HELP!!

[Guest]

a baseball diamond is a square 90ft an each side. a player runs from first base to second at a rate of 16ft/sec
a) at what rate is the player's distnace form thrid base changing when the player is 30ft from first base
b) at what rates are angles theta and theta 2 changing at that time

 

[arthur ohlsten]

let x be distance runner is from 1st base
let h be distance from 3rd base to runner

from the right triangle 3rd base , 2nd base , runner we have
h^2=90^2+[90-x]^2 take derivative with respect to t
2h dh/dt = 2[90-x] [-dx/dt] divide both sides by 2h
dh/dt = [x-90]/h dx/dt dx/dt =16 ft /sec
dh/dt =16 [x-90]/h ft/sec but when x=30 90-x=60 and h is

h^2=60^2+90^2
h^2=10^2[ 6^2+9^2]
h^2=10^2[36+81]
h^2=10^2[117]
h^2=10^2 [3*3*17
h^2=10^2* 3^2 *17 take positive square root
h= 10*3 sqrt 17
h=30 sqrt17

dh/dt = 16 [30-90] /[ 30 sqrt17]
dh/dt = - 960/[30 sqrt17]
dh/dt = -12/sqrt17 ft/sec

please check for errors

I do not know how theta is measured
Arthur