Question

how can i solve this problem?

Answer 1

would it be dy/dx=(dy/dθ)/(dx/dθ)?

Answer 2

You have to solve this quadratic for x
\[

L^2=a^2-2 a x \cos (\theta )+x^2\\
x^2 -2 a x \cos (\theta ) +(a^2-L^2)
\]

Answer 3

you get
\[
x=\frac{1}{2} \left(2 a \cos (\theta ) \pm\sqrt{4 a^2 \cos ^2(\theta )-4
\left(a^2-L^2\right)}\right)
\]

Answer 4

If we take
\[
x=\sqrt{a^2 \cos ^2(\theta
)-\left(a^2-L^2\right)}+a \cos (\theta )\\

\frac{\partial \left(\sqrt{a^2 \cos ^2(\theta
(t))-\left(a^2-L^2\right)}+a \cos (\theta
(t))\right)}{\partial t}=\\-\frac{a^2 \theta '(t)
\sin (\theta (t)) \cos (\theta (t))}{\sqrt{a^2
\cos ^2(\theta (t))-a^2+L^2}}-a \theta '(t) \sin
(\theta (t))
\]

Answer 5

i should use this formula then?

Answer 6

@skullofreak You should derive this formula yourself and use it as you need it with this as a guide. Don't just "use" formulas without knowing what you're doing.

Answer 7

He's started out by using the law of cosines, but you want to isolate x so that you can take its derivative with respect to theta.

Answer 8

Personally I'd prefer to just take the derivative implicitly first then solve for x and plug it in afterwards since I hate square roots. Just a preference though, good luck.