University:
North Central College
Course:
MATH 152 | Calculus II
Academic year:

2024
Views:

215

Pages:

3
Author:

nesikhiyawl0

Question 1 A population like that of the United States with an age structure of roughly equal numbers in each of the age groups can be predicted to A) grow rapidly over a 30-year-period and then stabilize B) grow little for a generation and then grow rapidly C) fall slowly and steadily over many decades D) show slow and steady growth for some time into the future Answer Because the age structure has roughly equal numbers, the population will not grow rapidly, or decrease rapidly, but instead grow slowly and steadily. D) show slow and steady growth for some time into the future Question 2 Can you indefinite the integral of x²arcsin (x) please? Answer indefinite integral of x². arcsin (x), using integration by parts. we use the fact that ∫ arcsin(x) = x. arcsin(x) + (1 − x²)^1/2 ∫ x². arcsin(x)dx = x²{x. arcsin(x) + (1 − x²)^1/2 } - ∫{2x. x. arcsin(x) + 2x. (1 − x²)^1/2} dx ∫ x². arcsin(x)dx = x²{x. arcsin(x) + (1 − x²)^1/2 } - 2∫{x². arcsin(x) + x.(1 − x²)^1/2 }dx 3∫ x². arcsin(x)dx = x²{x. arcsin(x) + (1 − x²)^1/2 } - ∫2x.(1 − x²)^1/2 dx 3∫ x². arcsin(x)dx = x³arcsin(x) + x²(1-x²)^1/2 + 2/3 (1-x²)^3/2 3∫ x². arcsin(x)dx = x³arcsin(x) + x²(1-x²)^1/2 - (2/3) (1-x²)^3/2 - (2/15)(1-x²)^5/2 + c Question 3 Solve the initial value problem using Laplace transforms. y'' + y = f(t), y(0) = 0, y'(0) = 1 Here f(t) = {0 0 ≤ t <3π {1 t > 3π Answer Given differential equation y'' + y = f(t), y(0) = 0, y'(0) = 1 Here f(t) = {0 0 ≤ t <3π {1 t ≥ 3π To find: Solve the initial value problem using Laplace Transform We begin with taking Laplace transformation on both sides, L(y'') = L(f(t)) L(y'') + L(y) = L(f(t)) Note: L(y'') = s²L(y) – sy(0) – y'(0) Substitute in the equation, (s²L(y) - sy(0) - y'(0)) + L(y) = L(f(t)) We begin with finding the Laplace for RHS, L(f(t)) = ∫(0 to 3π) e^(-st) (0)dt + ∫(3π to ∞) e^(-st) (1)dt = ∫(3π to ∞) e^(-st) (1)dt = -e^(-st)/s | from 3π to ∞ = -e^(-∞)/s - (-e^(-3πs)/s) = e^(-3πs)/s Now, use the conditions given, s²L(y) - 1 + L(y) = L(f(t)) (s² + 1)L(y) - 1 = e^(-3πs)/s (s² + 1)L(y) = e^(-3πs)/s + 1 L(y) = e^(-3πs)/(s(s²+1)) + 1/(s²+1) Now, find inverse Laplace transform y(t) = L⁻¹{e^(-3πs)/(s(s²+1))} + L⁻¹{1/(s²+1)} Note: L⁻¹{F(s)} = f(t) L⁻¹{e^(-as)F(s)} = (t - a)u(t-a) Question 4 Using the derivative theorem and, find the Laplace transform of the following functions t cosh(t) Answer Laplase transform of t cosh(t) L(tcosh(t)) derivative of Laplase transform of cosh(t) L(cosh(t)) = s/(s² - 1) L(tcosh(t)) = -d/ds(s/(s²-1)) L(tcosh(t)) = -[(s²-1)(1)-s(2s)]/(s²-1)² L(tcosh(t)) = -(s²-1 - 2s²)/(s² - 1)² = (-s²-1)/(s²-1)² = -(s²+1)/(s²-1)² Question 5 Laplace transform u(t) is the unit step function Find the Laplace transform of the following continuous-time signal. x(t) = 4te^(-5t)u(t) Answer Step 1 The given signal is x(t) = 4te^(-5t)u(t) Step 2 The above signal can be compared to x(t) = f(t - c)u(t - c) with c=0 and f(t - c) = 4te^(-5t) but, the laplace transform L {f(t - c)u(t - c)} = e^(-as)F(s) where F(s) is the laplace transform of f(t). Here, f(t - 0) = f(t) = 4te^(-5t) Hence, L{x(t)} = L {f(t)u(t)} L{x(t)} = e^0L {f(t)} L{x(t)} = L {4te^(-5t)} L{x(t)} = 4L {te^(-5t)} L{x(t)} = 4*1/((s+5)²) [using: L{te^(at)} = 1/(s-a)²] Hence, the laplace transform of the given signal is X(s) = 4/(s+5)²