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?
Originally Posted by estabaaqui
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
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.