The tangent to an ellipse from polar coordinates

In summary, the speaker discusses their exploration of finding the slope of a tangent to an ellipse using both Cartesian and polar coordinates. They realize they made a mistake in their calculation and correct it, finding that the slope is actually (ecc*sin(Theta)**2/(1+Ecc*cos(Theta)) + cos(Theta)) / (ecc*sin(Theta)*cos(Theta)/(1+Ecc*cos(Theta)) - sin(Theta)) and not dr/r*dTheta as they originally thought.
  • #1
Rapidrain
31
0
I have an ellipse. Quite simple, ecc=0.60. And I'm doodling with calculus I learned 40 years ago.

I can find the tangent to the ellipse, that is, the slope of the tangent, using cartestian coordinates. At the point where the tangent skims the top of the minor axis (b) the slope is 0 and and at the semi-latus rectum slope = ecc, that is the way it should be.

Now I take the ellipse in polar coordinates around the right hand focus :

r = a*(1 - ecc**2)/(1 + ecc*cos(Theta))

to find the slope I differentiate and have :

dr/dTheta = a*ecc*(1 - ecc**2)*(sin(Theta)) / ((1 + ecc*cos(Theta))**2)

then I divide by r to get the slope of the tangent.

I am doing something wrong here because my two methods only agree at the ends of the major axis and the semi-latus rectum. At r = a, where the point is above the minor axis it has a non-zero slope.

What am I doing wrong?
 
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  • #2
Got it!

dy/dx is the slope, not dr/r*dTheta,

slope is (ecc*sin(Theta)**2/(1+Ecc*cos(Theta)) + cos(Theta)) /
(ecc*sin(Theta)*cos(Theta)/(1+Ecc*cos(Theta)) - sin(Theta))
 

1. What is the equation for the tangent to an ellipse from polar coordinates?

The equation for the tangent to an ellipse from polar coordinates is given by r = c / cos(theta - alpha), where c is the distance from the origin to the center of the ellipse, theta is the angle of the tangent line, and alpha is the angle of rotation of the ellipse.

2. How is the tangent to an ellipse from polar coordinates different from a regular tangent line?

The tangent to an ellipse from polar coordinates takes into account the eccentricity of the ellipse, whereas a regular tangent line only considers the slope of the curve at a given point. This means that the equation for the tangent to an ellipse from polar coordinates is more complex and involves parameters such as the distance from the origin to the center of the ellipse and the angle of rotation of the ellipse.

3. Can the tangent to an ellipse from polar coordinates be negative?

Yes, the tangent to an ellipse from polar coordinates can be negative. This occurs when the tangent line intersects the ellipse at a point where the slope of the curve is negative, resulting in a negative value for r in the equation r = c / cos(theta - alpha).

4. What is the significance of the angle of rotation in the equation for the tangent to an ellipse from polar coordinates?

The angle of rotation, alpha, in the equation for the tangent to an ellipse from polar coordinates determines the orientation of the ellipse in relation to the origin. It is important to include this angle in the equation in order to accurately calculate the tangent line.

5. How is the tangent to an ellipse from polar coordinates used in real-world applications?

The tangent to an ellipse from polar coordinates is used in various fields of science and engineering, such as astronomy, physics, and mechanics. It can be used to calculate the velocity of an object moving along an elliptical path, or to determine the trajectory of a planet or satellite orbiting around a central body. It is also used in the design of mechanical systems, such as gears and pulleys, where the shape of an ellipse may be involved.

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