University:
Montgomery College
Course:
MATH 181: Calculus I
Academic year:

2019
Views:

403

Pages:

2
Author:

maselarmwpa

Section 5.3 Worksheet Name: 1) Find the most general antiderivative of the following functions. a) f(x) = 12x^3 + 6x^2 - 5 ∫(12x^3 + 6x^2 - 5) dx = 12 * (x^4)/4 + 6 * (x^3)/3 - 5x + C = 3x^4 + 2x^3 - 5x + C b) f(x) = (2 - 1/√x)(2 + 1/√x) f(x) = 4 - 1/x ∫(4 - 1/x) dx = 4x - ln|x| + C c) f(t) = (t^3 + 2t^4)/√t f(t) = t^(5/2) + 2t^(7/2) ∫(t^(5/2) + 2t^(7/2)) dt = (2/7)t^(7/2) + (4/9)t^(9/2) + C d) f(x) = π^3 + 15x ∫(π^3 + 15x) dx = π^3x + (15/2)x^2 + C e) f(t) = e^t + sec(t)tan(t) + 2 ∫(e^t + sec(t)tan(t) + 2) dt = e^t + sec(t) + 2t + C f) f(t) = 1/(t^2 + 1) + 1/√(1 - t^2) F(t) = tan^(-1)t + sin^(-1)t + C The acceleration of an object dropped or thrown on Earth is -32 feet/sec^2 2) A ball is thrown directly upward at a speed of 40 feet per second from a cliff 100 feet above the ground. a) Find expressions for the velocity and height of the ball t seconds after it was released. V(t) = -32t + 40 ft/sec S(t) = -16t^2 + 40t + 100 ft b) At what time does the ball reach its highest point? When V(t) = 0 => t = 40/32 = 1.25 sec c) How high above the ground (from the base of the cliff) does the ball reach? S(1.25) = -16(1.25)^2 + 40(1.25) + 100 = 125 feet d) When does the ball strike the ground at the base of the cliff? S(t) = 0 => t = 4.045 sec e) What is its velocity at that instant (i.e. when the ball hits the ground)? V(4.045) = -32(4.045) + 40 = -89.44 ft/sec