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# Tutor profile: Jay H.

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Jay H.
PhD Student in Mathematics
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## Questions

### Subject:Linear Algebra

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Question:

Under what conditions does a system of $$m$$ equation in $$n$$ variables have exactly one solution? Infinite solutions? No solutions? Express your answer in terms of the coefficient matrix $$A$$ and the augmented matrix $$A'$$.

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Jay H.

We know that when we row reduce the augmented matrix $$A'$$, we do not change the solutions of any of the equations; this is because operations like multiplying equations by a constant or adding together equations do not change solutions. The key here is to row reduce $$A'$$ (and then the first $$n$$ of $$n+1$$ columns is indeed $$A$$, because when we row reduce we proceed column by column). We will look at the rank of the resulting matrix, or, the number of nonzero rows in its row echelon form. Recall the very important fact that row reducing does not change the rank of a matrix! The rank of an augmented matrix tells us something like the minimum number of non-redundant equations that express the same information. If the rank of $$A'$$ is exactly the rank of $$A$$, then there are two options. The first option is that there is a row of zeros at the bottom (the rank is not $$m$$, the number of equations). Since any solution will solve the equation $$0x=0$$, this gives us an infinite number of solutions. The second option if $$A'$$ has the same rank as $$A$$ is that the rank is $$m$$. In this case, every single row will give us a solution for a different variable; and this allows us to get exactly one solution for each variable. We covered the case where the rank of $$A'$$ is exactly the rank of $$A$$. Can $$A'$$ have a rank less than $$A$$? Well, no, because if $$A'$$ has a zero row, certainly $$A$$ does as well, since $$A'$$ is simply $$A$$ with an extra column tacked on. The only remaining possibility, then, is that $$A'$$ is greater than the rank of $$A$$. But in this case, if we take a zero row of $$A$$ in which the corresponding row of $$A'$$ is nonzero, that gives us an equation that looks like, for example, $$0x=1$$. This has clearly no solution. To summarize; if the rank of $$A'$$ is greater than the rank of $$A$$, no solution. If the rank of $$A'$$ is equal to the rank of $$A$$, but is less than $$m$$, there are infinite solutions. Finally, if the rank of $$A'$$ is equal to the rank of $$A$$ and both are equal to $$m$$, then there is exactly one solution.

### Subject:Calculus

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Question:

Compute the following integral: $$\int_0^{2\pi} \sin^2(x)dx$$

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Jay H.

The hard way to compute this answer is to use the double angle identity: $$\cos 2x = 1-2\sin^2 x$$. Using this identity, we can rewrite the integrand as $$\frac{1}{2}(1-\cos 2x)$$. The rest of the integral is fairly standard; recall that the antiderivative of $$\frac{1}{2} \cos 2x$$ is $$\frac{1}{4}\sin 2x$$. However, we can also exploit symmetry here. Note that $$\sin^2 x + \cos^2 x = 1$$, so indeed integrating both sides we get: $$\int_0^{2\pi} \sin^2x dx + \int_0^{2\pi} \cos^2x dx = \int_0^{2\pi} dx = 2\pi$$. Furthermore, just by looking at the graphs, you can see that over a whole period of $$2\pi$$, $$\sin^2x$$ and $$\cos^2 x$$ have the same area underneath the curve! So indeed, since twice the quantity of our integral is $$2\pi$$, the computed integral is $$\pi$$.

### Subject:Physics

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Question:

A mass is thrown directly upwards at a speed of $$v_0$$ m/s. What is the maximum height reached by the object? (note: there is no dependence on the mass of the object!)

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Jay H.

First, we calculate the time at which the trajectory reaches its maximum height. We know from calculus that this is exactly when we have a local maximum, or equivalently the first derivative of height is zero. From kinematics, the trajectory of our object is $$y(t) = y_0 + v_0 t - \frac{1}{2}gt^2$$, where $$y_0$$ is the initial height, $$v_0$$ the initial velocity, and $$g$$ the gravitational constant. In this example, $$y_0=0$$ and $$v_0$$ is given. The equation for position is therefore $$y(t) = v_0 t - \frac{1}{2}gt^2$$. Taking the first derivative, we set $$0 = y'(t) = v_0 - gt$$, giving us $$t = \frac{v_0}{g}$$ at the maximum height. At this time, the position is given by: $$y(t) = v_0 t - \frac{1}{2}gt^2 = v_0 (\frac{v_0}{g}) - \frac{1}{2}g (\frac{v_0}{g})^2 = \frac{v_0^2}{2g}$$

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