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Dr/ dt rate/distance problem.

  1. Nov 8, 2007 #1
    1. The problem statement, all variables and given/known data
    A baseball diamond is a square 90 feet on a side. A player runs from first base to second at a rate of 16 ft/sec. At what rate is the player's distance from third base changing when the player is 30 ft from first base..



    2. Relevant equations
    dr/dt ds/dt

    dx/dt? a^2+b^2=C^2


    3. The attempt at a solution

    I have no idea... i dont know where to start, but it must be solved using the derivative.
     
  2. jcsd
  3. Nov 8, 2007 #2
    hint: they are asking for acceleration when they say "at what rate... blah blah blah... changing"

    and there are wel known aceleation, distance, time equations out there
     
  4. Nov 8, 2007 #3
    wouldn't they be asking for velocity or rate, not acceleration?
     
  5. Nov 8, 2007 #4
    when there is a rate of change that is synonymous with acceleration
     
  6. Nov 8, 2007 #5
    They're asking for the players rate of change of distance, which infers velocity, as opposed to acceleration, which is the rate of change of velocity.

    You have your equation set up correctly, why don't you begin by differentiating both sides.
     
  7. Nov 8, 2007 #6
    oops. my bad.

    change of rate wrt distance = velocity
    change of rate wrt time = acceleration

    so sue me.
     
  8. Nov 8, 2007 #7
    i dont care that you were wrong, i just have no idea where to start? and im not sure i understand the notation..
     
  9. Nov 8, 2007 #8
    Do you know how to differientiate both sides of the equation you posted? You'll want to include the derivative notation for each variable when you do this.

    For example, if I had [tex] 2x^2 = c^3 [/tex] and I differientiated both sides, I would have [tex] 4x \frac{dx}{dt} = 3c^2 \frac{dc}{dt} [/tex].
     
  10. Nov 8, 2007 #9
    i know how to differentiate both sides, but i dont even have two sides...
     
  11. Nov 8, 2007 #10
    In your opening post, you had the equation "a^2+b^2=C^2," right? There are two sides to this equation.
     
  12. Nov 8, 2007 #11
    i don't understand how that works? but it would be:

    60^2 + 90^2 = C^2
     
  13. Nov 8, 2007 #12
    You'll want to wait until the very end to plug in your values. Just begin by treating "a," "b," and "c" as variables, and take the derivative of both sides of the equation with respect to time.
     
  14. Nov 8, 2007 #13
    2a + 2b = 2c?
     
  15. Nov 8, 2007 #14
    it seems like the type of problem where first you do a derivative

    and then once you have a resultant equation you plug in 30 for one of the variables
     
  16. Nov 8, 2007 #15
    This is correct, and sprint is right, you'll want to wait until the very end to plug in values. Your equation is missing three things however, you'll want to include the da/dt, db/dt, and dc/dt terms next to their respective variables. So far so good?
     
    Last edited: Nov 8, 2007
  17. Nov 8, 2007 #16
    yea so da/dt(2a) + db/dt(2b) = dc/dt(2c)?
     
  18. Nov 8, 2007 #17
    Right, so now the 2's cancel out. The problem said that you're dealing with two sides of a square, so how can we relate "a" and "b" ? What's our new equation?
     
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