Calculus I

Find the point on the curve y=x^2 closest to point (0,1)

Here's what I have:

Sqrt {(x-0)^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!

1. 👍
2. 👎
3. 👁
1. Mandy, you messed up in your first line
you forgot to square the last term in the brackets

dist = √((x-0)^2 + (y-1)^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.

1. 👍
2. 👎
2. You started out doing it correctly, but made an algebraic error.

d^2 = x^2 + (x^2-1)^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

1. 👍
2. 👎
3. I seem to disagree with some of the others above, so check themn all for accuracy.

1. 👍
2. 👎
4. 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.)

1. 👍
2. 👎

Similar Questions

1. Math (repost, URGENT)

A curve is traced by a point P(x,y) which moves such that its distance from the point A(-1,1) is three times the distance from the point B(2,-1). Find the equation of the curve and identity.

2. AP Calculus

Consider the curve given by x^2+4y^2=7+3xy a) Show that dy/dx=(3y-2x)/(8y-3x) b) Show that there is a point P with x-coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P. c) Find the

3. calculus

The slope of a curve is at the point (x,y) is 4x-3. Find the curve if it is required to pass through the point (1,1). Work... 4(1)-3=1 y-1=1(x-1) y=x

4. Calculus

The line that is normal to the curve x^2=2xy-3y^2=0 at(1,1) intersects the curve at what other point? Please help. Thanks in advance. We have x2=2xy - 3y2 = 0 Are there supposed to be 2 equal signs in this expression or is it x2 +

1. Calculus 1

The curve y = |x|/(sqrt(5- x^2)) is called a bullet-nose curve. Find an equation of the tangent line to this curve at the point (2, 2)

2. Calculus

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.

3. math

Find the point P on the graph of the function y=sqrt{x} closest to the point (10,0)

4. calculus

Consider the curve given by the equation y^3+3x^2y+13=0 a.find dy/dx b. Write an equation for the line tangent to the curve at the point (2,-1) c. Find the minimum y-coordinate of any point on the curve. the work for these would

1. calculus

1. Given the curve a. Find an expression for the slope of the curve at any point (x, y) on the curve. b. Write an equation for the line tangent to the curve at the point (2, 1) c. Find the coordinates of all other points on this

2. AP AB Calculus

Linear approximation: Consider the curve defined by -8x^2 + 5xy + y^3 = -149 a. find dy/dx b. write an equation for the tangent line to the curve at the point (4,-1) c. There is a number k so that the point (4.2,k) is on the

3. Math-Calculus

Find the point on the line -2x + 4y + 3 =0 which is closest to the point (-2,1). Note: I have been struggling on this for hours!

4. calculus

The point on the curve for y = sqrt(2x+1) at which tangent is perpendicular to the line y = -3x + 6 is.... a. (4, 3) b. (0,1) c. (1, sqrt3) d. (4, -3) e. (2, sqrt 5) I'm not sure, but do you find the perpendicular slope (1/3) set