Calculus I
posted by Mandy on .
Find the point on the curve y=x^2 closest to point (0,1)
Here's what I have:
Sqrt {(x0)^2 + ((x^2)1)}
Sqrt {((x^2) + (x^2)1}
d= sqrt (x^4)1
y= (x^4)1
Fprime(x) = 4x^3
Domain = ARN
Critical Number x= 0
Any help or comments would help alot. Thanks!

Mandy, you messed up in your first line
you forgot to square the last term in the brackets
dist = √((x0)^2 + (y1)^2)
=√(x^2 + (x^2  1)^2)
=√(x^4  x^2 + 1)
then d(dist)/dx = 1/2(x^4  x^2 + 1)^(1/2)(4x^3  2x)
if we set this to zero we get
4x^3  2x = 0
for x=±1/√2 sub back to get y = 1/2
so the point is (1/√2,1/2) in the first quadrant
there is a second point in the second quadrant which is the same distance away because of the symmetry of the parabola. 
there is an easier way to do these kind of questions.
It uses the fact that at the "closest" point the tangent must be perpendicular to the line from the given point to that tangent
So let the point on the curve be (a,a^2)
form y=x^2
dy/dx = 2x, so at (a,a^2) the slope of the tangent is 2a
the slope of the line from (0,1) to the tangent is then (a^2  1)/a
from the perpendicular line slope property
(a^2  1)/a = 1/(2a)
to get a = ±1/√2 as above
(notice that we also get a=0, which is correct according to the condition we imposed on the equation, namely that the lines had to be perpendicular.) 
You started out doing it correctly, but made an algebraic error.
d^2 = x^2 + (x^21)^2
d is a minimum where d^2 is a minimum, so you don't have to deal with the square root in finding the minimum distance.
d/dx(d^2) = 2x + 2(x^2 1)* 2x
= 2x +4x^3 4x = 4x^3 3x
Set that = 0 and solve for x
One solution is x=0, where y=0
Another solution is
Another solution is where
4x^2  3 = 1
x = sqrt (3/4) = +or 0.866
y = 0.75
Not all of these three points represent minima. Try them out (compute d^2) or use the second derivative test to find the true minimum. I get the minimum d^2 to be at
x = sqrt (3/4) = +or 0.866
where y = 0.75 
I seem to disagree with some of the others above, so check themn all for accuracy.