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

Question

Isaiah Allen · Accepted Answer

Please see below.

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)#. By reducing the radicand, we can reduce the distance: #f(x) = (x+3)^2+((x-3)^3-3)^2#. Distinguish: #f'(x) = 2(x+3) + 2((x-3)^3-3)(3(x-3)^2)# Utilize technology or an approximation technique to obtain #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#

Cameron Alexander · Answer

#d ~~ 6.265#

Determine the curve's first derivative:

#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 normal line's slope is:

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

At the point of nomalcy, the y coordinate, y_1, is:

#y_1 = (x_1-3)^3#

The equation of the normal line can be found using the point-slope form of the equation of a line:

#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#

This fifth order equation was solved by me using WolframAlpha:

#x_1 ~~ 2.278#

The associated y-coordinate is:

#y_1 = (2.278-3)^3#

#y_1 ~~ -0.376#

Apply the formula for distance:

#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#

Charles Arocha · Answer

undefined To find the shortest distance from the point (-3,3) to the curve y = (x - 3)^3, we can use the distance formula. The distance between a point (x1, y1) and a point (x2, y2) is given by the formula:

Distance = √[(x2 - x1)^2 + (y2 - y1)^2]

Substituting (-3, 3) for (x1, y1) and (x, (x - 3)^3) for (x2, y2), we get:

Distance = √[(x - (-3))^2 + ((x - 3)^3 - 3)^2]

Now, to minimize the distance, we differentiate the distance formula with respect to x and set it equal to zero to find the critical points. After finding the critical points, we can determine which one yields the minimum distance.