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Tutor profile: Sergio M.

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Sergio M.
Been teaching college-level math for several years
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Questions

Subject: Basic Math

TutorMe
Question:

Is it possible that Anna will be twice as old as Ben in two years, but she was born two years before him?

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Sergio M.
Answer:

One option would be to try and guess, which is not too hard in this case. The systematic way would be: denote by $A$ Anna's current age, and by $B$ Ben's current age. The first bit of information means that $$ A + 2 = 2(B+2), $$ while the second one means that $$ A = B + 2. $$ But then, $A + 2 = B+4,$ from the second equation, so the first one reads $$ B + 4 = 2(B+2) = 2B + 4, $$ which forces $B=0$, so Ben is right now being born, and Anna's age is $A = 0 + 2 = 2.$ So it is indeed possible.

Subject: Calculus

TutorMe
Question:

A mountain has a circular base of radius $1$ km; assume the center of the circle coincides with the origin in the $xy$-plane. The elevation at point $(x,y)$ is $h(x,y) = 1-x^2-y^2$. Find the average elevation of the mountain.

Inactive
Sergio M.
Answer:

By definition of average of a function, the average elevation should be $$ \frac{1}{A(D)} \int{D} h(x,y) dx dy $$ where $D$ is the circular base of the mountain (a disk of radius $1$) and $A(D)$ is its area. Thus, we need to calculate the double integral $$ \frac{1}{\pi} \int_D (1-x^2-y^2)dx dy, $$ which is easier done in polar coordinates: $x = r \cos\theta,$ $y= r\sin\theta,$ $r^2 = x^2 + y^2$, and the area element $dxdy = r dr d\theta,$ so we have $$ \frac{1}{\pi} \int_0^{2\pi}\int_0^1 (1-r^2) r dr d\theta = 2 \int_0^1 (r-r^3) dr = 1/2. $$

Subject: Differential Equations

TutorMe
Question:

Find all solutions to the equation $$ x^2 y'' - 3x y' + y = 0. $$

Inactive
Sergio M.
Answer:

Notice that the coefficients of the equation have decreasing exponents, and differentiation of power functions decreases the exponent. One is tempted to look for a solution of the form $y = x^m.$ This is indeed called a Cauchy-Euler equation. If one tries $y= x^m$ into the equation, one gets, for the left-hand side: $$ x^2 m(m-1)x^{m-2} - 3x m x^{m-1} + x^m = m(m-1) x^m - 3m x^m + x^m = [m(m-1) - 3m + 1]x^m $$ so we should find $m$ that will make this zero, which means solving $$ m^2 - 4m + 1 = 0, $$ giving us $m = 2-\sqrt{3}$ and $m = 2+\sqrt{3}$. Thus, $$ y_1 = x^{2-\sqrt{3}}, \qquad y_2 = x^{2+\sqrt{3}} $$ are two solutions of the given equation. We can compute the Wronskian of this two functions, and we will get $$ W(y_1,y_2) = (2+\sqrt{3})x^{2+\sqrt{3} + 2 -\sqrt{3}} - (2 - \sqrt{3})x^{2+\sqrt{3} + 2 -\sqrt{3}}, $$ which is different from zero. So the two solutions are linearly independent. Since the equation is of second order, we have found all the solutions: $$ y = C_1 x^{2-\sqrt{3}} + C_2 x^{2+\sqrt{3}}. $$

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