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1. Recall: tan(x) = sin(x) cos(x) cot(x) = cos(x) sin(x) sec(x) = 1 cos(x) csc(x) = 1 sin(x) (a) Sketch the graph of each of these functions. (b) Using the di↵erentiation rules that you’ve learned, calculate the derivatives. (For this problem, simplifying your answer really helps.) Solution: (a) (b) 2. Find the equation to the tangent line of y = x + tan(x) at the point (⇡, ⇡). Solution: To find the slope of the tangent line, we must first take the derivative of y = x + tan(x) and evaluate it at x = ⇡ To that end: dy dy dy (x + tan(x)) = (x) + (tan(x)) dx dx dx 2 = 1 + sec (x) And so, at x = ⇡, the slope of the tangent line is 1 + sec2 (⇡) = 1 + ( 1)2 = 2. Given the slope of the tangent line is 2, and the point (⇡, ⇡) is on the line, we can calculate the equation for the tangent line: ⇡ = 2(⇡) + b ) b = ⇡ y = 2x ⇡ 2⇡ = ⇡ Thus, the equation for the tangent line is 2 3. A mass on a spring vibrates horizontally on a smooth level surface (see image). Its equation of motion is x(t) = 8sin(t), where t is in seconds and x is in centimeters. (a) Find the velocity and acceleration at time t. (b) Find the position, velocity, and acceleration of the mass at time t = direction is it moving in that time? 2⇡ . 3 In what Solution: (a) The velocity is the rate of change of distance/position with respect to time, and acceleration is the rate of change of velocity with respect to time. Thus, formally, we can write that: v(t) = x0 (t) a(t) = v 0 (t) = x00 (t) Thus, for x(t) = 8sin(t), we have that the velocity at time t is given by: v(t) = x0 (t) = 8cos(t) and the acceleration at time t is given by: a(t) = x00 (t) = v 0 (t) = 3 8sin(t) (b) At t = 2⇡ , 3 we have that the position is given by: ✓ 2⇡ x 3 ◆ ✓ 2⇡ = 8sin 3 ◆ ✓p ◆ p 3 =8 =4 3 2 The velocity is given by: ✓ 2⇡ v 3 ◆ =x 0 ✓ 2⇡ 3 ◆ ✓ 2⇡ = 8cos 3 ◆ =8 ✓ 1 2 ◆ = 4 The acceleration is given by: ✓ 2⇡ a 3 ◆ =x 00 ✓ 2⇡ 3 ◆ Notice that the velocity at t = negative direction. ✓ 2⇡ 8sin 3 = 2⇡ 3 ◆ = ✓p ◆ 3 8 = 2 p 4 3 is negative; hence, the mass is moving in the 4 4. Find constant A and B such that the function y = Asin(x) + Bcos(x) satisfies the di↵erential equation y 00 + y 0 2y = sin(x) Solution: y are. In order to solve this problem, we first need to figure out what y 0 and 00 For y = Asin(x) + Bcos(x) we have: y 0 = Acos(x) Bsin(x) And y 00 = Asin(x) Bcos(x) Thus, y 00 + y 0 2y = ( Asin(x) Bcos(x)) + (Acos(x) Bsin(x)) = ( 3A B)sin(x) + (A 3B)cos(x) 2(Asin(x) + Bcos(x)) In order for y = Asin(x) + Bcos(x) to satisfy the di↵erential equation y 00 + y 0 sin(x), we must have: 3A B = 1 A 3B = 0 Solving this linear system of equations, we get that A = 5 3 10 and B = 1 . 10 2y =