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Related rates and baseball

  1. Apr 29, 2006 #1

    F.B

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    Can someone check if i did these two questions right please?

    The questions are:

    1. A baseball diamond is a perfect square with each side measuring 90 feet
    in length. A player runs from the first base to second base at a speed of
    25 ft/s. How fast is she moving from home plate when she is 30 ft away from
    first base.

    2. A fish is being reeled in a rate of 2 m/s (i.e, the fishing line is
    being shortened by 2 m/s by a man. If he is sitting 30 m above water on a
    dock, how fast is the fish moving through water when the line is 50 m long.

    My answers are these:

    1. A drew my diamond and figured that i will be using half of the diamond.

    Let the distance between home plate and first base be x
    Let the distance from first base to second base by y

    Given
    y = 90 ft
    dy/dt = 25 ft/s
    x = 90 ft - 30ft
    =60 ft
    dx/dt = ?

    x^2 + y^2 = z^2
    90^2 + 90^2 = z^2
    z = 127.3 ft

    x^2 + y^2 = z^2
    2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
    sine dz/dt = 0
    2(60)(dx/dt) + 2(90)(25) = 0
    dx/dt = -37.5 ft/s

    Is this correct? and why is my answer negative shouldn't it be positive
    because you increasing the distance from home plate or is my answer
    negative because the distance to first base is decreasing.


    2. Given
    y = 30 m
    z = 50 m
    dz/dt = - 2m/s
    x = ?
    dx/dt = ?

    x^2 + y^2 = z^2
    x^2 = 50^2 - 30^2
    x = 40 m

    x^2 + y^2 = z^2
    2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
    since dy/dt = 0
    2(40)(dx/dt) = 2(50)(-2)
    dx/dt = -2.5 m/s

    Is this question correct because im not sure when to make my numbers
    negative, i know we have to when something is decreasing but im not sure.

    So can you please check if i did this two questions right?
     
  2. jcsd
  3. Apr 29, 2006 #2
    For the first one, y isn't 90. Also, you won't find x by subtracting. It's a right triangle. How can you find an unknown side of a right triangle? The baseball player is 30 ft. from second base, not 90 ft. You had the general setup right though.

    Using your defined variables.

    [tex]x^2=90^2+y^2[/tex]

    [tex]x\frac{dx}{dt}=y\frac{dy}{dt}[/tex]

    You know y, dy/dt, and you can solve for x using the Pythagorean Theorem. Now just find dx/dt.
     
  4. Apr 29, 2006 #3
    The second one looks good to me. Your sign is correct. This question is worded weirdly. It asks how fast the fish is moving through the water and I interpret that to mean what speed, which is [tex]|\vec{v}|[/tex], where v is velocity. I think going by the question it's more correct to put the positive value, only showing magnitude and not direction, but you should ask your teacher.
     
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