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

2024
Views:

170

Pages:

3
Author:

nesikhiyawl0

Question 1 When in standard position, the terminal side of an angle intersects the unit circle at (5/13, 12/13). Describe sin (θ). Answer If (x,y) is a point on a unit circle, then x = 5/13 and y = 12/13 so: sin(θ) = 12/13 Question 2 Find the exact value of the trigonometric function at the given real number. sin(5π/4) Answer We would like to find the exact value of sin(5π/4), First, we can find the reference angle of 5π/4 where 5π/4 = π + π/4 so the reference angle is π/4. Now the next step is to define the sign of sin(5π/4). We know that 5π/4 is in quadrant 3 which the sine function is negative in this quadrant, so the value of sin(5π/4) is negative and we can simplify it as follows: sin(5π/4) = sin(π + π/4) = -sin(π/4) Note that the first step was to find the reference angle and then was to define the sign of sin(5π/4). Now we can find the value of sin(5π/4) by knowing the value of sin(π/4) which equals √2 / 2 ∴ sin(5π/4) = -sin(π/4) = -√2 / 2 Question 3 Find the solution of the differential equation that satisfies the given initial condition. dy/dx = x/y, y(0) = -3 Answer Step 1 dy/dx = x/y We can integrate it and rewrite it in differential notation: xdx = ydy ∫x dx = ∫y dy 1/2 * x² + c = 1/2 * y² y² = x² + C y = ±√(x² + C) initial value of y is negative and square root is always positive. y = -√(x² + C) Since y(0) = -3 y(0) = -√(0² + C) = -3 C = 9 y = -√(x² + 9) Result y = -√(x² + 9) Question 4 Find the particular solution that satisfies the differential equation and the initial condition. f''(x) = sinx, f'(0) = 1, f(0) = 6 Answer Step 1 We have: f''(x) = sinx, f'(0) = 1, f(0) = 6 We integrate f''(x) = sinx (first) and f'(0) (second) to obtain a generalized solution: ∫f''(x)dx = ∫sinx dx = -cosx + C₁ Step 2 Now, we are applying the initial condition f'(0) = 1: -cos0 + C₁ = 1 -1 + C₁ = 1 C₁ = 1 + 1 C₁ = 2 So: f'(x) = -cosx + 2 Now, we integrate f'(x): ∫f'(x)dx = ∫(-cosx + 2)dx = -∫cosx dx + ∫2dx = -sinx + 2x + C₂ Now, we are applying the initial condition f(0) = 6: -sin0 + 2*0 + C₂ = 6 0 + C₂ = 6 C₂ = 6 So: f(x) = -sinx + 2x + 6 Finally, the particular solution is: f(x) = -sinx + 2x + 6 Step 3 Explanation: (1): f''(x) = sinx (2): ∫sinxdx = -cosx + C (3): f'(x) = -cosx + 2 (4): ∫cosx dx = sinx + C Result f(x) = -sinx + 2x + 6 Question 5 What is the exact value of cos 135°? Answer 1 Explanation: cos 135° = cos(45° + 90°) = -cos 45° = -1/√2 or cos 135° = cos(180° - 45°) = -cos 45° = -1/√2 Answer 2 Cos 135° is an angle in the second quadrant. In the second quadrant, cos is negative. cosθ = π/4 cos 135 = cos(180 - 45) = -cos 45° An angle of 45° is found in a right-angled triangle of sides 1:1:√2 cos 45° = 1/√2 cos 135° = -cos 45° = -1/√2 Note that √2 is an irrational number and cannot be given as an exact decimal.