University:
University of Michigan
Course:
MATH 116 | Calculus II
Academic year:

2023
Views:

86

Pages:

1
Author:

A Powell

Math 116 / Exam 2 (November 18, 2009) 5. [12 points] Consider the graph of the curve r = f (θ) = 2 + 2 cos(θ). a. [5 points] What is the minimal value of x taken by the points on the graph of f (θ)? Show appropriate work to justify your answer. Solution: The x-coordinates corresponding to the point (θ, r(θ)) is x(θ) = r(θ) cos(θ) = 2 cos(θ) + 2 cos2 (θ). To minimize, we take derivatives and find that x′ (θ) = −2 sin(θ) − 4 cos(θ) sin(θ) = −2 sin(θ)(1 + 2 cos(θ)). This last expression is zero when sin(θ) = 0, or θ = nπ for all integers n, or cos(θ) = −1/2. By inspection of the graph, it is these zeroes that give the minimum value of x, corresponding to the angles θ = 2π/3 and 4π/3, with a minimum x-value of −1/2. b. [3 points] Consider the vertical line x = a, where a is the value you found in part (a). Sketch the region bounded by x = a and the graph of f (θ). Be sure to label your axes and graph appropriately. Solution: 3 2 r = f(1) y 1 0 Shaded Area 11 12 13 11 0 1 2 x 3 4 5 c. [4 points] Find inequalities for θ and r that describe the region bounded in part (b). Solution: The region is bounded between the angles θ = 2π/3 and θ = 4π/3, the graph of 2 + 2 cos(θ) and the line x = −1/2. We have to convert this last inequality into polar coordinates. Doing this, we have r cos(θ) = −1/2, or r = −1/(2 cos(θ)). Therefore the inequalities describing the region are 2π 4π ≤θ≤ , 3 3 2 + 2 cos(θ) ≤ r ≤ − 1 2 cos(θ)