Question

could anybody please explain how to find pedal equation of
of : r=a(e^thetacot(α))

Answer 1

Do you know what pedal equation is?

Answer 2

yes

Answer 3

If you explain what it is it might help to figure the question out.

Answer 4

do u know about tangents and normals in polar coordinates

Answer 5

Yes.

Answer 6

it is a equation of curve in terams of r and p only where p =rsinΦ and Φ is angle b/w the radius vector of a point and the tangent at that point

Answer 7

"terams" ---- terms

Answer 8

The equation you wrote is ambiguous.

Answer 9

Is p not \(\rho\) from spherical coordinates?

Answer 10

Strange notation nonetheless.

Answer 11

i am posting a photograph of an example in the book

Answer 12

\[
r=a(e^\theta \cot(\alpha))
\]Is this where we are starting?

Answer 13

The fact that we have \(\alpha\) is a bit annoying.

Answer 14

Is the \(\cot\) supposed to be in the exponent?

Answer 15

yes cot is in exponent and α is constant

Answer 16

First thing you do is differentiate \(r\) with respect to \(\theta\)

Answer 17

\[
\frac{dr}{d\theta }=a\cot(\alpha)e^{\theta \cot(\alpha)}
\]

Answer 18

\[
\tan(\phi)=\frac{ae^{\theta \cot\alpha }}{a\cot\alpha e^{\theta \cot\alpha }}=\frac{1}{\cot\alpha }=\tan \alpha
\]

Answer 19

So I guess \(\phi =\alpha\)?

Answer 20

yes

Answer 21

tell further

Answer 22

\[
\ln(r)-\ln(a) = \theta \cot\alpha
\]

Answer 23

\[
\cot \alpha = \frac{\cos \alpha}{\sin\alpha } = \frac{\sqrt{1-\sin^2\alpha }}{\sin \alpha }= \frac{\sqrt{1-\sin^2\phi}}{\sin \phi}
\]

Answer 24

\[
r\cot \alpha =\frac{\sqrt{1-P^2}}{P}
\]

Answer 25

\[
r\ln r-r\ln a = \theta \frac{\sqrt{1-P^2}}{P}
\]

Answer 26

That theta doesn't seem to be going anywhere.

Answer 27

\(\color{blue}{\text{Originally Posted by}}\) @wio
\[
r\cot \alpha =\frac{\sqrt{1-P^2}}{P}
\]
\(\color{blue}{\text{End of Quote}}\)
This should be \[
r\cot \alpha =\frac{\sqrt{r^2-P^2}}{P}
\]