A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Section 6.11 Antiderivatives and Differential Equations

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

17. (a) $v(t)=-10t+5\text{;}$ (b) $s(t)=-5t^2+5t+30\text{;}$ (c) $t=6$ s; (d) $|v(6)|=5$ m/s.

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

19. (a) $\ds v(t)=\frac{3}{2}t^2+6\text{.}$ (b) 4 seconds. Solve $v(t)=30\text{.}$

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

21. (a) 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. (b) $\ds v=-\sqrt{2Hg}=-10\cdot 7=-70$ m/sec.

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

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

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

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

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

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

    7. $\ds y=2-e^{-t}\text{.}$

23. (a) $\ds y=\frac{3}{2}\sin 2x-\frac{1}{4}\exp (-4x)+\frac{5}{4}\text{.}$ (b) $\ds F(x)=\frac{1}{2}\ln (2x+1)+C\text{.}$

24. $v(0)=80$ ft/s.

25. 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.