Results 1 to 9 of 9

Math Help - distance for curve to point help

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    16

    distance for curve to point help

    i have completely forgot how to work this problem if anyone could give step by step way to solve. Thak you

    What is the (closest) distance from the curve y=x^2 to point

    x, y =
    (16,1 2) ?





    Moderator edit: This thread was moved from the Pre-algebra and Algebra subforum to Pre-calculus subforum. In light of subsequent posts by the OP, the thread has now been moved to the Calculus subforum.
    Last edited by mr fantastic; April 10th 2011 at 02:06 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    This question is in the Pre-Calculus sub-forum, are you allowed to use calculus?
    Last edited by mr fantastic; April 9th 2011 at 02:29 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,395
    Thanks
    1481
    Awards
    1
    Quote Originally Posted by jessmari86 View Post
    i have completely forgot how to work this problem if anyone could give step by step way to solve. Thak you

    What is the (closest) distance from the curve y=x^2 to point

    x, y =
    (16,1 2) ?
    I agree that you really need calculus for this, (see reply #2)
    But the shortest distance is a perpendicular distance.
    Find a point on y=x^2, (a,a^2) so that the slope \dfrac{12-a^2}{16-a}=\dfrac{-1}{2a}.

    Now you need calculus to know why that works.
    BTW: the point is in quadrant I.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2011
    Posts
    16
    ok so where does the a come from? is that supposed to represent any point on that is the curve?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,807
    Thanks
    116
    Quote Originally Posted by jessmari86 View Post
    ok so where does the a come from? is that supposed to represent any point on that is the curve?
    Yes.

    And therefore the points on the parabola have the coordinates P(a, aČ) as Plato wrote.

    The slope of the parabola at P has the value of 2a.

    Since you need the perpendicular direction of the parabola's slope you have to use the property:

    2 lines l_1, l_2 are perpendicular if their slopes m_1,m_2 satisfy the equation: m_1 \cdot m_2 = -1

    So if the first slope is 2a the 2nd slope is -\frac1{2a}.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2011
    Posts
    16
    Quote Originally Posted by earboth View Post
    Yes.

    And therefore the points on the parabola have the coordinates P(a, aČ) as Plato wrote.

    The slope of the parabola at P has the value of 2a.

    Since you need the perpendicular direction of the parabola's slope you have to use the property:

    2 lines l_1, l_2 are perpendicular if their slopes m_1,m_2 satisfy the equation: m_1 \cdot m_2 = -1

    So if the first slope is 2a the 2nd slope is -\frac1{2a}.

    ok i got this far now
    distance = sqrt[(16-x)^2 + (1/2 - y)^2]

    = sqrt[(16-x)^2 + (1/2 -
    x^2)^2]

    = sqrt(256 - 32x + x^2 + 1/4 - x^2 + x^4)

    =
    (x^4 - 32x + 256.25)^(1/2)

    now when i checked on another site the next step does not make any since i cannot figure out where they get the first part from.
    d '(x) = (1/2)(4x^3 - 32)(x^4 - 32x + 256.25)^(-1/2)

    set = 0

    4x^3 - 32 = 0

    where does the 4x^3 - 32 come from???
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by jessmari86 View Post
    ok i got this far now
    distance = sqrt[(16-x)^2 + (1/2 - y)^2]

    = sqrt[(16-x)^2 + (1/2 -
    x^2)^2]

    = sqrt(256 - 32x + x^2 + 1/4 - x^2 + x^4)

    =
    (x^4 - 32x + 256.25)^(1/2)

    now when i checked on another site the next step does not make any since i cannot figure out where they get the first part from.
    d '(x) = (1/2)(4x^3 - 32)(x^4 - 32x + 256.25)^(-1/2)

    set = 0

    4x^3 - 32 = 0

    where does the 4x^3 - 32 come from???
    You were asked in post #2: Are you allowed to use calculus? Please answer this question.

    And if you have not learned calculus then you should not be asking questions about a solution that uses calculus. (And more to the point, if you are asking the original question at another website and getting replies you don't understand, then that website is the place where you should be asking your question).

    You have been given a non-calculus method in earlier posts (and people have spent their time giving it to you). If this is not the method you want to use, please say so. Otherwise, use the method given in those earlier posts and ask questions about it if you're still stuck.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Mar 2011
    Posts
    16
    Quote Originally Posted by mr fantastic View Post
    You were asked in post #2: Are you allowed to use calculus? Please answer this question.

    And if you have not learned calculus then you should not be asking questions about a solution that uses calculus. (And more to the point, if you are asking the original question at another website and getting replies you don't understand, then that website is the place where you should be asking your question).

    You have been given a non-calculus method in earlier posts (and people have spent their time giving it to you). If this is not the method you want to use, please say so. Otherwise, use the method given in those earlier posts and ask questions about it if you're still stuck.
    yes calculus can be used.
    Last edited by mr fantastic; April 10th 2011 at 01:55 PM. Reason: Insulting tone has been deleted.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by jessmari86 View Post
    ok i got this far now
    distance = sqrt[(16-x)^2 + (1/2 - y)^2]

    = sqrt[(16-x)^2 + (1/2 -
    x^2)^2]

    = sqrt(256 - 32x + x^2 + 1/4 - x^2 + x^4)

    =
    (x^4 - 32x + 256.25)^(1/2)

    now when i checked on another site the next step does not make any since i cannot figure out where they get the first part from.
    d '(x) = (1/2)(4x^3 - 32)(x^4 - 32x + 256.25)^(-1/2) Mr F says: This is the derivative. You get it using the chain rule.

    set = 0

    4x^3 - 32 = 0 Mr F says: The derivative is equated to zero to find the x-coordinates of the stationary points of the distance function. Note: If a/b = 0 then a = 0. That is, since the derivative is equal to zero, the numerator of the derivative must be equal to zero.

    where does the 4x^3 - 32 come from???
    There is an extensive amount of material you need to review if you are to understand questions like this one. I suggest you take the time (perhaps a week or 2 ...) to so so. Then show what progress you have made with the question and, if necessary, ask for more help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 23rd 2011, 01:58 AM
  2. Replies: 6
    Last Post: May 16th 2011, 04:57 AM
  3. Replies: 1
    Last Post: April 6th 2011, 06:44 AM
  4. Minimum Distance from Point to Curve
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: March 26th 2010, 04:38 AM
  5. Replies: 3
    Last Post: February 20th 2008, 10:17 AM

Search Tags


/mathhelpforum @mathhelpforum