 ## MATH1003 Quizzes

Quiz 10: Solving Differential Equations; Modelling
Question 1 Questions
Find the general solution to $\frac{dx}{dt}=1+t-x-xt$.
(In each option, $C$ is an arbitrary constant.) Exactly one option must be correct)
 a) $x=C{e}^{-t}$ b) $x=1-C{e}^{t+{t}^{2}∕2}$ c) $x=1-C{e}^{-t-{t}^{2}∕2}$ d) $x={e}^{-t-{t}^{2}∕2}\left(t+\frac{{t}^{2}}{2}+C\right)$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Find the general solution to ${x}^{2}\frac{dy}{dx}+2xy={cos}^{2}x$.
(In each option, $C$ is an arbitrary constant.) Exactly one option must be correct)
 a) $y=\frac{1}{2x}+\frac{sin2x}{4{x}^{2}}+\frac{C}{{x}^{2}}$ b) $y=\frac{1}{2x}+\frac{sin2x}{4{x}^{2}}+C$ c) $y=\frac{{sin}^{2}x+C}{{x}^{2}}$ d) $y=\frac{x}{2}+\frac{sin2x}{4}+C$

Choice (a) is correct!
Choice (b) is incorrect
Add the arbitrary constant before dividing by ${x}^{2}$.
Choice (c) is incorrect
Choice (d) is incorrect
A cup of coffee has a temperature of $9{5}^{\circ }$C and is in a room where the temperature is $2{0}^{\circ }$C.
Let $T=$ the temperature of the coffee after $t$ minutes.
Assuming Newton’s Law of Cooling, which of the following options describes $T$ ? Exactly one option must be correct)
 a) $\frac{dT}{dt}=k\left(T-20\right)$, $k$ a positive constant. b) $\frac{dT}{dt}=-k\left(T-20\right)$, $k$ a positive constant. c) $\frac{dT}{dt}=-k\left(T-95\right)$, $k$ a positive constant. d) $\frac{dT}{dt}=k\left(T-95\right)$, $k$ a positive constant.

Choice (a) is incorrect
Note that $\frac{dT}{dt}$, must be negative, since the coffee is cooling.
Choice (b) is correct!
Choice (c) is incorrect
The rate of change of $T$ is proportional to the difference between $T$ and the temperature of the surroundings.
Choice (d) is incorrect
The rate of change of $T$ is proportional to the difference between $T$ and the temperature of the surroundings.
A learning curve is the graph of a function $P\left(t\right)$, the performance of someone learning a skill as a function of the training time. Let $M$ be the maximum level of performance of which the learner is capable. Suppose that the rate at which the performance improves is proportional to the difference between the maximum level and the current level. Write down the differential equation which models the learning. Exactly one option must be correct)
 a) $\frac{dM}{dt}=k\left(M-P\right)$, $\phantom{\rule{1em}{0ex}}k$ a positive constant. b) $\frac{dP}{dt}=kM-P$, $k$ a positive constant. c) $\frac{dP}{dt}=-k\left(M-P\right)$, $k$ a positive constant. d) $\frac{dP}{dt}=k\left(M-P\right)$, $k$ a positive constant.

Choice (a) is incorrect
$M$ is a constant. It is $P$ that is changing.
Choice (b) is incorrect
Try again.
Choice (c) is incorrect
$\frac{dP}{dt}$ must be positive.
Choice (d) is correct!
Suppose that a tumour in a rat is approximately spherical, and that its rate of growth is proportional to its diameter. If the tumour has diameter 5 mm when detected, and 8 mm three months later, what will the diameter be after another three months?

Correct!
Water leaks out of a barrel at a rate proportional to the square root of the depth of the water at the time. If the water level starts at 36 cm and drops to 35 cm in one hour, how long will it take for all the water to leak out?

Correct!
Start with the differential equation $\frac{dh}{dt}=k\sqrt{h}$, where $h$ is the depth of the water at time $t$.
A tank holds 1000 litres of water, in which 15 kg of salt is dissolved. Pure water enters the tank at the rate of 10 litres per minute. The solution is kept thoroughly mixed and is drained from the tank at the same rate. If $m$ is the mass of salt in the tank at time $t$, which of the following options describes the rate of change of the mass of salt in the tank? Exactly one option must be correct)
 a) $\frac{dm}{dt}=15-\frac{m}{100}$ b) $\frac{dm}{dt}=-\frac{m}{100}$ c) $\frac{dm}{dt}=-\frac{15}{100}$ d) $\frac{dm}{dt}=\frac{15-m}{1000}$

Choice (a) is incorrect
No salt is being added to the tank.
Choice (b) is correct!
Choice (c) is incorrect
The rate at which salt is removed is not constant; it depends on $m$.
Choice (d) is incorrect
Try again. Remember that $\frac{dm}{dt}=$ rate at which salt is being added $-$ rate at which it is being removed.
A tank holds 1000 litres of pure water. Brine which contains 0.05 kg of salt per litre enters the tank at the rate of 5 litres per minute. The solution is kept thoroughly mixed and is drained from the tank at the rate of 5 litres per minute. If $m$ is the mass of salt in the tank at time $t$, which of the following options describes the rate of change of the mass of salt in the tank? Exactly one option must be correct)
 a) $\frac{dm}{dt}=-\frac{m}{200}$ b) $\frac{dm}{dt}=0.25$ c) $\frac{dm}{dt}=\frac{50-m}{200}$ d) $\frac{dm}{dt}=0.05-\frac{m}{200}$

Choice (a) is incorrect
Salt is being added to the tank, as well as being removed.
Choice (b) is incorrect
Salt is added at the rate of $0.25$ kg per minute, but it is also being removed.
Choice (c) is correct!
Choice (d) is incorrect
Salt is being added at the rate of $5×0.05$ kg per minute.
For the tank described in Question 8, determine the mass $m$ of salt in the tank at time $t$ as a function of $t$. (That is, solve the differential equation that is the correct answer to Question 8.) Exactly one option must be correct)
 a) $m=50-A{e}^{t∕200}$, A an arbitrary constant. b) $m=50\left(1-{e}^{t∕200}\right)$. c) $m=50-A{e}^{-t∕200}$, A an arbitrary constant. d) $m=50\left(1-{e}^{-t∕200}\right)$.

Choice (a) is incorrect
$\int \frac{1}{50-m}\phantom{\rule{0.3em}{0ex}}dm=-ln\left(50-m\right)$. Also, an initial condition is given, so the value of $A$ should be found.
Choice (b) is incorrect
$\int \frac{1}{50-m}\phantom{\rule{0.3em}{0ex}}dm=-ln\left(50-m\right)$.
Choice (c) is incorrect
An initial condition is given, so the value of $A$ should be found.
Choice (d) is correct!
A large tank (with capacity 500 litres) contains 100 litres of fresh water. A solution with a salt concentration of 0.4 kg per litre is added at a rate of 5 litres per minute. The solution is kept mixed and is drained from the tank at the rate of 3 litres per minute.
Find the concentration of salt in the tank after $20$ minutes. Give your answer in kg/litre, correct to two decimal places.

Correct!
Well done!
If $y$ is the amount of salt (in kg) in the tank after $t$ minutes, the differential equation to be solved is $\frac{dy}{dt}=2-\frac{3y}{100+2t}$. Find the particular solution to this equation corresponding to the fact that there is no salt in the tank initially, and then find the value of $y$ when $t=20$.