###### tags: `MATA35-2021`
# MATA35: Final Exam - Practice w/ Solutions
**Department of Computer & Mathematical Sciences
University of Toronto Scarborough**

**MATA35H3Y: Calculus II for Biological Sciences
Final Examination Practice Sheet**

**Examiner**: Yun William Yu
**Date**: August 16, 2021, 8am
**Duration**: 2 hours

## Notes:
* No calculators, or any electronic aid is permitted.
* As this exam will be invigilated over Zoom (and recorded), please ensure that your webcam is on throughout the course of the exam, and that your writing surface and face is maximally visible for the entire duration of the exam.
* Please also bring an ID (ideally a University of Toronto T-card, but other forms of ID are also acceptable); at the start and end of the exam, please show your ID card to the webcam.
* If you finish within the first 90 minutes, you may send a private message to an invigilator to ask permission to be allowed to scan, upload, and leave early. Once this permission is granted, you must immediately scan and upload your questions, and you will not be allowed to make any further edits to your responses.
* If you finish in the last 30 minutes of the 2-hour period, please simply wait until the end of the exam to scan and upload.
* After the 2-hour exam period is over, we ask that you immediately scan and upload to Crowdmark responses to each question, as well as your cover sheet. We will allocate 20 minutes after the 2-hour exam period for you to scan and upload. If you have technical difficulties, please **immediately** inform an invigilator.

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## Cover sheet
Please make sure you include the following information on your cover sheet, to be scanned to Crowdmark:

Student information:
* Family Name
* Given Names
* Student Number
* Signature

Course information
* MATA35H3Y - UTSC - Summer 2021
* Exam time: August 16, 2021, 8-10am

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## 1. Leslie matrix **[20pts]**

A species of bird has three life stages, hatching (H), juvenile (J), and adult (A). After some study, you build a Leslie Age-Structured model capturing the yearly change in the population with the following Leslie diagram:

1. **[5pts]** Write down the Leslie matrix for this model.
**Hint**: the determinant of the Leslie matrix is 0.3. You may wish to check that you have the correct Leslie matrix before doing moving on.
2. **[5pts]** If right now, the population consists of 90 hatchlings, 40 juveniles, and 60 adults, estimate the population one year into the future. Estimate the population two years into the future.
3. **[10pts]** Using the same model, estimate the population one year ago?

> 

> 
> The population one year into the future will be 160 hatchlings, 18 juveniles, and 70 adults.
> 
> The population two years into the future will be 158 hatchlings, 32 juveniles, and 53 adults.

> To estimate the population one year ago, you need to invert the matrix multiplication problem. One method is to find the matrix inverse, but this is probably overkill. Instead, we simply need to solve the following:
> 
> We can do that either as a system of equations or using Gaussian elimination.
> **Gaussian elimination:**
> 
> **System of equations:**
> 
> Either way, our estimate for the population in the previous year is 200 hatchlings, 50 juveniles, and 20 adults.

## 2. Multivariable critical points **[10pts]**
Let $f(x,y)= x^3 + y^3 - 3xy$.
1. **[5pts]** Find all the (real) critical points of $f$.
2. **[5pts]** Identify whether the critical points found above are relative minimums, relative maximums, or saddle points.

> 

> To classify the saddle points, we can either use the D-test or look at the eigenvalues of the Hessian. I personally prefer eigenvalues of the Hessian.
> 
> Since $H(0,0)$ has opposite sign eigenvalues, it is a saddle point.
> Since $H(1,1)$ has positive eigenvalues, it is a relative minimum.

## 3. Phase line problem **[10pts]**
You are an epidemiologist from the University of Toronto who was recently tasked with understanding the Cooties. To this end, you observe children at a local primary school, and come to several conclusions:

- The disease has no adverse side-effects. In fact, if it weren't for the children loudly proclaiming they or another person caught the Cooties, you wouldn't be able to determine their infection status.
- No one is immune from the Cooties.
- Cooties are transmitted at a rate proportional to both the number of Susceptible individuals and the number of Infected individuals.
- Infected individuals seem to recover spontaneously at a high random rate.

After weeks of painstaking observation, you gather enough information to build a two-compartment model:
- There are a total of $N=100 = S(t) + I(t)$ kids at the school.
- Kids are either Susceptible (S) or Infected (I)
- $0.01SI$ Susceptible individuals become Infected each day.
- Infected individuals recover and become Susceptible again at a rate of 15% per day.

1. Draw a 2-compartment model for $S$ and $I$.
2. Convert the model to a one-variable model counting only infected individuals using the fact that $N = S + I$.
3. Find the equilibrium values.
4. Classify the equilibria by type and stability.
5. What is the long-term behavior of the system.

> 
> In the long-term, we have the endemic disease equilibrium, where asymptotically 85 students will have Cooties at any one time.

## 4. Solving single-variable ODEs **[30pts]**
Solve each of the following problems as generally as possible.
1. Solve the initial value problem: $y' = 2x + xy$, where $y(0)=1$.
2. Find the general solution: $y'' + 4y' + 4y = 4+e^x$.
3. Give an implicit solution to the following initial value problem: $(ye^{xy} + 3x^2 + x)dx + (xe^{xy}- 3e^y + y)dy=0$, where $y(0)=0$.
4. Extra practice for a harder separable equation. Solve the initial value problem, $y'=xy+xy^2$, where $y(0)=1$.

> Part 1 is separable.
> $\frac{dy}{dx} = x(y+2)$
> $\frac{dy}{y+2}= x dx$
> $\ln|y+2| = \frac{1}{2}x^2 + C$
> $y+2 = Ce^{0.5x^2}$
> $y = -2+Ce^{0.5x^2}$
> Substitute in $y(0)=1$, and we get $C=3$
> $y=-2+3e^{0.5x^2}$

> Part 2 is inhomogeneous linear with constant coefficients.
> 

> Part 3 is exact, so we can assume there exists some $f(x,y)$ such that
> 
> Then, we simply need to integrate both pieces holding the other variable constant to get $f(x,y)
> 
> Then, we need to substitute in the initial conditions $y=0$ when $x=0$ to solve for $C$.
>

> The extra separable practice problem in part 4 resembles the logistic equation.
> 

## 5. First-order linear system **[10pts]**
Consider the first-order linear system
\begin{align}
\dot{x} &= 7x + 8y \\
\dot{y} &= -4x - 5y
\end{align}
Find the general solution. Classify the type of equilibrium at the origin, and determine its stability.

> Either the reduction method or the eigenbasis method will work for solving the equation. Using eigenvectors is somewhat easier whenever it is applicable.
> 
> The only equilibrium is at the origin (0,0), and the equilibrium is a saddle point, which is unstable.

> **Reduction method**
> $\ddot{x} = 7\dot{x} + 8\dot{y}$
> $\ddot{x} = 7\dot{x} + 8(-4x - 5y)$
> $\ddot{x} = 7\dot{x} + -32x - 40y$
> Note that $y = \frac{1}{8}(\dot{x}-7x)$
> $\ddot{x} = 7\dot{x} + -32x - 5(\dot{x}-7x)$
> $\ddot{x} = 7\dot{x} + -32x - 5\dot{x} + 35x$
> $\ddot{x} = 2\dot{x} + 3x$
> $\ddot{x} - 2\dot{x} - 3x = 0$
> Characteristic equation $\lambda^2 - 2\lambda - 3 = (\lambda-3)(\lambda+1)=0$
> Thus, $\lambda_1 = -1$, $\lambda_2 = 3$.
> Thus $x = c_1 e^{-t} + c_2 e^{3t}$
> And thus $\dot{x} = -c_1 e^{-t} + 3c_2 e^{3t}$
> So $y = \frac{1}{8}(\dot{x}-7x) = \frac{1}{8}\left((-c_1 e^{-t} + 3c_2 e^{3t})-7(c_1 e^{-t} + c_2 e^{3t})\right)$
> $y = \frac{1}{8} (-8c_1 e^{-t} - 4c_2 e^{3t}) = -c_1 e^{-t} -\frac{1}{2}c_2 e^{3t}$
> In conclusion, we get the following:
> $x = c_1 e^{-t} + c_2 e^{3t}$
> $y = -c_1 e^{-t} -\frac{1}{2}c_2 e^{3t}$
> Notice that the $c_1$ and $c_2$ are slightly different than the ones we got above using the eigenbasis method. However, the answers are actually the same, by just sending the $c_2$ here to $-2c_2$, which you can do becuase it's an arbitrary constant.

## 6. Power series **[10pts]**
Find a cubic approximation of the following function using the first 4 terms of its Maclaurin Series (Taylor Series at 0).
$f(x) = e^{x^2} + x^3$.
**Hint**: $e^x = \sum_{i=0}^\infty \frac{x^i}{i!}$

> We can solve this either directly by using the Taylor Series formula, or we more easily by plugging $x^2$ in for $x$ in the formula for $e^x$.
> **Hard direct method**
> $f(x)\approx f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3 = 1 + x^2 + x^3$
> **Faster plugging-in method**
> 

## 7. Nonlinear system **[10pts]**
Consider the following system
\begin{align}
\frac{dx}{dt} &= y - x^2 \\
\frac{dy}{dt} &= x - y
\end{align}

1. Find the coordinates of the two real equilibria.
2. Classify the equilibria and determine their stability.

> Real equilibria are at $(0,0)$ and $(1,1)$.
> $J(0,0) = \begin{pmatrix}
0 & 1 \\
1 & -1
\end{pmatrix}$, which has eigenvalues $\lambda_{1,2} = \frac{1}{2}(-1 \pm \sqrt{5})$, which are opposite signs, so this is an unstable saddle point.

> $J(1,1) = \begin{pmatrix}
-2 & 1 \\
1 & -1
\end{pmatrix}$, which has eigenvalues $\lambda_{1,2} = \frac{-3 \pm \sqrt{5}}{2}$ by the quadratic equation, which are both negative, so it is an asymptotically stable node.