Answer
1) The point $(1,\pi/2)$ is on the curve.
2) The tangent line to the curve at $(1,\pi/2)$ is $$y=-\frac{\pi}{2}x+\pi$$
3) The normal line to the curve at $(1,\pi/2)$ is $$y=\frac{2}{\pi}x+\frac{\pi^2-4}{2\pi}$$
Work Step by Step
$$2xy+\pi\sin y=2\pi$$
1) Verify that the point $(1,\pi/2)$ is on the curve:
We substitute the coordinates of the point $(1,\pi/2)$ into the formula of the function to see if it equals $2\pi$ or not:
$$2\times1\times\frac{\pi}{2}+\pi\sin\frac{\pi}{2}=\pi+\pi=2\pi$$
Since it equals $2\pi$, the point is on the curve.
2) Find the tangent at $(1,\pi/2)$:
- Find the derivative of the function with implicit differentiation:
$$2(y+xy')+\pi(\cos y)y'=0$$ $$2y+2xy'+\pi y'\cos y=0$$ $$y'(2x+\pi\cos y)=-2y$$ $$y'=-\frac{2y}{2x+\pi\cos y}$$
- The slope of the tangent line at $(1,\pi/2)$ is
$$y'=-\frac{2\times\frac{\pi}{2}}{2\times1+\pi\cos\frac{\pi}{2}}=-\frac{\pi}{2+\pi\times0}=-\frac{\pi}{2}$$
- The tangent line to the given curve at $(1,\pi/2)$ is
$$y-\frac{\pi}{2}=-\frac{\pi}{2}(x-1)$$ $$y-\frac{\pi}{2}=-\frac{\pi}{2}x+\frac{\pi}{2}$$ $$y=-\frac{\pi}{2}x+\pi$$
3) Find the normal line at $(1,\pi/2)$:
The normal line at $(1,\pi/2)$ would be perpendicular with the tangent line at $(1,\pi/2)$, so the product of the slopes of these two lines equals $−1$.
So if we call the slope of the normal line at $(1,\pi/2)$ $k$, we would have
$$k\times\Big(-\frac{\pi}{2}\Big)=-1$$ $$k=\frac{2}{\pi}$$
- The normal line to the given curve at $(1,\pi/2)$ therefore is
$$y-\frac{\pi}{2}=\frac{2}{\pi}(x-1)$$ $$y-\frac{\pi}{2}=\frac{2}{\pi}x-\frac{2}{\pi}$$ $$y=\frac{2}{\pi}x+\frac{\pi}{2}-\frac{2}{\pi}=\frac{2}{\pi}x+\frac{\pi^2-4}{2\pi}$$
