# Thread: Angle between curves in parametric form

1. ## Angle between curves in parametric form

Hi,

In a problem i have to find the angle between two curves that are in parametric form. Those curves are:

(e^t*cos(t),e^t*sin(t)) and (R*cos(s),R*sin(s)) where s,t are in [0,2*Pi] and R>0

I can't even find their intersection point. I tried something but i had to exclude two values of s and t which are Pi/2 and 3*Pi/2 so i could obtain s=t

2. Hello, DBS!

I've made a little progress . . .

I have to find the angle between two parametric curves:

. . $\begin{Bmatrix}x &=& e^t\cos t \\ y &=& e^t\sin t \end{Bmatrix} \qquad \begin{Bmatrix}x &=& R\cos s \\ y &=& R\sin s\end{Bmatrix}\quad \text{ where }s,t \in [0,\,2\pi]\:\text{ and }\:R>0$

I can't even find their intersection point.

Equating $x$'s and $y$'s:

. . $\begin{array}{ccccccccc}
e^t\cos t &=& R\cos s & \Rightarrow & e^t &=& R\,\dfrac{\cos s}{\cos t} & [1] \\ \\[-3mm]
e^t\sin t &=& R\sin s & \Rightarrow & e^t &=& R\,\dfrac{\sin s}{\sin t} & [2]\end{array}$

Equate [1] and [2]: . $R\,\frac{\cos s}{\cos t} \:=\:R\,\frac{\sin s}{\sin t} \quad\Rightarrow\quad \sin s\cos t - \cos s\sin t \:=\:0$

We have: . $\sin(s-t) \:=\:0 \quad\Rightarrow\quad s-t \:=\:\begin{Bmatrix}0 \\ \pi \end{Bmatrix}$

. . Hence: . $s \;=\;\begin{Bmatrix}t \\ t+\pi \end{Bmatrix}$

It turns out that $s \:=\:t +\pi$ is an extraneous root.

Hence, the intersection occurs at $s \,=\,t$

Then [1] becomes: . $e^s \:=\:R\,\frac{\cos s}{\cos s} \quad\Rightarrow\quad e^s \:=\:R \quad\Rightarrow\quad s \:=\:\ln R$

Therefore, the curves intersect at: . $\bigg(R\cos(\ln R),\;R\sin(\ln R)\bigg)$

But check my work . . . please!
.

3. Thanks very much!

Actually what i find weird is that i got to s=t with a method but the problem is for that i got to tan(t)=tan(s) but i had to exclude the values that i told earlier...but your method seems correct enough without excluding values. So again, thank you.

4. Where can i learn to write like you did?...is it with latex by any chance? thx