Find the curvature if $x= \cos t$ and $y=\ln 2t$ at $t=\pi$

Question

Find the curvature of the following curve: $$ \begin{cases} x= \cos t \\ y=\ln(2t) \end{cases}$$ at $t=\pi$.

Can anyone help me with how to solve the question? Because I am confused if I need to put the $\pi$ earlier or after I solve the $x$ and $y$ equation by using the curvature formula.

Answer 1

By definition, the curvature is given by

$$\kappa = \frac{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}{(\dot{x}^2 + \dot{y}^2)^{3/2}}$$

Where $\dot{x}$ means derivative wrt $t$ (and similarly for $y$).

hence we have

$$\dot{x} = -\sin(t) ~~~~~~~~~~~ \ddot{x} = -\cos(t)$$ $$\dot{y} = \frac{1}{t} ~~~~~~~~~~~ \ddot{y} = -\frac{1}{t^2}$$

$$\kappa = \frac{\Big|\dfrac{\sin(t)}{t} + \dfrac{\cos(t)}{t^2}\Big|}{\left( \sin^2(t) + \dfrac{1}{t^2}\right)^{3/2}}$$

You don't neet to arrange, just plug $t = \pi$ and you get

$$\kappa = \pi$$

Answer 2

Hint.

First calculate $x'(t),y'(t),x''(t),y''(t)$. Then plug in $t=\pi$ to get $x'(\pi),y'(\pi),x''(\pi),y''(\pi)$. Then apply the curvature formula for plane curves.