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March 28, 2017

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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)

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