Math Problem Statement

On the same grid below, use your calculator (in radian mode) to sketch r = 2cosθ + 1 and r = 2. Using algebraic solving, find the intersections of the two graphs and indicate them on your graph. There are four locations on r = 2cosθ + 1 where the tangent line to the graph is horizontal. Indicate those four locations on the graphs at left. Using your graphing calculator, calculate the slope of r = 2cosθ + 1 at θ = π/2. Draw the tangent line and give its equation.

Solution

The task in the image is focused on polar coordinates and involves a graph of the equation r=2cos(θ)+1r = 2 \cos(\theta) + 1. There are several parts to the problem:

(a) Sketch r=2cos(θ)+1r = 2 \cos(\theta) + 1 and r=2r = 2

You are asked to sketch the two polar equations.

  1. The equation r=2cos(θ)+1r = 2 \cos(\theta) + 1 is a Limaçon shape with a dimpled appearance.
  2. The second equation r=2r = 2 is simply a circle with a radius of 2.

(b) Find the intersection points algebraically

To find the points where the graphs intersect, set the equations equal to each other: 2cos(θ)+1=22 \cos(\theta) + 1 = 2 Solving for cos(θ)\cos(\theta): 2cos(θ)=1cos(θ)=122 \cos(\theta) = 1 \quad \Rightarrow \quad \cos(\theta) = \frac{1}{2} Thus, θ=±π3\theta = \pm \frac{\pi}{3}. These are the angles where the two graphs intersect, and you can calculate the rr-coordinates to confirm.

(c) Four locations where the tangent line is horizontal

You are asked to identify four points where the tangent line to the curve r=2cos(θ)+1r = 2 \cos(\theta) + 1 is horizontal. These points can be found by checking where the slope (calculated using polar derivatives) equals zero. The slope is given by: dydx=rsin(θ)+rcos(θ)rcos(θ)rsin(θ)\frac{dy}{dx} = \frac{r' \sin(\theta) + r \cos(\theta)}{r' \cos(\theta) - r \sin(\theta)} You need to solve this for zero and find the appropriate values of θ\theta.

(d) Find the slope of the tangent line at θ=π2\theta = \frac{\pi}{2} and give its equation

This part asks you to use your calculator to find the slope of the tangent line at θ=π2\theta = \frac{\pi}{2} for the graph of r=2cos(θ)+1r = 2 \cos(\theta) + 1. You'll use derivative calculations here to find the slope and then write the equation of the tangent line.

Let me know if you'd like help solving any specific parts of this problem!


Do you have any questions or need more details on a specific part of the problem? Here are five related questions to deepen your understanding:

  1. How do you find the derivative of a polar equation to calculate slope?
  2. What kind of shapes do the polar graphs of Limaçon equations typically take?
  3. Can you confirm the coordinates of the intersection points for r=2cos(θ)+1r = 2 \cos(\theta) + 1 and r=2r = 2?
  4. What is the geometrical significance of a horizontal tangent in polar graphs?
  5. How does the parameter cos(θ)\cos(\theta) affect the shape of the polar curve?

Tip: Always double-check your angle and radius values when solving polar intersection problems, as errors can easily occur when converting between rectangular and polar forms.

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Math Problem Analysis

Mathematical Concepts

Polar Coordinates
Graphing
Slopes of Tangents

Formulas

r = 2cosθ + 1
r = 2
Slope of tangent: (dr/dθ) * sinθ + r * cosθ

Theorems

Intersection points of polar graphs
Horizontal tangent condition

Suitable Grade Level

Grades 11-12