1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating Variable from a Derivative

  1. Apr 10, 2012 #1
    My answer for part (i) matches up with that of the book but solving for part (ii) has gone awry. Can anyone help me clear this up?
    Many thanks.

    1. The problem statement, all variables and given/known data

    Q. The given diagram (see attachment) shows part of the graph of y = x2. p(x,y) is a point on the curve & a(6,0). bp [itex]\perp[/itex] ab. (i) Express the coordinates of p in terms of x only, (ii) Find the value of x if the area of the triangle abp is a maximum & hence find this maximum area.

    2. Relevant equations

    3. The attempt at a solution

    (i) If y = x2 => p(x,y) => p(x,x2)

    (ii) [itex]\frac{1}{2}[/itex][(y2 - y1)(x1 - x3) - (x2 - x1)(y1 - y3)] => [itex]\frac{1}{2}[/itex][(0 - x2)(x - x) - (6 - x)(x2 - 0)] => [itex]\frac{1}{2}[/itex][0 - (6x2 - x3)] => [itex]\frac{1}{2}[/itex][-6x2 + x3]
    [itex]\frac{dA}{dx}[/itex] = [itex]\frac{1}{2}[/itex](-12x + 3x2) = 0 => -12x + 3x2 = 0 => 3x2 = 12x => x2 = 4x => x = 4
    [itex]\frac{d^2A}{dx^2}[/itex] = [itex]\frac{1}{2}[/itex](-12 + 6x) => [itex]\frac{1}{2}[/itex](-12 + 6(4)) => [itex]\frac{1}{2}[/itex] => [itex]\frac{1}{2}[/itex](12) => 6 > 0, making x = 4 a minimum.
    Area = [itex]\frac{1}{2}[/itex](-6x2 + x3) => [itex]\frac{1}{2}[/itex](-6(4)2 + 43) => [itex]\frac{1}{2}[/itex](-96 + 64) => [itex]\frac{1}{2}[/itex](-32) => -16, i.e. Area is 16

    Ans.: (ii) x = 2, Area = 8
     

    Attached Files:

  2. jcsd
  3. Apr 10, 2012 #2
    Let the point [itex]b(x_0, y_0) = (x, 0)[/itex].

    [itex]A = \frac{1}{2} (base)(height)[/itex]
    [itex]A = \frac{1}{2} (6 - x)(x^2 - 0)[/itex]
    [itex]A = \frac{1}{2} (6x^2 - x^3)[/itex]
    [itex]A = 3x^2 - \frac{1}{2}x^3[/itex]

    [itex]A' = 6x - \frac{3}{2} x^2[/itex]
    [itex]A' = \frac{3}{2} x (4 - x)[/itex]

    Setting [itex]A'= 0[/itex],
    [itex]\frac{3}{2} x (4 - x) = 0[/itex]
    [itex]x = 0, x = 4[/itex]

    Applying the first derivative test:
    On the interval [itex](0, 4), A' > 0[/itex], which indicates that [itex]A[/itex] is increasing on this interval.
    On [itex](4, 6), A' < 0[/itex], so [itex]A[/itex] is decreasing.
    The increase-decrease indicates the occurrence of a maximum at [itex]x = 4[/itex].

    And when [itex]x = 4, y = 4^2 = 16[/itex]. The area is then
    [itex]A = \frac{1}{2} (6 - 4) (16) = 16[/itex].

    If I haven't made any mistakes here, my best guess would be that it's a typo in your book.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Calculating Variable from a Derivative
  1. Calculate derivatives (Replies: 5)

Loading...