Exercises3.6Antiderivatives and Differential Equations

Solve the following problems.

1.

Suppose that $h$ is a function such that $h^\prime(x)=x^2+2e^x+3$ and $h(3)=0\text{.}$ What is $h(1)\text{?}$

$\ds -\frac{44}{3}+2e(1-6e^2)\text{.}$
2.

Give an expression of the most general function $g$ for which $\ds g^\prime(x)=\frac{5}{x}+\frac{1}{1+x^2}\text{.}$

$\ds g(x)=\ln|x|+\arctan x+c\text{.}$
3.

Find $f$ if $f^{\prime\prime}(t)=2e^t+3\sin t$ and $f(0)=0\text{,}$ $f^\prime(0)=0\text{.}$

$\ds f(t)=2e^t-3\sin t+t+2\text{.}$
4.

Find $f$ if $f^{\prime}(x)=2\cos(8x) +e^{3x}-1$ and $f(0)=1\text{.}$

$\ds f(x)=\frac{\sin 8x}{4}+\frac{1}{3}e^{3x}-x+\frac{2}{3}\text{.}$
5.

If $f^\prime(x)=kf(x)$ and $f(0)=A\text{,}$ where $k$ and $A$ are constants, what is $f(x)\text{?}$

$f(x)=Ae^{kx}\text{.}$
6.

Find $f$ if $f^\prime(x)=2\cos x+8x^3-e^x$ and $f(0)=7\text{.}$

$\ds f(x)=2\sin x+2x^4-e^x+8\text{.}$
7.

Find $g$ if $g^\prime(x)=\sin x+x^{-2}-e^x$ and $g(\pi )=1\text{.}$

$\ds g(x)=-\cos x -x^{-1}-e^x +\pi ^{-1}+e^{\pi }\text{.}$
8.

Suppose $f$ is a function such that $f'(x)=x^2+2x+3$ and $f(5)=1\text{.}$ What is $f(1)\text{?}$

$\ds f(x)=\frac{1}{3}x^3+x^2+3x-\frac{242}{3}$ and $\ds f(1)=-\frac{229}{3}\text{.}$
9.

Suppose $h$ is a function such that $h'(x)=x^2+2e^x+3$ and $h(3)=0\text{.}$ What is $h(1)\text{?}$

$\ds h(1)=2e(1-e^2)-\frac{44}{3}\text{.}$
10.

Find the general form for the following antiderivative: $\ds \int \frac{z}{z^2+9}dz$

$\ds F(z)=\frac{1}{2}\ln (z^2+9)\text{.}$
11.

Find a curve $y=f(x)$ with the following properties:

• $\displaystyle \ds \frac{d^2y}{dx^2}=6x$

• Its graph passes through the point $(0,1)$ and has a horizontal tangent there.

It is given that $f(0)=1$ and $f'(0)=0\text{.}$ Thus $f(x)=x^3+1\text{.}$

12.

Find $\ds \int \frac{dx}{x+x\ln x}\text{.}$

$\ds \int \frac{dx}{x(1+\ln x)}=\ln(1+\ln x)+C\text{.}$

For each case compute the indefinite integral.

13.

$\ds \int (1-x)^8d$

$\ds F(x)=-\frac{1}{9}(1-x)^9$
14.

$\ds\int \tan ^2xdx$

$\ds \int \tan ^2xdx=\int (\sec ^2x-1)dx=\tan x-x+C$
15.

$\ds \int \frac{dx}{2+x^2}$

$\ds F(x)=\frac{1}{\sqrt{2}}\arctan \frac{x}{\sqrt{2}}+C$
16.

$\ds \int e^{2x}\cosh xdx$

$\ds F(x)=\frac{1}{6}e^{3x}+\frac{1}{2}e^x+C$
17.

Find $f$ if $f^{\prime \prime}(t)=2e^t+3\sin t$ and $f(0)=0\text{,}$ $f^\prime (0)=0\text{.}$

$f(t)=2e^t-3\sin t+t-2\text{.}$
18.

Given the acceleration of a particle is $a(t)=2e^t+3\sin t\text{,}$ and $s(0)=0\text{,}$ $s(\pi)=0\text{,}$ find the position function $s(t)$ of the particle.

$s(t)=2e^t-3\sin t-(2e^{\pi}+3)t-2\text{.}$
19.

A particle moves in a straight line and has an acceleration given by $a(t)=\sin t+3\cos t\text{.}$ Its initial velocity $v(0)=2$ cm/s and its initial displacement is $s(0)=0$ cm. Find its position function $s(t)\text{.}$

$s(t)=-\sin t-3\cos t+3t+3\text{.}$
20.

A stone is thrown downward off a building 30 metres high. Suppose that the stone has an initial velocity of $5$ m/s, and that it falls with a constant acceleration due to gravity of $10$ m/s$^2\text{.}$

1. Find the velocity function $v(t)$ for the falling stone.

2. Find the displacement function $s(t)$ for the falling stone.

3. How many seconds does it take for the stone to hit the ground?

4. How fast is the stone travelling when it hits the ground?

1. $v(t)=-10t+5\text{.}$

2. $s(t)=-5t^2+5t+30\text{.}$

3. $t=6$ s.

4. $|v(6)|=5$ m/s.

21.

A particle starts from rest (that is with initial velocity zero) at the point $x=10$ and moves along the $x$-axis with acceleration function $a(t)=12t\text{.}$ Find the resulting position function $x(t)\text{.}$

$x(t)=2t^3+10\text{.}$
Solution

It is given that $x(0)=10\text{,}$ $x'(0)=v(0)=0$ and $x''(t)=12t\text{.}$ Hence $x(t)=2t^3+10\text{.}$

22.

At time $t=0$ a car is moving at 6 m/s. and driver smoothly accelerates so that the acceleration after $t$ seconds is $a(t)=3t$ m/s$^2\text{.}$

1. Write a formula for the speed $v(t)$ of the car after $t$ seconds.

2. How far did the car travel between during the time it took to accelerate from 6 m/s to 30 m/s?

1. $\ds v(t)=\frac{3}{2}t^2+6\text{.}$

2. 4 seconds. Solve $v(t)=30\text{.}$

23.

The skid marks made by a car indicate that its breaks were fully applied for a distance of 160 ft before it came to a stop. Suppose that the car in question had a constant deceleration of 20 ft/s$^2$ under the conditions of the skid. How fast was the car traveling when its breaks were applied?

$v(0)=80$ ft/s.
24.
1. You are standing at ground level, and you throw a ball upward into the air with an initial velocity of 64 feet per second. The acceleration due to gravity is 32 feet per second squared (towards the ground). How much time is the ball in the air for? Use antiderivatives.

2. What is the velocity of the ball at the time that it hits the ground?

1. 4 seconds.

2. $-64$ ft/sec$^2\text{.}$

Solution
1. Let $s(t)$ be the height of the ball after $t$ seconds. It is given that $s(0)=0\text{,}$ $s'(0)=v(0)=64$ ft/sec and $s"(0)=v'(0)=a(0)=-32$ ft/sec$^2\text{.}$ Thus $s(t)=-16t^2+64t=16t(4-t)\text{.}$ From $s(4)=0$ it follows that the ball is in the air for 4 seconds.

2. $v(4)=s'(4)=-64$ ft/sec$^2\text{.}$

25.

In 1939, Joe Sprinz, a professional baseball player, attempted to catch a ball dropped from a blimp. This was done for the purpose of breaking the record for catching a ball dropped from the greatest height set the previous year by another player on another team.

1. Ignoring air resistance, give an expression for the time it takes for a ball to drop $H$ metres (from rest) under a constant gravitational acceleration $g\text{.}$

2. Give an expression for the velocity of the ball after it has fallen $H$ metres. What is this value if $H=245$ m and $g=-10$ m/s$^2\text{?}$ Would you try to catch this ball? (Even with air resistance, the ball slammed Sprinz's glove into his face, fractured his upper jaw at 12 places, broke five teeth and knocked him unconscious. Worse still, he dropped the ball!)

1. $\ds\sqrt{\frac{2H}{g}}$s.

2. $-70$ m/sec.

Solution
1. From the fact that the velocity of an falling object is approximated by $v(t)=-gt+v(0)$ and the fact that, in the given case, $v(t)=0\text{,}$ we conclude that the distance $y=y(t)$ between the ball and the surface of the Earth at time $t$ is given by $\ds \frac{dy}{dt}=-gt\text{.}$ Hence, $\ds y=-\frac{gt^2}{2}+H\text{,}$ where $H$ is the height of the blimp at the moment when the ball was dropped. At the moment when the ball hits surface we have that $\ds 0=-\frac{gt^2}{2}+H$ which implies that it takes $\ds t=\sqrt{\frac{2H}{g}}$ seconds for a ball to drop $H$ metres.

2. $\ds v=-\sqrt{2Hg}=-10\cdot 7=-70$ m/sec.

Solve the initial value problem:

26.

$\ds \frac{dy}{dx}=2\sin 3x+x^2+e^{3x}+1\text{,}$ $y(0)=0\text{.}$

$\ds y=\frac{1}{3}\cdot(-2\cos 3x +x^3+e^{3x}+1)+x\text{.}$
27.

$\ds \frac{dy}{dx}=\sqrt{1-y^2}\text{,}$ $y(0)=1\text{.}$

$\ds y=\sin \left( x+\frac{\pi }{2}\right)\text{.}$
28.

$\ds \frac{dy}{dx}=y^2+1\text{,}$ $y(\pi /4)=0\text{.}$

$\ds y=\tan \left( x-\frac{\pi }{4}\right)\text{.}$
29.

$\ds \frac{dy}{dx}=1+y\text{,}$ $y(0)=3\text{.}$

$\ds y=4e^x-1\text{.}$
30.

$\ds \frac{dx}{dt}=\frac{36}{(4t-7)^4}\text{,}$ $x(2)=1\text{.}$

$\ds x(t)=-3(4t-7)^{-3}+4\text{.}$
31.

$\ds \frac{dy}{dx}=e^{-y}\text{,}$ $y(0)=2\text{.}$

$\ds y=\ln (x+e^2)\text{.}$
32.

$\ds \frac{dy}{dt}=2-y\text{,}$ $y(0)=1\text{.}$

$\ds y=2-e^{-t}\text{.}$
33.
1. Find the solution to $\ds \frac{dy}{dx}=3\cos 2x+\exp (-4x)$ such that $y(0)=1\text{.}$

2. Use the change of variables $u=2x+1$ to find the general form for the antiderivative of $\ds f(x)=\frac{1}{2x+1}\text{.}$

1. $\ds y=\frac{3}{2}\sin 2x-\frac{1}{4}\exp (-4x)+\frac{5}{4}\text{.}$
2. $\ds F(x)=\frac{1}{2}\ln (2x+1)+C\text{.}$
The price that maximizes the profit is $\ds p=-\frac{0.10\cdot 1000}{50}+4.15=2.15$ CAD.
Let $x$ be the number of towels sold per week at the price $p=p(x)\text{.}$ Let $C=C(x)$ be the cost of manufacturing $x$ towels. It is given that $\ds \frac{dC}{dx}=0.15$ CAD/towel and $\ds \frac{dp}{dx}=-\frac{0.10}{50}$ CAD/towel. Hence $C(x)=0.15x+a$ and $\ds p(x)=-\frac{0.10x}{50}+b\text{,}$ for some constants $a$ and $b$ (in CAD). Then the profit is given by $\ds P=P(x)=\mbox{Revenue} -\mbox{Cost} =x\cdot p(x)-C(x)=-\frac{0.10x^2}{50}+bx-0.15x-a\text{.}$ The quantity that maximizes revenue is $x=1000$ towels and it must be a solution of the equation $\ds \frac{dP}{dx}=-\frac{0.10x}{25}+b-0.15=0\text{.}$ Hence $\ds -\frac{0.10\cdot 1000}{25}+b-0.15=0$ and $b=4.15$ CAD. The price that maximizes the profit is $\ds p=-\frac{0.10\cdot 1000}{50}+4.15=2.15$ CAD.