Question

a baseball diamond is a square 90 f on a side. A player runs from first base to second at a rate of 16 ft/sec. a) At what rate is the player's distance from third base changing when the player is 30 ft from first base?
b) At what rate are angles theta 1 and theta 2 changing at that time?

Answer 1

From where are we measuring theta 1 and theta 2?

Answer 2

a) We can draw a right triangle with sides a, b, c, 'c' being the hypotenuse, 'a' being the the side of the right triangle that goes from the first base to the second base, and b is just 90, which is the distance between the third base and the second base. We are also given that the 'a' side of the triangle is DECREASING at a rate of -16 ft/sec. They say it's 16ft/sec but because as the player runs, the side length decreases, so we write the rate as a negative value. We are looking for dc/dt at a = 60, b = 90 and c = sqrt(11700). Those values are the ones you get when you differentiate a^2 + b^2 = c^2 by plugging in the values provided. After rearranging to solve for dc/dt, you get:

\[\frac{ dc }{ dt } \approx -8.875ft/\sec\]Therefore the rate at which the distance from the player and third base is changing when the player is 30 ft from the bast is approximately -8.875ft/sec.

Now the angles. If we keep the same right triangle, and label the angle at the first base theta 1 and the angle at the third base theta 2, we can find the rates at which these change from the info we got. We know theta 1 and theta add to 90 because this is a right triangle. And we also know that d(theta1)/dt + d(theta2)/dt = 0. We can know find the either d(theta1)/dt or d(theta2)/dt and we can plug it in to that equation to solve for the other one. I found d(theta1)/dt. Use Sine, of this angle, we know that sin(theta1) = 90/c. After differentiating, we plug in the value of dc/dt we got from solving part a) and plug in cos(theta1) by finding the sides of this triangle when the runner is 30 ft from base 1 and using adjacent/hypotenuse. We then solve for d(theta1)/dt and get this:\[\frac{ d \theta _{1} }{ dt }=0.08 rad/\sec \rightarrow \frac{ d \theta _{1} }{ dt } + \frac{ d \theta _{2} }{ dt }=0 \implies \frac{ d \theta _{2} }{ dt }=-0.08rad/\sec\]And we are done.
@success16

Answer 3

@success16

Does this make sense to you and is this similar to what you got?

Answer 4

a is similar to what i have, I haven't tried b yet using your method but i'm working on it now. thanks.