## HMMT 1998: Calculus Solutions 1. Problem: Farmer Tim is lost in the densely-forested Cartesian plane. Starting from the origin he walks a sinusoidal path in search of home; that is, after $t$ minutes he is at position $(t, \sin t)$. Five minutes after he sets out, Alex enters the forest at the origin and sets out in search of Tim. He walks in such a way that after he has been in the forest for $m$ minutes, his position is $(m, \cos t)$. What is the greatest distance between Alex and Farmer Tim while they are walking in these paths? Solution: At arbitrary time $t$, Farmer Tim is at position $(t, \sin t)$ and Alex is at position $(t-5, \cos t)$. Hence at time $t$, the distance, $d$, between Tim and Alex is $d=\sqrt{(\sin t-\cos t)^{2}+25}$. To find the maximum value of $d$, we solve for $t$ such that $\frac{d d}{d t}=0$. $\frac{d d}{d t}=\frac{(\sin t-\cos t)(\cos t+\sin t)}{\sqrt{(\sin t-\cos t)^{2}+25}}$. Then $\frac{d d}{d t}=0 \Rightarrow \sin ^{2} t-\cos ^{2} t=0 \Rightarrow \sin ^{2} t=\cos ^{2} t$. Equality happens if $t$ is any constant multiple of $\frac{\pi}{4}$. Notice that to maximize $d$, we need to maximize $(\sin t-\cos t)^{2}$. This is achieved when $\cos t=-\sin t$. Because we determined earlier that $t$ is a constant multiple of $\frac{\pi}{4}$, then with this new condition, we see that $t$ must be a constant multiple of $\frac{3 \pi}{4}$. Then $(\sin t-\cos t)^{2}=2 \Rightarrow d=\sqrt{29}$. 2. Problem: A cube with sides 1 m in length is filled with water, and has a tiny hole through which the water drains into a cylinder of radius 1 m . If the water level in the cube is falling at a rate of $1 \mathrm{~cm} / \mathrm{s}$, at what rate is the water level in the cylinder rising? Solution: The magnitude of the change in volume per unit time of the two solids is the same. The change in volume per unit time of the cube is $1 \mathrm{~cm} \cdot \mathrm{~m}^{2} / \mathrm{s}$. The change in volume per unit time of the cylinder is $\pi \cdot \frac{d h}{d t} \cdot m^{2}$, where $\frac{d h}{d t}$ is the rate at which the water level in the cylinder is rising. Solving the equation $\pi \cdot \frac{d h}{d t} \cdot m^{2}=1 \mathrm{~cm} \cdot m^{2} / \mathrm{s}$ yields $\frac{1}{\pi} \mathrm{~cm} / \mathrm{s}$. 3. Problem: Find the area of the region bounded by the graphs of $y=x^{2}, y=x$, and $x=2$. Solution: There are two regions to consider. First, there is the region bounded by $y=x^{2}$ and $y=x$, in the interval $[0,1]$. In this interval, the values of $y=x$ are greater than the values of $y=x^{2}$, thus the area is calculated by $\int_{0}^{1}\left(x-x^{2}\right) d x$. Second, there is the region bounded by $y=x^{2}$ and $y=x$ and $x=2$, in the interval [1,2]. In this interval, the values of $y=x^{2}$ are greater than the values of $y=x$, thus the area is calculated by $\int_{1}^{2}\left(x^{2}-x\right) d x$. Then the total area of the region bounded by the three graphs is $\int_{0}^{1}\left(x-x^{2}\right) d x+\int_{1}^{2}\left(x^{2}-x\right) d x=1$. 4. Problem: Let $f(x)=1+\frac{x}{2}+\frac{x^{2}}{4}+\frac{x^{3}}{8}+\ldots$, for $-1 \leq x \leq 1$. Find $\sqrt{e^{\int_{0}^{1} f(x) d x}}$. Solution: Observe that $f(x)$ is merely an infinite geometric series. Thus $f(x)=\frac{1}{1-\frac{x}{2}}=\frac{2}{2-x}$. Then $\int_{0}^{1} \frac{2}{2-x}=2 \ln 2$. Then $\sqrt{e^{2 \ln 2}}=\sqrt{2^{2}}=2$. 5. Problem: Evaluate $\lim _{x \rightarrow 1} x^{\frac{x}{\sin (1-x)}}$. ![](https://cdn.mathpix.com/cropped/2025_01_24_0575deb0934c1721855fg-1.jpg?height=90&width=1529&top_left_y=2285&top_left_x=292) $\lim _{x \rightarrow 1} \ln x^{\frac{x}{\sin (1-x)}}=\lim _{x \rightarrow 1}\left(\frac{x}{\sin (1-x)} \ln x\right)$. Because direct calculation of the limit results in indeterminate form $\left(\frac{1}{0} \cdot 0\right)$, we can use L'Hopital's rule to evaluate the limit. By L'Hopital's rule, $\lim _{x \rightarrow 1}\left(\frac{x}{\sin (1-x)} \ln x\right)=$ $\lim _{x \rightarrow 1} \frac{\ln x+1}{-\cos (1-x)}$. This limit is simply -1. Hence $\lim _{x \rightarrow 1} e^{\ln x^{\frac{x}{\sin (1-x)}}}=e^{-1}=\frac{1}{e}$. 6. Problem: Edward, the author of this test, had to escape from prison to work in the grading room today. He stopped to rest at a place 1,875 feet from the prison and was spotted by a guard with a crossbow. The guard fired an arrow with an initial velocity of $100 \mathrm{ft} / \mathrm{s}$. At the same time, Edward started running away with an acceleration of $1 \mathrm{ft} / \mathrm{s}^{2}$. Assuming that air resistance causes the arrow to decelerate at $1 \mathrm{ft} / \mathrm{s}^{2}$ and that it does hit Edward, how fast was the arrow moving at the moment of impact (in $\mathrm{ft} / \mathrm{s}$ )? Solution: We use the formula for distance, $d=\frac{1}{2} a t^{2}+v t+d_{0}$. Then after $t$ seconds, Edward is at location $1875+\frac{1}{2}(1)\left(t^{2}\right)$ from the prison. After $t$ seconds, the arrow is at location $\frac{1}{2}(-1)\left(t^{2}\right)+100 t$ from the prison. When the arrow hits Edward, both objects are at the same distance away from the tower. Hence $1875+\frac{1}{2}(1)\left(t^{2}\right)=\frac{1}{2}(-1)\left(t^{2}\right)+100 t$. Solving for $t$ yields $t^{2}-100 t+1875=0 \Rightarrow t=25$ or $t=75$. Then it must be $t=25$, because after the arrow hits Edward, he will stop running. After 25 seconds, the arrow is moving at a velocity of $100-25(1)=75 \mathrm{ft} / \mathrm{s}$. 7. Problem: A parabola is inscribed in equilateral triangle $A B C$ of side length 1 in the sense that $A C$ and $B C$ are tangent to the parabola at $A$ and $B$, respectively. Find the area between $A B$ and the parabola. Solution: Suppose $A=(0,0), B=(1,0)$, and $C=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. Then the parabola in question goes through $(0,0)$ and $(1,0)$ and has tangents with slopes of $\sqrt{3}$ and $-\sqrt{3}$, respectively, at these points. Suppose the parabola has equation $y=a x^{2}+b x+c$. Then $\frac{d y}{d x}=2 a x+b$. At point $(0,0), \frac{d y}{d x}=b$. Also the slope at $(0,0)$, as we determined earlier, is $\sqrt{3}$. Hence $b=\sqrt{3}$. Similarly, at point $(1,0), \frac{d y}{d x}=2 a+b$. The slope at $(1,0)$, as we determined earlier, is $-\sqrt{3}$. Then $a=-\sqrt{3}$. Since the parabola goes through $(0,0), c=0$. Hence the equation of the parabola is $y=-\sqrt{3} x^{2}+\sqrt{3} x$. The desired area is simply the area under the parabolic curve in the interval $[0,1]$. Hence $\int_{0}^{1}\left(-\sqrt{3} x^{2}+\sqrt{3} x\right) d x=\frac{\sqrt{3}}{6}$. 8. Problem: Find the slopes of all lines passing through the origin and tangent to the curve $y^{2}=x^{3}+39 x-35$. Solution: Any line passing throug the origin has equation $y=m x$, where $m$ is the slope of the line. If a line is tangent to the given curve, then at the point of tangency, $(x, y), \frac{d y}{d x}=m$. First, we calculate $\frac{d y}{d x}$ of the curve: $2 y d y=3 x^{2} d x+39 d x \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}+39}{2 y}$. Substituting $m x$ for $y$, we get the following system of equations: $$ \begin{aligned} m^{2} x^{2} & =x^{3}+39 x-35 \\ m & =\frac{3 x^{2}+39}{2 m x} \end{aligned} $$ Solving for $x$ yields the equation $x^{3}-39 x+70=0 \Rightarrow(x-2)(x+7)(x-5)=0 \Rightarrow x=2$ or $x=-7$ or $x=5$. These solutions indicate the $x$-coordinate of the points at which the desired lines are tangent to the curve. Solving for the slopes of these lines, we get $m= \pm \frac{\sqrt{51}}{2}$ for $x=2$, no real solutions for $x=-7$, and $m= \pm \frac{\sqrt{285}}{5}$ for $x=5$. Thus $m= \pm \frac{\sqrt{51}}{2}, \pm \frac{\sqrt{285}}{5}$. 9. Problem: Evaluate $\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^{n-1}}$. Solution: Note that if we take the integral of $f(x)$ in problem 4, we get the function $F(x)=x+\frac{x^{2}}{2 \cdot 2}+$ $\frac{x^{3}}{3 \cdot 2^{2}}+\ldots$. Evaluating this integral in the interval $[0,1]$, we get $1+\frac{1}{2 \cdot 2}+\frac{1}{3 \cdot 2^{2}}+\ldots$, which is the desired sum. Hence $\int_{0}^{1} \frac{2}{2-x} d x=2 \ln 2$. 10. Problem: Let $S$ be the locus of all points $(x, y)$ in the first quadrant such that $\frac{x}{t}+\frac{y}{1-t}=1$ for some $t$ with $0