the motorcycle travels with constant speed $v_0$ along the path that, for a short distance, takes the form…

Explanation:

Step1: Recall the relationship between velocity components

The magnitude of the velocity is $v_0$. The velocity vector $\vec{v}$ has components $v_x$ and $v_y$. We know that $v_0^2=v_x^2 + v_y^2$. Also, the slope of the curve $y = c\sin(\frac{\pi}{L}x)$ gives the ratio $\frac{v_y}{v_x}$. First, find the derivative of the curve.

Step2: Differentiate the curve

Differentiate $y = c\sin(\frac{\pi}{L}x)$ with respect to $x$. Using the chain - rule, if $u=\frac{\pi}{L}x$, then $\frac{dy}{dx}=c\cos(\frac{\pi}{L}x)\cdot\frac{\pi}{L}=\frac{c\pi}{L}\cos(\frac{\pi}{L}x)$. And we know that $\frac{v_y}{v_x}=\frac{dy}{dx}$. So $v_y = v_x\cdot\frac{c\pi}{L}\cos(\frac{\pi}{L}x)$.

Step3: Substitute into the velocity - magnitude equation

Substitute $v_y$ into $v_0^2=v_x^2 + v_y^2$. We get $v_0^2=v_x^2+v_x^2\cdot(\frac{c\pi}{L})^2\cos^2(\frac{\pi}{L}x)=v_x^2\left[1 + (\frac{c\pi}{L})^2\cos^2(\frac{\pi}{L}x)\right]$.

Step4: Solve for $v_x$

$v_x=\frac{v_0}{\sqrt{1 + (\frac{c\pi}{L})^2\cos^2(\frac{\pi}{L}x)}}$.

Step5: Solve for $v_y$

Since $v_y = v_x\cdot\frac{c\pi}{L}\cos(\frac{\pi}{L}x)$, substituting $v_x$ we get $v_y=\frac{v_0\cdot\frac{c\pi}{L}\cos(\frac{\pi}{L}x)}{\sqrt{1 + (\frac{c\pi}{L})^2\cos^2(\frac{\pi}{L}x)}}$.

Answer:

$v_x=\frac{v_0}{\sqrt{1 + (\frac{c\pi}{L})^2\cos^2(\frac{\pi}{L}x)}}$, $v_y=\frac{v_0\cdot\frac{c\pi}{L}\cos(\frac{\pi}{L}x)}{\sqrt{1 + (\frac{c\pi}{L})^2\cos^2(\frac{\pi}{L}x)}}$