(down) 0 (up)

Thanks Steve204 for your previous help with my Calculus questions. I have a question regarding related rates and is a step beyond the usual answer given.
⚑ Flag

by JDavid at July 15, 2011

I can work a related rates question ok, but have been wondering if the answer given as a rate would change if the specified distance would change. I can use a baseball diamond as an example. A baseball diamond is 90 feet square, and the pitcher's mound is at the center of the square. If a pitcher throws a baseball at 100mph (146.7 ft/sec), how fast is the distance y between the ball and first base changing as the ball crosses home plate? The home plate is approximately 63.64 feet from the pitcher's mound and if x is the distance the ball is from the pitcher's mound then dy/dt at home plate is 103.7 ft/sec. I am wondering what dy/dt will be at not only home plate but half way to home and beyond home plate, etc (assuming a continuous straight line path of the baseball; unless this is not assumed in the previously mentioned problem) . Will dy/dt be different rates as x changes and thus involve an acceleration? Thanks for any help. JDavid

Answers

(up) -1 (down)

JDavid -It's summer vacation ... why aren't you out having fun! I guess this is mental fun, lol.You are correct, dy/dt is increasing as x increases. Here is a simple way to prove it.From your previous comments, we know this is the differential equation:2x(dx/dt) = 2y(dy/dt)If you solve for dy/dt you get"dy/dt = x(dx/dt)/yIn this equation, dx/dt is fixed at 103.7, so let's only worry about the variables x and y. Actually, let's put y in terms of x.If you sketched the diamond and labeled x and y, you will see that this equation is true:(63.64)^2 + x^2 = y^2Now solve for y:y = sqrt[(63.64)^2 + x^2]Plug this value for y into this equation:dy/dt = x(dx/dt)/ydy/dt = 103.7x ÷ sqrt[(63.64)^2 + x^2]Over the interval 0 ≤ x ≤ 63.64 this function increases. Graph it on your TI-84 (note: window Y from 0 to 1)So, what we proved is that as the ball gets farther away from the pitcher's mound, the rate of change of y with respect to t increases.Hope that helps

⚑ Flag

19846

Steve204 July 15, 2011

Join Game Changers!

Flag this because it ...

Thanks Steve204 for your previous help with my Calculus questions. I have a question regarding related rates and is a step beyond the usual answer given. ⚑ Flag

Answers

Thanks Steve204 for your previous help with my Calculus questions. I have a question regarding related rates and is a step beyond the usual answer given.
⚑ Flag