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

1) The point $(3,-4)$ is on the curve. 2) The tangent line to the curve at $(3,-4)$ is $$y=\frac{3}{4}x-\frac{25}{4}$$ 3) The tangent line to the curve at $(3,-4)$ is $$y=-\frac{4}{3}x$$
$$x^2+y^2=25$$ 1) Verify that the point $(3,-4)$ is on the curve We substitute the coordinates of the point $(3,-4)$ into the formula of the function to see if it equals $25$ or not: $$3^2+(-4)^2=9+16=25$$ Since it equals $25$, the point is on the curve. 2) Find the tangent at $(3,-4)$: - Find the derivative of the function with implicit differentiation: $$2x+2yy'=0$$ $$x+yy'=0$$ $$yy'=-x$$ $$y'=-\frac{x}{y}$$ - The slope of the tangent line at $(3,-4)$ is $$y'=-\frac{3}{-4}=\frac{3}{4}$$ - The tangent line to the given curve at $(3,-4)$ is $$y+4=\frac{3}{4}(x-3)$$ $$y+4=\frac{3}{4}x-\frac{9}{4}$$ $$y=\frac{3}{4}x-\frac{25}{4}$$ 3) Find the normal line at $(3,-4)$: The normal line at $(3,-4)$ would be perpendicular with the tangent line at $(3,-4)$, so the product of the slopes of these two lines equal $-1$. So if we call the slope of the normal line at $(3,-4)$ $k$, we would have $$k\times\frac{3}{4}=-1$$ $$k=-\frac{4}{3}$$ - The normal line to the given curve at $(3,-4)$ therefore is $$y+4=-\frac{4}{3}(x-3)$$ $$y+4=-\frac{4}{3}x+4$$ $$y=-\frac{4}{3}x$$