# Tutor profile: Patrick C.

## Questions

### Subject: Linear Algebra

Find the characteristic equation of $(A = \begin{bmatrix} 5 & -2 & 6 & -1 \\ 0 & 3 & -8 & 0 \\ 0 & 0 & 5 & 4 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$)

This problem is rather straightforward to solve using $$A-\lambda*I$$ and the triangularity of our matrix $$A$$. First, we will find the determinant of $$A-\lambda*I$$: $(det(A - \lambda * I) = det\begin{bmatrix} 5 -\lambda & -2 & 6 & -1 \\ 0 & 3 -\lambda& -8 & 0 \\ 0 & 0 & 5 -\lambda& 4 \\ 0 & 0 & 0 & 1 -\lambda\\ \end{bmatrix}$). Thanks to our knowledge of triangular matrices, we know that the determinant of this matrix equals the product of our diagonal entries: $(=(5-\lambda)(3-\lambda)(5-\lambda)(1-\lambda)$). Thus our characteristic equation is this polynomial set equal to $$0$$: $((5-\lambda)^{2}(3-\lambda)(1-\lambda) = 0$).

### Subject: Calculus

Let's minimize the amount of fencing needed to enclose some area of a pasture: A farmer wants to build a 150-square-meter rectangular arena for his cows, but needs to reduce the cost spent on fencing as much as he can. To do this, one of the sides of his arena will be a wall of the barn that's 30 meters long. The cost of fencing is $20/meter. What dimensions should he make the arena?

First, we need to formalize our word problem into equations, and the we can optimize and solve from there. We know that the arena has three sides made from fencing, two of which must be the same length: let $$w$$ be the length of the parallel sides extending from the barn, and let $$l$$ be the length of the fencing parallel to the barn wall. Also, we should note that $$l \leq 30$$ for this given problem, since the barn wall has a length of $$30$$. Therefore, the total cost of fencing, $$C$$, will equal our unit cost of fencing, $20/meter, times the total length of fencing used by the three sides, which will equal $$2w + l$$, thus yielding $(C = 20(2w + l)$). Now we will incorporate the fact that we know the surface area that we wish to enclose in our arena to be $$150$$, and that this area is equal to $$w * l$$, since the arena is rectangular, and thus arrive at the equation $(150 = wl$). We can rewrite this second equation to be $(w = \frac{150}{l}$) and substitute this into our first equation to get $(C = 20(2(\frac{150}{l}) + l)$). This simplifies algebraically to be $(C = \frac{6000}{l} + 20l$), which equals $(C = 6000l^{-1} + 20l$). Now, to optimize this last equation, we will take its derivative with respect to $$l$$ and set it equal to $$0$$. By doing this, what we are searching for is a minimal value of the cost function, and then finding the value of $$l$$ that gets us this cost. However, we are going to have to check that our value of $$l$$ satisfies the constraint $$0 < l \leq 30$$, since we can't have $$l$$ less than or equal to $$0$$, and it cannot surpass the length of the barn wall, $$30$$. Proceeding with this plan, we will take the derivative of $$C$$ with respect to $$l$$: $(\frac{dC}{dl} = \frac{d}{dl} (6000l^{-1} + 20l)$) $(= -6000l^{-2} + 20$). Now we will set this equal to $$0$$ and see what result we get: $(0 = -6000l^{-2} + 20$) $(6000l^{-2} = 20$) $(6000 = 20l^2$) $(300 = l^2$) $(\sqrt{300} = l$). Noting that $$\sqrt{300} \approx 17.32$$ (we will ignore the negative counterpart) and that $$0 < 17.32 \leq 30$$, we have found an acceptable value of $$l$$. Now we need to find $$w$$ to get the complete dimensions of the arena that we want. The easiest way to find $$w$$ is to plug our value of $$l$$, $$\sqrt{300}$$ into our equation for the area of the arena, $$150 = wl$$, and solve, which we will proceed with next: $(150 = w * \sqrt{300}$) $(\frac{150}{\sqrt{300}} = w$) $(w \approx 8.66$). Therefore, we have found that the optimal dimensions of our farmer's arena are $$w \approx 8.66$$ meters and $$l \approx 17.32$$ meters.

### Subject: Algebra

A train leaves Dallas, TX at 11 a.m. traveling west at a speed of 40 mph. Two hours later, another train leaves Dallas going 50 mph on a parallel track in the same direction as the first. Assuming no stops, at what time would the second train catch up to the first? And how far away from Dallas will both trains be at that time?

To solve this problem, we need to first organize our information into mathematical expressions. Let's start with defining our variables. Let $$v_1$$ be the velocity of train 1, and $$d_1$$ be its distance away from Dallas. Likewise, let $$v_2$$ be the velocity of train 1, and $$d_2$$ be its distance from Dallas. We'll save our time variable, $t$, to be defined just a little later. At 11 a.m., the first train is just leaving Dallas, so $$v_1 = 40$$ miles/hr and $$d_1 = 0$$ miles at that time. Two hours later, train 2 leaves Dallas, so we want to set $$v_2 = 50$$ miles/hour and $$d_2 = 0$$ miles at that time, but for the first train, $$v_1$$ still equals $$40$$ and has for the past 2 hours, which makes our distance from Dallas equal to $(2*40 = 80$) miles. Therefore, $$d_1 = 80$$ miles at 1 p.m. Now we need to take this information and find when the trains will be the same distance away from Dallas, or in other words, when $$d_1 = d_2$$. We know that at 1 p.m., $(d_1 = 80 + 40t$), where $$t$$ is the time, in hours, after 1 p.m. Likewise, for the second train, we have that $(d_2 = 50t$). By setting our time variable, $$t$$, to "start" at 1 p.m., this same time variable can be used for both trains, which will be crucial to solving this problem smoothly. Thus, in order to find when the trains converge, we set $$d_1 = d_2$$ and find $$t$$: $(d_1 = d_2$) $(80 + 40t = 50t$) Subtracting $$40t$$ from both sides, we get $(80 = 10t$), which simplifies to $(8 = t$). Therefore, it will be $$8$$ hours after 1 p.m. that both trains will be the same distance away from Dallas. This puts us at approximately 9 p.m. This is the answer to our first question. To answer the second question (how far away from Dallas are the trains at the time they converge), all we need to do is plug our found time into either $$d_1$$ or $$d_2$$. Let's use $$d_2$$: $(d_2 = 50t$) $(d_2 = 50 * 8$) $(d_2 = 400$) Therefore, trains 1 & 2 will be $$400$$ miles from Dallas when they converge on their respective parallel tracks.