## December 2012 Mission Math Tournament: Calculus Round Solutions

I contributed a total of 11.5 problems for the December 2012 Mission Math Tournament (this year also S/BMTTP/ST, thus altogether making an MSJHSS/BMTTP/ST(A)MT), 10 of which were Calculus Round problems. Since that round was mainly written by me, I’ll post solutions to the problems in that round here.

1. What is the maximum possible area of a rectangle of perimeter 16?

Solution without calculus: If one side of the rectangle has length $x$, then the other would have length $8-x$, since the perimeter of a rectangle is twice the sum of the side lengths. By the AM-GM inequality, $\sqrt{x(8-x)} \leq \frac{x+(8-x)}{2}$, so $x(8-x) \leq \boxed{16}$. Since the expression on the left hand side in the previous inequality is the expression for the area of a rectangle, we have shown the maximum possible area of such a rectangle to be 16, and all we need to do now is find a case where the area actually is 16. We easily see that $x=4$ gives this result.

Solution with calculus: If one side of the rectangle has length $x$, then the other would have length $8-x$, since the perimeter of a rectangle is twice the sum of the side lengths. The area of such a triangle is $x(8-x)$. To maximize this expression, we find that its derivative is $8-2x$. Setting this equal to zero to find the x value of the maximum, we get $x=4$. The area of a rectangle with $x=4$ is $4(8-4)=\boxed{16}$.

2. Find all y that satisfy the equation $\int_0^y (x^2-4x+4) \,dx=\int_y^4 (x^2-4x+4) \,dx$.

We visualize the parabolic graph of $x^2-4x+4=(x-2)^2$, and find that solving for y is equivalent to finding a point between 0 and 4 such that it divides the area under the parabola between $x=0$ and $x=4$ into two equal parts. Note that for y less than 0 or greater than 4, one integral will be nonnegative whereas the other will be nonpositive, with no instance of both being zero at the same time, so no values of y outside the $[0,4]$ range will work. Since a vertically opening parabola is reflectively symmetric over the vertical axis through its vertex, $y=\boxed{2}$ satisfies the equation, and since $x^2-4x+4 \geq 0$ for all $x$, any value to the left or right of $x=2$ will cause one area to be greater than the other.

3. Evaluate $\int_0^{3\pi/2} \sin{x}\cos{x} \,dx$.

This problem was written by J. Yuan.

Solution 1: $\int^{3\pi/2}_0 \sin{x}\cos{x}\,dx$ can be rewritten as $\frac{1}{2} \int^{3\pi/2}_0\sin{2x}\,dx$. Evaluating, $-\frac{1}{4}\cos{2x}|^{3\pi/2}_0=-\frac{1}{4}(-1-1)=\boxed{\frac{1}{2}}$.

Solution 2: Substitute $u = \sin{x}$: $\int^{3\pi/2}_0 \sin{x} \cos{x} \,dx = \int^{-1}_0 u \,du = \frac{u^2}{2} |^{-1}_{0} = \boxed{\frac{1}{2}}$.

4. A man is absentmindedly trying to draw a circle with a compass. The compass can be represented by an isosceles triangle with a vertex angle of $\theta$ and two legs of length 9 cm. The length of the base of the triangle is the radius r of the arc that is drawn. When the man starts drawing, $\theta = \pi / 4$ radians, and the compass turns at 1 radian per second (so that he would draw a full circle in $2 \pi$ seconds). Due to the man’s absentmindedness and the poor construction of the compass, $\theta$ steadily increases by 1/24 radians per second, and the man finally notices and stops drawing $2\pi$ seconds later. The length in cm of the partial spiral that he drew can be expressed in the form $a(\sqrt{b}+\sqrt{c})-d\sqrt{e+\sqrt{f}}$, where a, b, c, d, e, and f are positive integers, $b\geq c$, and b, c, and f are square-free. Find the ordered hextuple (a, b, c, d, e, f).

This problem is written by J. Wu.

Where v i velocity, $v=18\sin(\frac{\pi}{8}+\frac{t}{48})$. Since distance is the integral of velocity, we integrate to get our answer: $\int_{\pi/8}^{\pi/6} 18*48\sin u \,du=-864(\cos(\pi/6)-\cos(\pi/8))=-864(\frac{\sqrt{3}}{2}-\frac{\sqrt{2+\sqrt{2}}}{2})$. There are two possible answers: $\boxed{(216, 2, 6, 432, 2, 2)}$ and $\boxed{(216, 6, 2, 432, 2, 2)}$.

5. Find $\int_{-1}^{-3} \frac{1}{x^2+4x+5} \,dx$.

$\int_{-1}^{-3} \frac{1}{x^4+4x+5} \,dx=\int_{-1}^{-3} \frac{1}{(x+2)^2+1} \,dx=\arctan(x+2)\,|_{-1}^{-3}=-\frac{\pi}{4}-\frac{\pi}{4}=\boxed{-\frac{\pi}{2}}$.

6. Find $\int_1^e \frac{e^x (1+x\ln x)}{x} \,dx$.

$\int_1^e \frac{e^x (1+x\ln x)}{x} \,dx=\int_1^e (\frac{e^x}{x}+e^x\ln x) \,dx$. Notice that this integrand is in the form of the result of applying a product rule. The antiderivative of the integrand is thus $e^x \ln x$, which evaluates to 0 at $x=1$. Thus, the definite integral is simply the above antiderivative evaluated at $x=e$, or $\boxed{e^e}$.

7. Find the equation of the tangent line to the graph of the 19th derivative of $y=\prod\limits_{n=1}^{57} (x-n)^n$ at x=38.

Consider the product rule. The derivative of an expression containing the factor $(x-a)$ to a certain power $m$ will in each resulting term of the derivative via the product rule have at least $m-1$ as its exponent, or, applying this logic to multiple derivatives, $m-d$, where $d$ is the order of the derivative. At the 19th derivative, therefore, the $(x-38)$ factor will in each term that results from applying the product rule 19 times have an exponent of at least $38-19=19$. After taking the derivative yet one more time, the exponent will still be at least $18$. In both of these cases, the $(x-38)$ factor still exists in every term, and thus the entire expression will evaluate to 0 at $x=38$. Since both this function and the derivative of this function evaluate to 0 here, the tangent line is the x-axis, $\boxed{y=0}$.

8. Find a function f(x) of real numbers with no trigonometric function terms (i.e. sin, cos, sec, sinh, etc.) such that the sum of f’(x) and one of the antiderivatives of f(x) is $\frac{x^2\ln x+1}{x}-x$.

With x’s in the denominator and a natural log in the expression, we consider the function to have a natural log in it. In fact, the function is just $\boxed{\ln x}$, since its derivative is $\frac{1}{x}$ and its antiderivative is $x\ln x-x$. Note that there are other functions for which the derivative and an antiderivative sum to $\frac{x^2\ln x+1}{x}-x$, but the problem specifies that there are to be no trigonometric function terms.

9. Evaluate $\int_{-2012}^{2012} x^{2011} \sec x \,dx$.

$x^{2011} \sec x$ is an odd function, and thus has 180-degree rotational symmetry about the origin. The area under the curve above the x-axis on the positive-x half of the graph is thus the area above the curve under the x-axis in the negative-x half of the graph. Since these two areas are equal, the former counts positively in the integral, and the latter counts negatively in the integral, the integral evaluates to $\boxed{0}$.

10. Given that $17^{2013}-12^{2013}\approx7.8285\times10^{2476}$, find $\lfloor\log_{10} \sum\limits_{n=1}^{2012} \frac{5}{2012} (12+\frac{5n}{2012})^{2012}\rfloor$.

$\frac{5}{2012} (12+\frac{5n}{2012})^{2012}$ is the area of a rectangle of width $\frac{5}{2012}$ and height $(12+\frac{5n}{2012})^{2012}$. Considering the limits of summation, there is one of these rectangles for each $\frac{5}{2012}$ increment from 12 to 17. Thus, this series of rectangles is very well approximated by the integral of $x^{2012}$ from 12 to 17. $\int_{12}^{17} x^{2012}=\frac{x^{2013}}{2013}|_{12}^{17}=\frac{17^{2013}-12^{2013}}{2013}$. Since $2<7$, this quantity has three less digits (the power of 10 in the number divided by in the above expression, 2013) than $7.8285\times10^{2476}$. Its power of 10 in exponential notation is thus $2476-3=\boxed{2473}$.

11. Evaluate $\sum\limits_{n=0}^\infty \frac{1}{2n^2+7n+6}$.

$\sum\limits_{n=0}^\infty \frac{1}{2n^2+7n+6}=\sum\limits_{n=0}^\infty \frac{1}{2(n+2)^2-(n+2)}=\frac{1}{6}+\frac{1}{15}+\frac{1}{28}+\frac{1}{45}+...=\frac{2}{3\cdot4}+\frac{2}{5\cdot6}+\frac{2}{7\cdot8}+\frac{2}{9\cdot10}+...=2(\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+\frac{1}{7\cdot8}+\frac{1}{9\cdot10}+...)=2(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}...)=2(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}...)-2(1-\frac{1}{2})=\boxed{2\ln 2-1}$

12. Evaluate $\int_1^\infty \frac{\sqrt{x^2-1}}{x^3} \,dx$.

Assign $u=\frac{1}{x}$. Then, $x=\frac{1}{u}$ and $dx=-\frac{1}{u^2} \,du$. Thus, $\int_1^\infty \frac{\sqrt{x^2-1}}{x^3} \,dx$ is equivalent to $\int_1^0 u^3\sqrt{\frac{1}{u^2}-1} (-\frac{1}{u^2}) \,du=\int_0^1 u^3\sqrt{\frac{1-u^2}{u^2}} (\frac{1}{u^2}) \,du=\int_0^1 u^3\frac{\sqrt{1-u^2}}{u} (\frac{1}{u^2}) \,du=\int_0^1 \sqrt{1-u^2}\,du$. This is the area of a quarter-circle of radius 1, or $\boxed{\frac{\pi}{4}}$.