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

Find the point on the graph of y = x^2 + 1 that’s closest to the point 8, 1.5. Hint: Remember the distance formula.

- Calculus -
**drwls**, Saturday, December 17, 2011 at 5:30am
Distance 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 6:35am
Note 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 8:35am
Thanks Steve for noticing my error

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