Find Tangent Lines to Polar Graph: r=2-3sin(θ)

  • Context: MHB 
  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Graph Polar
Click For Summary
SUMMARY

The discussion focuses on finding the polar coordinates for horizontal and vertical tangent lines of the polar graph defined by the equation r=2-3sin(θ). The parametrization of the curve is established with x and y coordinates derived from the polar equation. The derivatives dx/dθ and dy/dθ are calculated, leading to the identification of tangent lines at specific angles. The final results yield vertical tangent angles of approximately 1.1043, 2.0373, 3.7358, and 5.689, and horizontal tangent angles of approximately 0.3398, 1.5708, 2.8016, and 4.7124.

PREREQUISITES
  • Understanding of polar coordinates and their graphical representation
  • Knowledge of differentiation in parametric equations
  • Familiarity with the quadratic formula and its application
  • Ability to interpret trigonometric functions and their inverses
NEXT STEPS
  • Study the properties of polar curves and their tangent lines
  • Learn about the application of the quadratic formula in trigonometric contexts
  • Explore advanced topics in calculus related to parametric equations
  • Investigate the graphical representation of polar functions using software tools like Desmos or GeoGebra
USEFUL FOR

Students and educators in mathematics, particularly those focusing on calculus and polar coordinates, as well as anyone interested in graphing and analyzing polar functions.

MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

How do I find the polar coordinates for the horizontal and vertical tangent lines?

For r=2-3sin(θ)

With decimal approximations to 4 places for the angles at which the tangent lines occur?

I keep messing up when I try to do this on my own, much appreciation to any who help

I have posted a link there to this topic so the OP can see my work.
 
Physics news on Phys.org
Hello Uncorrupted.Innocence,

The first thing I would do is look at a polar plot of the given function:

View attachment 1018

Next, let's parametrize the curve:

$$x=r\cos(\theta)=\left(2-3\sin(\theta) \right)\cos(\theta)$$

$$y=r\sin(\theta)=\left(2-3\sin(\theta) \right)\sin(\theta)$$

Hence:

$$\frac{dx}{d\theta}=\left(2-3\sin(\theta) \right)\left(-\sin(\theta) \right)+\left(-3\cos(\theta) \right)\cos(\theta)=6\sin^2(\theta)-2\sin(\theta)-3$$

Equating this to zero, and applying the quadratic formula, we find:

$$\theta=\sin^{-1}\left(\frac{1\pm\sqrt{19}}{6} \right)$$

Because $\sin(\pi-\theta)=\sin(\theta)$, we also have:

$$\theta=\pi-\sin^{-1}\left(\frac{1\pm\sqrt{19}}{6} \right)$$

We may add integral multiples of $2\pi$ as needed to place $0\le\theta<2\pi$.

So, rounded to four places, the angles at which the tangent lines are vertical are:

$$\theta\approx1.1043,\,2.0373,\,3.7358,\,5.689$$

$$\frac{dy}{d\theta}=\left(2-3\sin(\theta) \right)\left(\cos(\theta) \right)+\left(-3\cos(\theta) \right)\sin(\theta)=2\cos(\theta)\left(1-3\sin(\theta) \right)$$

Equating each factor in turn to zero, we find:

$$\cos(\theta)=0$$

$$\theta=\frac{\pi}{2},\,\frac{3\pi}{2}$$

$$1-3\sin(\theta)=0$$

$$\theta=\sin^{-1}\left(\frac{1}{3} \right),\,\pi-\sin^{-1}\left(\frac{1}{3} \right)$$

So, rounded to four places, the angles at which the tangent lines are horizontal are:

$$\theta\approx0.3398,\,1.5708,\,2.8016,\,4.7124$$
 

Attachments

  • uipolar.jpg
    uipolar.jpg
    5.3 KB · Views: 93

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 9 ·
Replies
9
Views
8K
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K