Results 1 to 3 of 3

Math Help - Related Rates, urgent

  1. #1
    estabaaqui
    Guest

    Related Rates, urgent

    I did some of the work but didn't have time to put it up here. I'm really confused...can someone help me.

    A baseball diamond has the shape of a square with sides 90 feet long. A player hits the ball and runs toward first base with a speed of 24 feet per second. At what rate is the player’s distance s from third base changing when the player is halfway to first base?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by estabaaqui View Post
    I did some of the work but didn't have time to put it up here. I'm really confused...can someone help me.

    A baseball diamond has the shape of a square with sides 90 feet long. A player hits the ball and runs toward first base with a speed of 24 feet per second. At what rate is the player’s distance s from third base changing when the player is halfway to first base?

    As with all related rates problems, you want to begin by drawing a diagram to see what's going on. I have attached one to show you how I am interpreting the problem. Then you want to write down what you know and don't know. I know the value for dD/dt but not ds/dt.

    Let the distance of the batter from home as he runs be D. Let distance he is away from the first base be s. The rate at which the batter is running from home, dD/dt, is equal to 24 feet/sec. Notice the rate is positive since the distance is increasing between the runner and home base. Also notice that we have s and D changing, and not the distance between home and third, so we can keep the latter distance at a constant 90 feet.

    From the diagram, we realize that we have a right angled triangle connecting the runner, 3rd and home base. So our main formula will be Pythagoras' formula. Here goes.

    s^2 = 90^2 + D^2
    differentiate implicitly, we obtain:

    2s ds/dt = 2D dD/dt
    => ds/dt = (2D dD/dt)/2s
    The above is the formula we will be using, but we have to make sure that the only unknown in it is what we want to find, ds/dt.

    By pythagoras again, we realize, when the runner is halfway to 1st, he is 45 feet away from home. So we can find s, by s^2 = 45^2 + 90^2 => D = 100.62 feet.

    Now we know everything we need.
    Using D = 45, s = 100.62, dD/dt = 24, we get:

    ds/dt = [2(45)(24)]/[2(100.62)]
    = 10.73 feet per sec
    Attached Thumbnails Attached Thumbnails Related Rates, urgent-baseball.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Don't be overwhelmed by the volume of writing you see, i was just trying to explain how to do the problem. On a test your answer would only be a few lines.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need urgent help with Related Rates
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 22nd 2009, 10:45 AM
  2. Rates and Related Rates!!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 2nd 2008, 10:53 AM
  3. Urgent help on related rates:
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2007, 07:41 AM
  4. Related rates: urgent
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 6th 2007, 09:25 PM
  5. Related rates: urgent
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 6th 2007, 05:00 PM

Search Tags


/mathhelpforum @mathhelpforum