Polar Equation Question
The figure above shows the graph of the polar curve r=1−2cosθ for 0≤θ≤π and the unit circle r=1.
(a) Find the area of the shaded region in the figure.
Question 2
(b) Find the slope of the line tangent to the polar curve r=1−2cosθ at the point where x=−2. Show the computations that lead to your answer.
Question 3
(c) A particle moves along the polar curve r=1−2cosθ so that dθdt=2. Find the value of drdt at θ=2π3, and interpret your answer in terms of the motion of the particle.
Can’t post diagram because it is a url
Q1: Why did the polar curve bring a blanket to the party?
A1: Because it wanted to stay warm in the shaded region!
Q2: Why did the line tangent to the polar curve decide to become a math teacher?
A2: Because it knew all the slopes!
Q3: Why did the particle along the polar curve need an interpreter?
A3: Because it had trouble speaking in terms of motion!
To solve these polar equation questions, you need to follow a step-by-step approach. Here's how you can find the answers to each question:
Question (a): To find the area of the shaded region, you need to find the area enclosed by the curve r=1−2cosθ for 0≤θ≤π and the unit circle r=1.
1. First, find the points of intersection between the polar curve and the unit circle by setting the two equations equal to each other: 1 - 2cosθ = 1.
Solve for θ: 2cosθ = 0 ⇒ cosθ = 0 ⇒ θ = π/2 or 3π/2.
2. Set up the integral to find the area: A = ∫[0 to π/2] (1−2cosθ)^2/2 dθ - ∫[π/2 to 3π/2] (1−2cosθ)^2/2 dθ.
3. Simplify and evaluate the integrals, and then subtract the result of the second integral from the first to find the area.
Question (b): To find the slope of the line tangent to the polar curve r=1−2cosθ at the point where x=−2, you need to find the derivative of r with respect to θ and evaluate it at the given point.
1. Differentiate r with respect to θ using the chain rule: dr/dθ = d(1-2cosθ)/dθ.
2. Simplify the derivative expression to get dr/dθ = 2sinθ.
3. Find the value of sinθ at the point where x = -2, using the fact that x = rcosθ: -2 = (1-2cosθ)cosθ.
4. Solve the equation for cosθ and then find sinθ based on the value of cosθ.
5. Substitute the obtained value of sinθ into the derivative expression dr/dθ = 2sinθ to find the slope of the tangent line at the given point.
Question (c): To find dr/dt at θ = 2π/3, given that dθ/dt = 2, you need to use the chain rule and polar to rectangular conversion.
1. Use the chain rule and polar to rectangular conversion to get dr/dt = dr/dθ * dθ/dt.
2. Plug in the given values of dθ/dt = 2 and evaluate dr/dθ at θ = 2π/3 by differentiating the polar curve equation r = 1 - 2cosθ with respect to θ.
3. Once you have both values, multiply them together to find dr/dt at θ = 2π/3.
Interpretation: The value of dr/dt represents the rate of change of the radial distance r with respect to time. In this case, it tells you how fast the distance from the origin is changing as the particle moves along the polar curve.
(a) recall that A = ∫ 1/2 r^2 dθ
(b) dy/dx = (dy/dθ) / (dx/dθ)
= (r' sinθ + r cosθ)/(r'cosθ - r sinθ)
so find r and θ when x=2, and plug and chug.
(c) dr/dt = 2sinθ dθ/dt = 2(√3/2)(2)