Find dydx and the slopes of the tangent lines shown on the graph of the polar equation. (If an answer does not exist, enter DNE.)r = 2(1 − sin(𝜃))The polar coordinate system is overlaid on the rectangular coordinate system. A curve and three lines are on the graph.Stated points for the curve are given as (x, y) in the rectangular coordinate plane, the cardioid has a dimple at the top of the graph at (0, 0), then proceeds clockwise through the points (2, 0), (0, −4), (−2, 0), and back to the starting point.The first line is tangent to the curve at the point with polar coordinates (2, 0) which is located on the positive horizontal axis, the line begins in the first quadrant, goes down and right, touches the curve and ends in the fourth quadrant.The second line is tangent to the curve at the point with polar coordinates (4, 3𝜋⁄2) which is located on the negative vertical axis, the line begins in the third quadrant, goes horizontally right, touches the curve and ends in the fourth quadrant.The third line is tangent to the curve at the point with polar coordinates (3, 7𝜋⁄6) which is located in the third quadrant, the line begins in the second quadrant, goes vertically down, touches the curve and ends in the third quadrant.dydx= at (2, 0) dydx= at 3, 7𝜋6dydx= at 4, 3𝜋2
Question
Find dydx and the slopes of the tangent lines shown on the graph of the polar equation. (If an answer does not exist, enter DNE.)r = 2(1 − sin(𝜃))The polar coordinate system is overlaid on the rectangular coordinate system. A curve and three lines are on the graph.Stated points for the curve are given as (x, y) in the rectangular coordinate plane, the cardioid has a dimple at the top of the graph at (0, 0), then proceeds clockwise through the points (2, 0), (0, −4), (−2, 0), and back to the starting point.The first line is tangent to the curve at the point with polar coordinates (2, 0) which is located on the positive horizontal axis, the line begins in the first quadrant, goes down and right, touches the curve and ends in the fourth quadrant.The second line is tangent to the curve at the point with polar coordinates (4, 3𝜋⁄2) which is located on the negative vertical axis, the line begins in the third quadrant, goes horizontally right, touches the curve and ends in the fourth quadrant.The third line is tangent to the curve at the point with polar coordinates (3, 7𝜋⁄6) which is located in the third quadrant, the line begins in the second quadrant, goes vertically down, touches the curve and ends in the third quadrant.dydx= at (2, 0) dydx= at 3, 7𝜋6dydx= at 4, 3𝜋2
Solution
To find the slope of the tangent line to a polar curve, we first need to find the derivative dy/dx. The derivative dy/dx in polar coordinates is given by the formula:
dy/dx = (dr/dθ * sin(θ) + r * cos(θ)) / (dr/dθ * cos(θ) - r * sin(θ))
Given the polar equation r = 2(1 - sin(θ)), we can find dr/dθ = -2cos(θ).
Now, we can substitute these values into the formula for dy/dx to find the slopes at the given points.
- At (2, 0):
dy/dx = ((-2cos(0) * sin(0) + 2(1 - sin(0)) * cos(0)) / (-2cos(0) * cos(0) - 2(1 - sin(0)) * sin(0)) = 0
So, the slope of the tangent line at (2, 0) is 0.
- At (3, 7π/6):
dy/dx = ((-2cos(7π/6) * sin(7π/6) + 3 * cos(7π/6)) / (-2cos(7π/6) * cos(7π/6) - 3 * sin(7π/6))
This simplifies to dy/dx = -√3.
So, the slope of the tangent line at (3, 7π/6) is -√3.
- At (4, 3π/2):
dy/dx = ((-2cos(3π/2) * sin(3π/2) + 4 * cos(3π/2)) / (-2cos(3π/2) * cos(3π/2) - 4 * sin(3π/2))
This simplifies to dy/dx = DNE (Does Not Exist).
So, the slope of the tangent line at (4, 3π/2) does not exist.
Similar Questions
r = 2(1 - sin(𝜃))(a) Find dy/dx (if it exists). (Use theta for 𝜃 as necessary.) If it does not exist, enter NONE.dy/dx = (b) Find the slope of the tangent line at (2, 0). If it does not exist, enter NONE.(c) Find the slope of the tangent line at (4, 3𝜋/2). If it does not exist, enter NONE.
Find the points of vertical tangency to the polar curve.
Find the points of horizontal tangency to the polar curve.r = a sin(𝜃) 0 ≤ 𝜃 < 𝜋, a > 0
Consider the polar curve defined below.r = 9a sin(𝜃)(a) Find the points of horizontal tangency to the polar curve. (Use pi for 𝜋 as necessary.)( , ) (smaller r value)( , ) (larger r value)(b) Find the points of vertical tangency to the polar curve.( , ) (smaller 𝜃 value)( , ) (larger 𝜃 value)
A graph in polar coordinates is of the form: r = -2 cosθ - 4 sinθ. Identify the graph of this function. Hint: Multiply both sides by r and then convert to Cartesian form. a. Circle b. Ellipse c. Hyperbola d. Line e. Parabola
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.