Question

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

Answer 1

`d ~~ 6.265`

Explanation:

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`

Answer 2

Please see below.

Explanation:

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`