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March 29, 2015

March 29, 2015

Posted by **Mishaka** on Friday, December 16, 2011 at 9:51pm.

- Calculus -
**drwls**, Saturday, December 17, 2011 at 5:30amDistance to point, squared, is:

R^2 = (x-8)^2 + (y-1.5)^2

= (x-8)^2 + (x^2 -0.5)^2

Solve for the x value when d(R^2)/dx = 0

d/dx [x^2 -16x +64 + x^4 -x^2 + 1/4] = 0

2x -16 +3x^3 -2x = 0

3x^3 = 16

x = 1.747

y = 4.053

- Calculus - typo -
**Steve**, Saturday, December 17, 2011 at 6:35amNote that d/dx x^4 = 4x^3 not 3x^3

From there on, we have

2x - 16 + 4x^3 - 2x = 0

4x^3 = 16

x = cbrt(4)

y = cbrt(16)+1

Or, looking at things in another way, the normal line from point (p,q) on the curve will have the shortest distance to (8,1.5)

At any point (p,q) on the curve, the slope is 2p, so the normal line has slope -1/2p

Now we have a point and a slope:

(y-q)/(x-p) = -1/2p

y - p^2 - 1 = (p-x)/2p

1.5 = (p-8)/2p + p^2 + 1

3p = p - 8 + 2p^3 + 2p

2p^3 = 8

p^3 = 4

p = cbrt(4)

q = cbrt(16)+1

- Thanks -
**drwls**, Saturday, December 17, 2011 at 8:35amThanks Steve for noticing my error

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