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Homework Help: Finding y prime (derivative)

  1. Sep 28, 2004 #1
    What is the y prime of:
    A 15 foot board rests against a vertical wall. If the bottom of the board slides away from the wall at the rate of 3 feet per second, how fast is the area of the triangle formed by the board, the wall and the ground changing at the instant the bottom of the board is 9 feet from the wall?

    I got this far:
    y= square root of (225 - x^2)

    I'm stuck after this step.

    Can the answer for this be [(-9 yprime /2) + 18] square feet per second.
    Last edited: Sep 28, 2004
  2. jcsd
  3. Sep 28, 2004 #2
    Hint x in terms of t. The displacement x away from the wall is equal to what in terms of t? It will be easy to then sub in this for x.

    Area of a triangle is 1/2 * base * height
    Base is x height is sqrt(225-x^2) right?

    so A = 1/2 * x * sqrt(225-x^2)

    However you need to plug in t to this I just didn't want to give away that much work.

    Differentiate with respect to t in the new equation you find using the product rule (gets a little messy). Since 3t = 9, t = 3 goes into this new derivative and bam you got your answer.
    Last edited by a moderator: Sep 28, 2004
  4. Sep 28, 2004 #3


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    I'm not sure what you mean by the question "what is the y prime of....?". If anything, the question seems to be asking for A'(t), where A is the area of the right triangle formed by the ladder, the wall, and the ground.

    Draw a diagram, label the horizontal distance between the bottom of the ladder and wall x, and the corresponding vertical distance y


    [itex] \frac{dx}{dt} [/itex] = 3 ft/s

    This is a related rates problem. A is a function of x, so given the rate at which x changes with time, you should be able to find the rate at which A changes with time using the functional relationship:

    [tex] A = \frac{1}{2}xy = \frac{1}{2}x\sqrt{225 - x^2} [/tex]

    You identified this relationship correctly! :smile:

    Now, please relate the rates:

    [tex] \frac{dA}{dt} = \frac{dA}{dx}\frac{dx}{dt} [/tex]

    [tex] = \frac{1}{2}\frac{d}{dx}(x\sqrt{225 - x^2})(3 \text{ft/s}) [/tex]

    Once you solve that, evaluate it at the value specified. I think that's right. It's kinda late here, so if any of this is bogus, please let me know.

    EDIT: right, vsage, I forgot that you need x in terms of t.
    Last edited: Sep 28, 2004
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