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

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Graph Polar
Click For Summary
To find the tangent lines for the polar graph r=2-3sin(θ), the curve is first parametrized into Cartesian coordinates. The derivative dx/dθ is calculated and set to zero to determine vertical tangents, resulting in angles approximately at 1.1043, 2.0373, 3.7358, and 5.689. For horizontal tangents, dy/dθ is also set to zero, yielding angles around 0.3398, 1.5708, 2.8016, and 4.7124. These calculations provide the necessary angles for both horizontal and vertical tangent lines.
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.
 
Mathematics 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
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 9 ·
Replies
9
Views
8K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K