What is the shortest distance from the point #(-3,3)# to the curve #y=(x-3)^3#?

Compute the first derivative of the curve:

#dy/dx = 3(x-3)^2 #

The slope of the tangent line at any given x coordinate at the point of tangency, #x_1#, is

#m = 3(x_1-3)^2#

The slope of the normal line is:

#n = -1/m = -1/(3(x_1-3)^2)#

The y coordinate, y_1, at the point of nomalcy is:

#y_1 = (x_1-3)^3#

Using the point-slope form of the equation of a line, the equation of the normal line is:

#y = -1/(3(x_1-3)^2)(x-x_1)+ (x_1-3)^3#

We want the normal line to contain the point #(-3,3)#:

#3 = -1/(3(x_1-3)^2)(-3-x_1)+ (x_1-3)^3#

I used WolframAlpha to solve this 5th order equation:

#x_1 ~~ 2.278#

The corresponding y coordinate is:

#y_1 = (2.278-3)^3#

#y_1 ~~ -0.376#

Use the distance formula:

#d = sqrt((x_1-x_0)^2+(y_1-y_0)^2#

#d = sqrt((2.278+3)^2+(-0.376-3)^2)#

#d ~~ 6.265#

Every point on the curve has coordinates #(x, (x-3)^3)# and the distance between such a point and the point #(-3,3)# is:

#sqrt((x+3)^2+((x-3)^3-3)^2)#.

We can minimize the distance by minimizing the radicand:

#f(x) = (x+3)^2+((x-3)^3-3)^2#.

Differentiate:

#f'(x) = 2(x+3) + 2((x-3)^3-3)(3(x-3)^2)#

Use some technology or approximation method to get

#x ~~ 2.278#. (The other 4 solutions are imaginary.)

There cannot be a maximum. There is a minimum at this #x#.

Using the distance above, we find a minimum distance of approximately #6.266#