## University Calculus: Early Transcendentals (3rd Edition)

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 $$y=-\frac{x}{2\pi}+\frac{1}{2\pi}$$
$$y=2\sin(\pi x-y)$$ 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 $$y-0=-\frac{1}{2\pi}(x-1)$$ $$y=-\frac{x}{2\pi}+\frac{1}{2\pi}$$