Answer
1) The point $(1,0)$ is on the curve.
2) The tangent line to the curve at $(1,0)$ is $$y=2\pi x-2\pi$$
3) The normal line to the curve at $(1,0)$ is $$y=-\frac{x}{2\pi}+\frac{1}{2\pi}$$
Work Step by Step
$$y=2\sin(\pi x-y)$$
1) Verify that the point $(1,0)$ is on the curve:
We substitute the coordinates of the point $(1,0)$ into the formula of the function to see if 2 sides are equal or not:
- Left side: $$y=0$$
- Right side: $$2\sin(\pi\times1-0)=2\sin\pi=2\times0=0$$
Since 2 sides are equal to each other, the point is on the curve.
2) Find the tangent at $(1,0)$:
- Find the derivative of the function with implicit differentiation:
$$y'=2\cos(\pi x-y)(\pi x-y)'$$ $$y'=2\cos(\pi x-y)(\pi -y')$$ $$y'=2\pi\cos(\pi x-y)-2y'\cos(\pi x-y)$$ $$y'+2y'\cos(\pi x-y)=2\pi\cos(\pi x-y)$$ $$y'=\frac{2\pi\cos(\pi x-y)}{1+2\cos(\pi x-y)}$$
- The slope of the tangent line at $(1,0)$ is
$$y'=\frac{2\pi\cos(\pi\times1-0)}{1+2\cos(\pi\times1-0)}=\frac{2\pi\cos\pi}{1+2\cos\pi}=\frac{2\pi(-1)}{1+2(-1)}=\frac{-2\pi}{-1}=2\pi$$
- The tangent line to the given curve at $(1,0)$ is
$$y-0=2\pi(x-1)$$ $$y=2\pi x-2\pi$$
3) Find the normal line at $(1,0)$:
The normal line at $(1,0)$ would be perpendicular with the tangent line at $(1,0)$, so the product of the slopes of these two lines equals $−1$.
So if we call the slope of the normal line at $(1,0)$ $k$, we would have
$$k\times2\pi=-1$$ $$k=-\frac{1}{2\pi}$$
- The normal line to the given curve at $(1,0)$ therefore is
$$y-0=-\frac{1}{2\pi}(x-1)$$ $$y=-\frac{x}{2\pi}+\frac{1}{2\pi}$$
