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Troy H.
Bachelor of Science in Physics, Minor in Math from the University of Texas at Austin
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Calculus
TutorMe
Question:

The following is a problem from the AP Calculus BC release test: Consider the differential equation: $(\frac{dy}{dx}=x^2 - \frac{1}{2}y$) (a) Find $$\frac{d^2y}{dx^2}$$ in terms of x and y (b) Let $$y = f(x)$$ be the particular solution to the given differential equation whose graph pases through the point (-2,8). Does the graph of f have a relative minimum, a relative maximum, or neither at the point (-2,8)? Justify your answer. (c) Let $$y=g(x)$$ be the particular solution to the given differential equation with $$g(-1)=2$$. Find $$\lim_{x\rightarrow -1}\frac{g(x)-2}{3(x+1)^2}$$. Show work that leads to your answer. (d)Let $$y=h(x)$$ be the particular solution to the given differential equation with $$h(0)=2$$. Use Euler's method, starting at $$x=0$$ with two steps of equal size, to approximate h(1).

Troy H.

(a) First, it would be important to make sure the student understands what it is that the question is asking and how to properly rewrite the question. A common mistake here would be to differentiate with respect to x twice and consider that the final answer, so being certain of notation is of the utmost importance. Differentiating and substituting the variables would give us: $(\frac{d^2y}{dx^2}=2x-\frac{1}{2}\frac{dy}{dx}=2x-\frac{1}{2}(x^2-\frac{1}{2}y)$) (b) Now that we have solved for $$\frac{d^2y}{dx^2}$$ in part (a), this problem becomes a matter of calculation and interpretation. We will solve both $$\frac{dy}{dx}$$ and $$\frac{d^2y}{dx^2}$$ at the given point (-2,8), which gives us: $(\frac{dy}{dx}|_{(x,y)=(-2,8)}=(-2)^2-\frac{1}{2}(8)=0$) $(\frac{d^2y}{dx^2}|_{(x,y)=(-2,8)}=2(-2)-\frac{1}{2}((-2)^2-\frac{1}{2}(8))=-4<0$) Therefore, we can conclude that f has a relative minimum at the given point. (c) First, I would have the student take the limits of both $$g(x)-2$$ and $$3(x+1)^2$$ as x approaches -1. Taking those limits gives us: $(\lim_{x\rightarrow-1}(g(x)-2))=0$) $(\lim_{x\rightarrow-1}(3(x+1)^2)=0$) At this point, the student may either be tempted to say the answer is zero or they may even become more confused initially, so if that is the case, I would discuss the situation with them in the hope that they realize that there is indeed method to further solve the question, and that method involved using L'Hospital's Rule, which involves taking the derivative of the numerator and denominator and again finding the limit. Using L'Hospital's Rule gives us: $(\lim_{x\rightarrow -1}\frac{g(x)-2}{3(x+1)^2} \Rightarrow \lim_{x\rightarrow -1}\frac{g'(x)}{6(x+1)}$) Again, taking these limits gives us $$\frac{0}{0}$$, so we use L'Hospital's Rule a second time to give us: $(\lim_{x\rightarrow -1}\frac{g'(x)}{6(x+1)} \Rightarrow \lim_{x\rightarrow -1}\frac{g''(x)}{6}=\frac{-2}{6}=-\frac{1}{3}$) Note: since we are given at the beginning that what we have now defined as $$g'(x)=x^2-\frac{1}{2}y$$, the second derivative becomes $$g''(x)=2x$$ which is what allows us to say that $$\lim_{x\rightarrow -1}g''(x)=-2$$ (d) First, since we are told to use Euler's Method, I would have the student write out the equation for Euler's Method, which is: $(y_n \approx y_{n-1}(x) + y'_{n-1}(x)*h$) Where h is our step size. Now, using the given information and the appropriately defined variables, we can substitute for the following equation: $(h(\frac{1}{2}) \approx h(0) + h'(0)(\frac{1}{2})=2+[0^2-\frac{1}{2}(2)](\frac{1}{2})=\frac{3}{2}$) Repeating this process for the second step gives us: $(h(1) \approx h(\frac{1}{2}) + h'(\frac{1}{2})(\frac{1}{2}) \approx \frac{3}{2}+[(\frac{1}{2})^2-\frac{1}{2}(\frac{3}{2})](\frac{1}{2})=\frac{5}{4}$)

Algebra
TutorMe
Question:

Simplify: $\frac{x^2-x-30}{x^2-12x+36}$

Troy H.

In my experience, many students who struggle with Algebra, especially questions like this one, have trouble finding the correct place to start and often times get in their own heads, which prevents them from solving the problem correctly, so I would first ask the student what they think the first step is. In this problem, the first step is to factor both the numerator and denominator, which would give us: $\frac{x^2-x-30}{x^2-12x+36} \Rightarrow \frac{(x-6)(x+5)}{(x-6)^2}$ At this point, I would hope the student begins to feel more relaxed with both this problem, and problems like it, and realize the next step is to cancel the like terms, which leaves us with the final answer of: $\frac{(x+5)}{(x-6)}$

Physics
TutorMe
Question:

The following is a problem from an AP Physics release exam: Two small blocks, each of mass m, are connected by a string of constant length 4h and negligible mass. Block A is placed on a smooth tabletop and block B hangs over the edge of the table. The tabletop is a distance 2h above the floor. Block B is then released from rest at a distance h above the floor at time t = 0. Express all algebraic answers in terms of h, m, and g. (a) Determine the acceleration of block B as it descends. (b) Block B strikes the floor and does not bounce. Determine the time at which block B strikes the floor. (c) Describe t motion of block A from time t = 0 to the time when block B strikes the floor. (d) Describe the motion of block A from the time block B strikes the floor to the time block A leaves the table. (e) Determine the distance between the landing points of the two blocks.

Troy H.

(a) First, I would have the student draw the force diagrams for each block. For block A, there would be the gravitational force pointing downward, the corresponding force from Newton's 2nd Law pointing upwards, and the force of tension from the string. Block B would have the gravitational force pointing downwards and the same tension force pointing upwards. For block A, we would have the following equation: $ma = T$ For block B, we would have the following equation: $ma = mg - T$ By substituting for T in the block B equation from the block A equation, we get the following: $ma = mg - ma \Rightarrow 2ma = mg \Rightarrow a = \frac{g}{2}$ (b) I would start by having the student write out the relevant equations from kinematics as well as the values for our known variables, which would yield the following: Relevant Equation: $x = x_0 + v_0t+\frac{1}{2}at^2$ Known Variables: $x_0 = 0, x=h, v_0=0, a=\frac{g}{2},$ Plugging this information into our relevant equation gives us: $h = 0 + (0)t + \frac{1}{2}(\frac{g}{2})t^2 \Rightarrow h = (\frac{g}{4})t^2$ Solving this for t gives us: $t = 2\sqrt\frac{h}{g}$ (c) This is a purely conceptual question, so I would first ask the student what they believe the answer is and why. Depending on their answer, I would ask relevant follow up questions to make sure they address the entire question. The kind of final answer I would be looking for is something that would address the left-to-right movement of block A and that the magnitude of the acceleration of block A would be equivalent to the acceleration of block B, which was found in part (a). In the event that the student is hung up on the concept of friction, which would be the most likely issue in my experience, I would remind them that they have been told that they are dealing with a smooth table, so we can ignore the affects of friction. (d) Much like part (c), I would discuss the concept with the student first. The kind of final answer I would be looking for is something that addresses the continued motion of block A after block B hits the ground; however, the block would be no longer accelerating and would instead continue at a constant velocity across the table until it reaches the end of the table. (e) Much like part (b), I would start by having the student write out the relevant equations and known variables. The relevant equations from kinematics are: $v = v_0 + at$ $x = vt$ $y = y_0 + v_{0,y}t+\frac{1}{2}at^2$ The known variables are: $v_0 = 0$ $a = \frac{g}{2]$ $t = 2\sqrt\frac{h}{g}$ $y = 2h$ $y_0 = 0$ $v_{0,y} = 0$ First, we would solve for our velocity, v, by plugging in our known variables: $v = 0 + (\frac{g}{2})(2\sqrt\frac{h}{g}) \Rightarrow v=\sqrt\frac{hg^2}{g} \Rightarrow v = \sqrt{hg}$ We then plug in the relevant variables into the 3rd kinematic equation for y and solve for t, which gives us: $y = y_0 + v_{0,y}t+\frac{1}{2}at^2\Rightarrow 2h = 0 + (0)t + \frac{1}{2}gt^2\Rightarrow t=2\sqrt\frac{h}{g}$ We then plug in these equations for both v and t into the kinematic equation for x to get our final answer: $x = \sqrt{hg}(2\sqrt\frac{h}{g})\Rightarrow x=2h$

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