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

1) The point $(-1,3)$ is on the curve. 2) The tangent line to the curve at $(-1,3)$ is $$y=3x+6$$ 3) The normal line to the curve at $(-1,3)$ is $$y=-\frac{1}{3}x+\frac{8}{3}$$
$$x^2y^2=9$$ 1) Verify that the point $(-1,3)$ is on the curve We substitute the coordinates of the point $(-1,3)$ into the formula of the function to see if it equals $9$ or not: $$(-1)^2\times3^2=1\times9=9$$ Since it equals $9$, the point is on the curve. 2) Find the tangent at $(-1,3)$: - Find the derivative of the function with implicit differentiation: $$(2x)y^2+x^2(2yy')=0$$ $$2xy^2+2x^2yy'=0$$ $$xy^2+x^2yy'=0$$ $$x^2yy'=-xy^2$$ $$y'=-\frac{xy^2}{x^2y}=-\frac{y}{x}$$ - The slope of the tangent line at $(-1,3)$ is $$y'=-\frac{3}{-1}=3$$ - The tangent line to the given curve at $(-1,3)$ is $$y-3=3(x+1)$$ $$y-3=3x+3$$ $$y=3x+6$$ 3) Find the normal line at $(-1,3)$: The normal line at $(-1,3)$ would be perpendicular with the tangent line at $(-1,3)$, so the product of the slopes of these two lines equals $-1$. So if we call the slope of the normal line at $(-1,3)$ $k$, we would have $$k\times3=-1$$ $$k=-\frac{1}{3}$$ - The normal line to the given curve at $(-1,3)$ therefore is $$y-3=-\frac{1}{3}(x+1)$$ $$y-3=-\frac{1}{3}x-\frac{1}{3}$$ $$y=-\frac{1}{3}x+\frac{8}{3}$$