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AP Calculus help

  1. Sep 27, 2007 #1
    AP Calculus help!!!

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

    consider y^2=2+xy

    a. show that dy/dx = y/2y-x
    b. find all pints (x,y) on the curve where the line tangent to the curve has slope 1/2
    c. show that there are no points (x,y) on the curve where the line tangent to the curve is horizontal
    d. Let x and y be functions of time t that are related by the equation y^2=2+xy. At time t=5, the value of y is 3 and dy/dt=6. Find the value of dx/dt at time t=5.

    2. Relevant equations
    chain rule, implicit differentiation


    3. The attempt at a solution

    for this did part a, but I am really confused for b, c and d


    Thanks in advance for all your help.

     
    Last edited: Sep 27, 2007
  2. jcsd
  3. Sep 27, 2007 #2
    Parts B and C:
    some hints:
    How would you interpret the derivative of a function at a point geometrically?
    Notice also that in part a you have shown dy/dx to be a function of x and y. How can you relate this function to the geometrical interpretation of the derivative?
     
  4. Sep 27, 2007 #3

    dynamicsolo

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    Homework Helper

    Set dy/dx = 1/2 . What do you find?

    Set dy/dx = 0. What do you find? (Do you find anything?)

    This looks like a job for the Chain Rule! Differentiate the equation for the curve with respect to the parameter t. I believe you otherwise have all the information you need to answer the question.
     
  5. Sep 27, 2007 #4
    Ok so check me if I am wrong:

    for part b:

    y/(2y-x) = 1/2
    2y-2y= x
    0 =x so the points would be (0, ± √2)

    for part c:
    y/(2y-x) = 0
    y = 0

    then 0 = 2 + x0
    0 = 2 so it doesnt exist, therefore there is no point where the tangent is horizontal.


    for d:

    2y dy/dt = 0 + y.dx/dt + x.dy/dx
    36 = 3.dx/dt + 16x
    36 = 6x+3dx/dt x= 22/3

    36 = 14 + 3dx/dt


    22/3 = dx/dt
     
  6. Sep 27, 2007 #5

    dynamicsolo

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    I concur. :-)
     
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