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

1) The point $(0,\pi)$ is on the curve. 2) The tangent line to the curve at $(0,\pi)$ is $y=\pi$. 3) The normal line to the curve at $(0,\pi)$ is $x=0$.
$$x^2\cos^2y-\sin y=0$$ 1) Verify that the point $(0,\pi)$ is on the curve: We substitute the coordinates of the point $(0,\pi)$ into the formula of the function to see if it equals $0$ or not: $$0^2\cos^2\pi-\sin\pi=0-0=0$$ Since it equals $0$, the point is on the curve. 2) Find the tangent at $(0,\pi)$: - Find the derivative of the function with implicit differentiation: $$2x\cos^2y+x^2(2\cos y)(\cos y)'-(\cos y)y'=0$$ $$2x\cos^2y+2x^2\cos y(-\sin y)y'-(\cos y)y'=0$$ $$2x\cos^2y-(2x^2\sin y\cos y)y'-(\cos y)y'=0$$ $$(2x^2\sin y\cos y)y'+(\cos y)y'=2x\cos^2y$$ $$y'(2x^2\sin y\cos y+\cos y)=2x\cos^2y$$ $$y'=\frac{2x\cos^2y}{2x^2\sin y\cos y+\cos y}$$ - The slope of the tangent line at $(0,\pi)$ is $$y'=\frac{2\times0\cos^2\pi}{2\times0^2\sin\pi\cos\pi+\cos\pi}=\frac{0}{0-1}=0$$ - The tangent line to the given curve at $(0,\pi)$ is $$y-\pi=0(x-0)$$ $$y-\pi=0$$ $$y=\pi$$ 3) Find the normal line at $(0,\pi)$: Since the tangent line to the curve at $(0,\pi)$ is the horizontal line $y=\pi$, the normal line at this point, being perpendicular with the tangent, will be a vertical line with the form $x=x_0$. And as the normal line is at $(0,\pi)$, we have $x_0=0$. The normal line in need to find, hence, is $$x=0$$