# calculus

posted by .

Find the point on the curve y=x^(1/2) that is a minimum distance from the point (4,0).

My book says you use the distance formula.

Then you let L = D^2 because the minimum value of D^2 will occur at the same value of x as the minimum value of D.

What is L, though.

• calculus -

L is the distance^2. You don't have to do that way, as I will demonstrate.

D^2=(4- x)^2+(0-y)^2 that comes from the distance formula.

Doing it the way the L=D^2 did:

L= ..
dL/dx=0=2(4-x)+2(y)dy/dx

but dy/dx = d(sqrt x)/dx= 1/2sqrtx
so 0=-2x+2sqrtx/2sqrtx or
2x=2
x= 1/2, y= 1/sqrt2

Now, lets do it without the L substitution:
D^2=(4- x)^2+(0-y)^2 that comes from the distance formula.

2D dD/dx=0=2(4-x)+2(y)dy/dx
again, dy/dx= d(sqrtx)/dx= 1/(2sqrtx)
so 0=-2x+2sqrtx/2sqrtx
and again x=1/2, y= 1/sqrt2

• calculus -

They are saying, let D^2 = L
so when later on you differentiate
the result for L is simpler than that for D^2

They are using the property that if a > b
then a^2 > b^2.

let the closest point be P(x,y)
then
L = D^2 = (x-4)^2 + (y-0)^2
= (x-4)^2 + (x^(1/2))^2
= (x-4)^2 + x
dL/dx = 2(x-4) + 1 = 0 for a min distance
2x - 8 + 1 = 0
x = 7/2

if x=7/2 , then y = √(7/2) = √7/√2 = √14/2

the closest point is ((7/2 , √14/2)

## Similar Questions

1. ### calculus

Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2 + 1 and connect the tangent point to the x-axis. If the tangent point is close to the y-axis, the line segment is long. If the tangent …
2. ### calculus

Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2 + 1 and connect the tangent point to the x-axis. If the tangent point is close to the y-axis, the line segment is long. If the tangent …
3. ### calculus

Find the point on the function y= 1/ x, x>0 that is closest to the point (2,2) and then state the minimum distance.
4. ### 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 value …
5. ### 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 be …
6. ### Calculus

Find the minimum distance from the parabola x + y^2 = 0 to the point (0,-3).
7. ### calculus

Find the minimum distance from a point (6,3) to the parabola y^2=8x.
8. ### Math

1. Find all values of p such that 2(x+4)(x-2p) has a minimum value of -18. I know that y is the minimum value,I know x is -4. But, I am not sure how to find p. 2.The temperature of a point (x,y) in the plane is given by the expression …
9. ### calculus

find the minimum distance from the point (4,1) to the parabola y^2=4x.
10. ### Calculus

Find the minimum value of the following function: h(x)= x-(12(x+1)^(1/3) -12) on the interval [0, 26] Honestly, I have found by graphing that the minimum value is at the coordinate point (9,-3), but I need to do this in a way that …

More Similar Questions