How do you find the Tangent line to a curve by implicit differentiation?

1 Answer
Sep 20, 2014

Let us this example:

Find the equation of the tangent line to the circle x^2+y^2=5^2x2+y2=52 at the point (3,4)(3,4). In order to identify a line, we need two pieces of information:
{("Point: " (x_1,y_1)=(3,4)), ("Slope: " m=?):}

Since the point is already provided, all you need is the slope m.
Let us find m by implicit differentiation.

By implicitly differentiating,

d/{dx}(x^2+y^2)=d/{dx}(5^2)Rightarrow 2x+2y{dy}/{dx}=0

by dividing by 2y,

{x}/{y}+{dy}/{dx}=0

by subtracting x/y,

{dy}/{dx}=-x/y

So, we can find m by evaluating {dy}/{dx} at (3,4).

m={dy}/{dx}|_{(3,4)}=-3/4

By Point-Slope Form: y-y_1=m(x-x_1),

Tangent Line: y-4=-3/4(x-3)