## MATH1013 Quizzes

Quiz 7: Exponential growth and decay
Question 1 Questions
Consider $y={e}^{kx}\phantom{\rule{0.3em}{0ex}}.$ Find $\frac{dy}{dx}\phantom{\rule{0.3em}{0ex}}.$
 a) $\frac{dy}{dx}={e}^{kx}$ b) $\frac{dy}{dx}=k{e}^{kx}+C$ c) $\frac{dy}{dx}=\frac{1}{k}{e}^{kx}+C$ d) $\frac{dy}{dx}=k{e}^{kx}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Consider $y=lnkx\phantom{\rule{0.3em}{0ex}}.$ Find $\frac{dy}{dx}\phantom{\rule{0.3em}{0ex}}.$
 a) $\frac{dy}{dx}=\frac{1}{x}$ b) $\frac{dy}{dx}=\frac{k}{x}$ c) $\frac{dy}{dx}=\frac{1}{kx}$ d) $\frac{dy}{dx}=\frac{1}{x}+\frac{1}{k}$

Choice (a) is correct!
Note that $y=lnkx=lnx+lnk\phantom{\rule{0.3em}{0ex}}.$ Hence $\frac{dy}{dx}=\frac{1}{x}+0\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find $y=\int \frac{k}{x}dx\phantom{\rule{0.3em}{0ex}}.$
 a) $y=lnx+kx$ b) $y=lnkx+C$ c) $y=klnx+C$ d) $y=-k{x}^{-2}+C$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Find $y=\int {e}^{4x}dx\phantom{\rule{0.3em}{0ex}}.$
 a) $y=4{e}^{4x}+C$ b) $y=\frac{1}{4}{e}^{4x}+C$ c) $y=A{e}^{4x}$ d) $y=\frac{A}{4}{e}^{4x}$

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Consider the expression $klnx+C$ where $k$ and $C$ are positive constants. Which of the following statements are true ?
 a) $klnx+C=lnA{x}^{k}$ where $A$ is a constant. b) $klnx+C=lnC{x}^{k}$ c) $klnx+C=ln{x}^{k}+C$ d) $klnx+C=klnCx$

There is at least one mistake.
For example, choice (a) should be True.
$klnx+C=ln{x}^{k}+C=ln{x}^{k}+lnA$, where $lnA=C$. So $klnx+C=lnA{x}^{k}$
There is at least one mistake.
For example, choice (b) should be False.
There is at least one mistake.
For example, choice (c) should be True.
There is at least one mistake.
For example, choice (d) should be False.
Correct!
1. True $klnx+C=ln{x}^{k}+C=ln{x}^{k}+lnA$, where $lnA=C$. So $klnx+C=lnA{x}^{k}$
2. False
3. True
4. False
Solve the equation ${e}^{2t}+5=20\phantom{\rule{0.3em}{0ex}}.$
 a) $t=\frac{ln4}{2}$ b) $t=\frac{ln25}{2}$ c) $t=ln7.5$ d) $t=\frac{ln15}{2}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
${e}^{2t}=15⇒2t=ln15⇒t=\frac{ln15}{2}\phantom{\rule{0.3em}{0ex}}.$
Solve the equation $ln2t+ln3=8\phantom{\rule{0.3em}{0ex}}.$
 a) $t=\frac{{e}^{8}}{6}$ b) $t=\frac{{e}^{8}}{2}-\frac{3}{2}$ c) $t={e}^{4}-\frac{3}{2}$ d) $t=\frac{{e}^{2}}{\sqrt[3]{2}}$

Choice (a) is correct!
$ln2t+ln3=ln6t=8⇒6t={e}^{8}⇒t=\frac{{e}^{8}}{6}\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Consider a population $P$ where $P={P}_{0}^{}{e}^{0.02t}$ and $t$ is measured in years. How long will it take for the population to double ?
 a) $1.96$ years. b) $34.66$ years. c) $2.3$ years. d) There is not enough information given.

Choice (a) is incorrect
Choice (b) is correct!
$P=2{P}_{0}^{}={P}_{0}^{}{e}^{0.02t}⇒{e}^{0.02t}=2⇒0.02t=ln2$
$⇒t=\frac{ln2}{0.02}=34.66\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Choice (d) is incorrect
The amount of a given radioactive isotope at time $t$ (in years) is given by $P={P}_{0}^{}{e}^{-kt}$ where $k=5×1{0}^{-6}\phantom{\rule{0.3em}{0ex}}.$ What is the half-life of the isotope ?
 a) $139000$ years. b) $6.1$ years c) $12.9$ years d) There is not enough information given.

Choice (a) is correct!
$P=\frac{1}{2}{P}_{0}^{}={P}_{0}^{}{e}^{-kt}⇒{e}^{-kt}=\frac{1}{2}⇒-kt=ln\frac{1}{2}$
$⇒kt=ln2⇒t=\frac{ln2}{5×1{0}^{-6}}=138629.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
A population $P$ grows exponentially. If it takes 15 years for the population to double, how long does it take for the population to treble ?
 a) $22.50$ years. b) $23.77$ years c) $16.48$ d) There is not enough information given.

Choice (a) is incorrect
Choice (b) is correct!
Since $P$ grows exponentially $P={P}_{0}^{}{e}^{kt}$ for some $k$ a positive constant.
$P=2{P}_{0}^{}={P}_{0}^{}{e}^{15k}⇒{e}^{15k}=2⇒k=\frac{ln2}{15}$
$P=3{P}_{0}^{}={P}_{0}^{}{e}^{kt}⇒{e}^{kt}=3⇒\frac{ln2}{15}\phantom{\rule{0.3em}{0ex}}t=ln3⇒t=15\phantom{\rule{0.3em}{0ex}}\frac{ln3}{ln2}=23.77\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Choice (d) is correct!