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=5^2# at the point #(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)#