Calculus
posted by Mishaka .
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.

Distance to point, squared, is:
R^2 = (x8)^2 + (y1.5)^2
= (x8)^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 
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:
(yq)/(xp) = 1/2p
y  p^2  1 = (px)/2p
1.5 = (p8)/2p + p^2 + 1
3p = p  8 + 2p^3 + 2p
2p^3 = 8
p^3 = 4
p = cbrt(4)
q = cbrt(16)+1 
Thanks Steve for noticing my error
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