Results 1 to 3 of 3

Math Help - can someone poke a hole in this?

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    2

    can someone poke a hole in this?

    I take AP Calculus AB at my high school and I think I have found a flaw in my textbook. In a section about the length of curves, there is this problem asking the reader to find a curve based on a given integral and the part b of the problem asks how many curves are possible. The back of the book says that there is only one curve possible, but I think I have found another.
    The problem is
    Find a curve through the point (1,1) whose length integral is
    L= \int_{1}^{4} \sqrt{1 + \frac {1}{4x}}dx

    so following the formula for the length of a curve, (dy/dx)^2=(1/4x) so (dy/dx) = + or - \frac {1}{2\sqrt{x}}. based on those dy/dx's, y is equal to either the square root of x plus a constant or the negative square root of x plus a constant, depending on whether you use the positive dy/dx or the negative one. After solving for constants I find two solutions that work:
    y= \sqrt{x}<br />
and y= -\sqrt{x} + 2

    but my text book only thinks the first one is an acceptable answer apparently. can someone help me out?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,628
    Thanks
    430
    Quote Originally Posted by helothar View Post
    I take AP Calculus AB at my high school and I think I have found a flaw in my textbook. In a section about the length of curves, there is this problem asking the reader to find a curve based on a given integral and the part b of the problem asks how many curves are possible. The back of the book says that there is only one curve possible, but I think I have found another.
    The problem is
    Find a curve through the point (1,1) whose length integral is
    L= \int_{1}^{4} \sqrt{1 + \frac {1}{4x}}dx

    so following the formula for the length of a curve, (dy/dx)^2=(1/4x) so (dy/dx) = + or - \frac {1}{2\sqrt{x}}. based on those dy/dx's, y is equal to either the square root of x plus a constant or the negative square root of x plus a constant, depending on whether you use the positive dy/dx or the negative one. After solving for constants I find two solutions that work:
    y= \sqrt{x}<br />
and y= -\sqrt{x} + 2

    but my text book only thinks the first one is an acceptable answer apparently. can someone help me out?
    textbook answers have been wrong on many occasions.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    2
    yeah i realize that i guess i was just wondering if i was missing something
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Pigeon-hole Principle
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: July 31st 2011, 05:14 AM
  2. Discontinuity Point Hole
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 25th 2009, 06:55 PM
  3. Pigeon Hole Principle
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: August 31st 2009, 07:59 AM
  4. volume of sphere with hole in it
    Posted in the Calculus Forum
    Replies: 6
    Last Post: August 9th 2009, 01:42 PM
  5. pigeon-hole
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: December 11th 2008, 01:41 PM

Search Tags


/mathhelpforum @mathhelpforum