{"year": "1998", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "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 $(t, \\sin t)$.\n\nFive 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 $(m, \\cos t)$.\n\nWhat 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 $\\frac{d d}{d t}=0$.\n$\\frac{d d}{d t}=\\frac{(\\sin t-\\cos t)(\\cos t+\\sin t)}{\\sqrt{(\\sin t-\\cos t)^{2}+25}}$. 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 $\\frac{\\pi}{4}$.\n\nNotice 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 $\\frac{3 \\pi}{4}$.\n\nThen $(\\sin t-\\cos t)^{2}=2 \\Rightarrow d=\\sqrt{29}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-12-1998-feb-calc-solutions.jsonl", "problem_match": "\n1. Problem: ", "solution_match": "\nSolution: "}} {"year": "1998", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "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.\n\nSolving the equation $\\pi \\cdot \\frac{d h}{d t} \\cdot m^{2}=1 \\mathrm{~cm} \\cdot m^{2} / \\mathrm{s}$ yields $\\frac{1}{\\pi} \\mathrm{~cm} / \\mathrm{s}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-12-1998-feb-calc-solutions.jsonl", "problem_match": "\n2. Problem: ", "solution_match": "\nSolution: "}} {"year": "1998", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "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 $\\int_{0}^{1}\\left(x-x^{2}\\right) d x$.\n\nSecond, 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 $\\int_{0}^{1}\\left(x-x^{2}\\right) d x+\\int_{1}^{2}\\left(x^{2}-x\\right) d x=1$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-12-1998-feb-calc-solutions.jsonl", "problem_match": "\n3. Problem: ", "solution_match": "\nSolution: "}} {"year": "1998", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "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 $\\sqrt{e^{2 \\ln 2}}=\\sqrt{2^{2}}=2$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-12-1998-feb-calc-solutions.jsonl", "problem_match": "\n4. Problem: ", "solution_match": "\nSolution: "}} {"year": "1998", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "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.\n\nAfter 25 seconds, the arrow is moving at a velocity of $100-25(1)=75 \\mathrm{ft} / \\mathrm{s}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-12-1998-feb-calc-solutions.jsonl", "problem_match": "\n6. Problem: ", "solution_match": "\nSolution: "}} {"year": "1998", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "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 $\\frac{d y}{d x}=2 a x+b$.\n\nAt 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 $a=-\\sqrt{3}$.\n\nSince 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 $[0,1]$.\n\nHence $\\int_{0}^{1}\\left(-\\sqrt{3} x^{2}+\\sqrt{3} x\\right) d x=\\frac{\\sqrt{3}}{6}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-12-1998-feb-calc-solutions.jsonl", "problem_match": "\n7. Problem: ", "solution_match": "\nSolution: "}} {"year": "1998", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "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, $(x, y), \\frac{d y}{d x}=m$.\n\nFirst, 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:\n\n$$\n\\begin{aligned}\nm^{2} x^{2} & =x^{3}+39 x-35 \\\\\nm & =\\frac{3 x^{2}+39}{2 m x}\n\\end{aligned}\n$$\n\nSolving 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 $m= \\pm \\frac{\\sqrt{51}}{2}, \\pm \\frac{\\sqrt{285}}{5}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-12-1998-feb-calc-solutions.jsonl", "problem_match": "\n8. Problem: ", "solution_match": "\nSolution: "}} {"year": "1998", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "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.\n\nHence $\\int_{0}^{1} \\frac{2}{2-x} d x=2 \\ln 2$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-12-1998-feb-calc-solutions.jsonl", "problem_match": "\n9. Problem: ", "solution_match": "\nSolution: "}} {"year": "1998", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "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