dfrac{3^{3}}{3}+3^{2}+3+C & \color{#0066A7}{{x=3,\quad y=-1}} \\[8pt] C&=-22 \end{array}$$

The particular solution of the differential equation with the boundary condition when $x=3,$ then $y=-1$, is $$y=\frac{x^{3}}{3}+x^{2}+x-22$$

NOW WORK

Problem 35.

Solve the differential equation $\dfrac{d^{2}y}{dx^{2}}=12x^{2}$ with the boundary conditions when $x=0,$ then $y=1$ and when $x=3,$ then $y=8$.

Solution All the antiderivatives of $\dfrac{d^{2}y}{dx^{2}}=12x^{2}$ are $$\hspace{12pt}\dfrac{dy}{dx}=4x^{3}+C_{1}$$

All the antiderivatives of $\dfrac{dy}{dx}=4x^{3}+C_{1}$ are $$y=x^{4}+C_{1}x+C_{2}$$

This is the general solution of the differential equation. To find $C_{1}$ and $C_{2}$ and the particular solution to the differential equation, we use the boundary conditions.

• When $x=0,\qquad 1=0^{4}+C_{1}( 0) +C_{2}\qquad \hbox{so}\quad C_{2}=1$
• When $x=3,\qquad 8=3^{4}+3C_{1}+1\qquad \hbox{so}\quad C_{1}=-\dfrac{74}{3}$

The particular solution with the given boundary conditions is $$y=x^{4}-\dfrac{74}{3}x+1$$

NOW WORK

Problem 39.

Find the distance $s$ of an object from the origin at time $t$ if its acceleration $a$ is $$a(t)=8t-3$$

and the initial conditions are $v_{0}=v(0)=4$ and $s_{0}=s(0)=1.$

Solution First we solve the differential equation $\dfrac{{d\kern-1ptv}}{dt} =a(t)=8t-3$ and use the initial condition $v_{0}=v( 0) =4$. $$\begin{eqnarray*} v(t) &=& 4t^{2}-3t+C_{1} \\[3pt] {v_{0}} &=& v( 0) =4( 0) ^{2}-3( 0) + C_{1}&=&4 \\[0pt] &&\qquad\quad\hspace{3.9pc}C_{1}&=&4 \end{eqnarray*}$$

The velocity of the object at time $t$ is $v(t)=4t^{2}-3t+4$.

The distance $s$ of the object at time $t$ satisfies the differential equation $$\frac{ds}{dt}=v(t)=4t^{2}-3t+4$$

Then $${s(t)={\frac{4}{3}}t^{3}-{\frac{3}{2}}t^{2}+4t+C_{2}}$$

Using the initial condition, $s_{0}=s(0)=1$, we have $$\begin{eqnarray*} {s_{0}=s(0)=0-0+0+C_{2}} &=&1 \\ {C_{2}} &=&1 \end{eqnarray*}$$

The distance $s$ of the object from the origin at any time $t$ is $$s=s(t) ={\frac{4}{3}}t^{3}-{\frac{3}{2}}t^{2}+4t+1$$

NOW WORK

Problem 41.

When the brakes of a car are applied, the car decelerates at a constant rate of $10\;{\rm{m/s}}^2$. If the car is to stop within $20 {\rm m}$ after the brakes are applied, what is the maximum velocity the car could have been traveling? Express the answer in miles per hour.

Solution Let $s(t)$ represent the distance $s$ in meters the car has traveled $t$ seconds after the brakes are applied. Let $v_{0}$ be the velocity of the car at the time the brakes are applied ($t=0$). Since the car decelerates at the rate of $10\;{\rm{m/s}}^2$, its acceleration $a,$ in meters per second squared, is $${a}(t) =\dfrac{{d\kern-1ptv}}{dt}={-10}$$

We solve the differential equation for $v$. $$v(t)=-10t+C_{1}$$

When $t=0$, $v(0)=v_{0}$, the velocity of the car when the brakes are applied, so $C_{1}=v_{0}.$ Then $${v(t)={\frac{ds}{dt}}}={-10t+v_{0}}$$

Now we solve the differential equation ${v(t)={\dfrac{ds}{dt}}}$ for $s$. $$s(t)= -5t^{2}+v _{0}t+C_{2}$$

Since the distance $s$ is measured from the point at which the brakes are applied, the second initial condition is $s( 0) =0$. Then $s( 0) =-5\cdot 0+v_{0}\cdot 0+C_{2}=0$, so $C_{2}=0$. The distance $s$, in meters, the car travels $t$ seconds after applying the brakes is $$s(t)=-5t^{2}+v_{0}t$$

The car stops completely when its velocity equals $0$. That is, when $$\begin{eqnarray*} v(t) ={-10t+v_{0}} &=&0 \\[3pt] t &=&{\frac{v_{0}}{10}} \end{eqnarray*}$$

This is the time it takes the car to come to rest. Substituting $\dfrac{v_{0}}{10}$ for $t$ in $s(t)$, the distance the car has traveled is $$s\left( {\frac{v_{0}}{10}}\right) =-5\left( {\frac{v_{0}}{10}}\right) ^{2}+v_{0}\left( {\frac{v_{0}}{10}}\right) =\frac{v_{0}^{2}}{20}\hbox{ }$$

If the car is to stop within 20 m, then $s\leq 20$; that is, $\dfrac{v_{0}^{2}}{20}\leq 20$ or equivalently $v_{0}^{2}\le 400$. The maximum possible velocity $v_{0}$ for the car is $v_{0}=20 \ {\rm m}/{\rm s}$.

To express this in miles per hour, we proceed as follows: $$\begin{eqnarray*} v_{0}=20\,{\rm{m}/{s}}&=&\left( \frac{20{{\rm m}}}{{{\rm s}}}\right)\, \left( \frac{1{\rm km}}{1000\,{{\rm m}}}\right)\, \left( \frac{3600 {{\rm s}} }{1\,{\rm h}}\right) \\[3pt] &=& 72{\rm km}/{\rm h}\approx \left( \frac{72\,{\rm km} }{{\rm h}}\right)\, \left( \frac{1\,{\rm mi}}{1.6\,{\rm km}}\right) =45 {\rm mi}/{\rm h} \end{eqnarray*}$$

The maximum possible velocity to stop within $20$ m is $45$ mi/h.

NOW WORK

Problem 53.

A rock is thrown straight up with an initial velocity of $19.6\;{\rm m}/{\rm s}$ from the roof of a building $24.5\;{\rm m}$ above ground level, as shown in Figure 65.

Figure 65

Solution To answer the questions, we need to find the velocity $v=v(t)$ and the distance $s=s(t)$ of the rock as functions of time. We begin measuring time when the rock is released. If $s$ is the distance, in meters, of the rock from the ground, then since the rock is released at a height of $24.5\;{\rm m}$, $s_{0}=s(0)=24.5\;{\rm m}$.

The initial velocity of the rock is given as $v_{0}=v(0)=19.6\;{\rm m}/{\rm s}$. If air resistance is ignored, the only force acting on the rock is gravity. Since the acceleration due to gravity is $-9.8\;{\rm m}/{\rm s}^{2}$, the acceleration $a$ of the rock is $$a=\dfrac{{d\kern-1ptv}}{dt}=-9.8$$

Solving the differential equation, we get $$v(t)=-9.8t+v_{0}$$

Using the initial condition, $v_{0}=v( 0) =19.6\;{\rm m}/{\rm s}$, the velocity of the rock at any time $t$ is $$v(t)=-9.8t+19.6$$

Now we solve the differential equation $\dfrac{ds}{dt}=v(t) =-9.8t+19.6$. Then $${s}(t)={-4.9t^{2}+19.6t+s_{0}}$$

Using the initial condition, $s( 0) =24.5\;{\rm m}$, the distance $s$ of the rock from the ground at any time $t$ is $$s(t)=-4.9t^{2}+19.6t+24.5$$

Now we can answer the questions.

The only meaningful solution is $t=5$. The rock is in the air for $5$ seconds.

NOW WORK

Problem 55.

THEOREM Newton’s First Law of Motion

If no force acts on a body, then a body at rest remains at rest and a body moving with constant velocity will continue to do so.

Proof

 The force $F$ acting on a body of mass $m$ is given by Newton’s Second Law of Motion $F=ma$, where $a$ is the acceleration of the body. If there is no force acting on the body, then $F=0$. In this case, the acceleration $a$ must be $0$. But $a=\dfrac{{d\kern-1ptv}}{dt}$, where $v$ is the velocity of the body. So, $${\dfrac{{d\kern-1ptv}}{dt}=0}\qquad \hbox{and}\qquad {v=}\hbox{ Constant}$$