HMMT 二月 1998 · CALC 赛 · 第 8 题
HMMT February 1998 — CALC Round — Problem 8
题目详情
1998 Harvard/MIT Math Tournament CALCULUS Answer Sheet Name: School: Grade: 1 6 2 7 3 8 4 9 5 10 TOTAL:
CALCULUS Question One . [3 points] 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? Question Two . [3 points] A cube with sides 1m in length is filled with water, and has a tiny hole through which the water drains into a cylinder of radius 1m. cm If the water level in the cube is falling at a rate of 1 , at what s rate is the water level in the cylinder rising? Question Three . [4 points] 2 Find the area of the region bounded by the graphs of y = x , y = x , and x = 2 . Question Four . [4 points] 1 f ( x ) dx 2 3 ∫ x x x 0 Let f ( x ) = 1 + + + +… , for − 1 ≤ x ≤ 1 . Find e . 2 4 8 Question Five . [5 points] x sin(1 − x ) Evaluate lim x . x → 1
Question Six . [5 points] 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. ft The guard fired an arrow with an initial velocity of 100 . At the s same time, Edward started running away with an acceleration of ft 1 . Assuming that air resistance causes the arrow to decelerate at 2 s ft 1 and that it does hit Edward, how fast was the arrow moving at 2 s ft the moment of impact (in )? s Question Seven . [5 points] A parabola is inscribed in equilateral triangle ABC of side length 1 in the sense that AC and BC are tangent to the parabola at A and B , respectively: Find the area between AB and the parabola. Question Eight . [6 points] Find the slopes of all lines passing through the origin and tangent to 2 3 the curve y = x + 39 x − 35. Question Nine . [7 points] ∞ 1 Evaluate ∑ n − 1 n ⋅ 2 n = 1 Question Ten . [8 points] Let S be the locus of all points ( x , y ) in the first quadrant such that x y
- = 1 for some t with 0< t <1. Find the area of S . t 1 − t
解析
- Problem: Find the slopes of all lines passing through the origin and tangent to the curve 2 3 y = x + 39 x − 35. Solution: Any line passing throug the origin has equation y = mx , where m is the slope of the line. If a dy line is tangent to the given curve, then at the point of tangency, ( x, y ), = m . dx 2 dy dy 2 3 x +39 First, we calculate of the curve: 2 ydy = 3 x dx + 39 dx ⇒ = . Substituting mx for y , we get dx dx 2 y the following system of equations: 2 2 3 m x = x + 39 x − 35 2 3 x + 39 m = 2 mx 3 Solving for x yields the equation x − 39 x + 70 = 0 ⇒ ( x − 2)( x + 7)( x − 5) = 0 ⇒ 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 √ 51 curve. Solving for the slopes of these lines, we get m = ± for x = 2, no real solutions for x = − 7, and 2 √ √ √ 285 51 285 m = ± for x = 5. Thus m = ± , ± . 5 2 5 ∞ ∑ 1