Can you help me with a Related Rates Baseball Problem?

Question

A baseball outfielder, standing 200 ft from home plate and 120 ft from the third base line, must attempt to throw out a runner going from third base to home plate at a speed of 25 ft/sec. When the runner is 60 ft from home plate, how fast is his distance from the outfielder changing?

Answer 1

I assume the distance from the third-base line is the perpendicular one. Let us call the point T the place where a line drawn from the outfielder to the third-base line and perpendicular to the third-base line meets the third-base line.

How far is T from home plate? Let's call that distance F.

F = sqrt(200^2 - 120^2)

Let's call the runner's distance from home plate H. Now, how far is the runner from point T? Let's call that distance R.

R = sqrt(200^2 - 120^2) - H

The distance between the runner and the outfielder is the hypotenuse of a smaller triangle whose sides are R and 120. The distance between the runner and the outfielder is thus

D = sqrt([sqrt(200^2 - 120^2) - H]^2 + 120^2)

We now have the distance between the runner and the outfielder, D, as a function of the distance between the runner and home plate, H.

Take dD/dH, put in 60 for H, and multiply the result by 25.

Answer 2

place the diamond on a variety airplane so as that domicile base is the muse (0, 0) and prevalent base B1 ?(ninety, 0). Then 2d base B2 ?(ninety, ninety) and 0.33 base B3 ?(0, ninety) Chris is at C ?(30 - 20t, ninety) and Sam is at S ?(0, 40 - 15t) |CS|² = ((30 - 20t) - 0)² + (ninety - (40 - 15t))² = (30 - 20t)² + (50 + 15t)² = 25(6 - 4t)² + 25(10 + 3t)² = 25(36 - 48t + 16t² + a hundred + 60t + 9t²) = 25(136 + 12t + 25t²) So CS = 5?(136 + 12t + 25t²) As t will improve (136 + 12t + 25t²) will improve So CS will improve. ie the gap between them is increasing dCS/dt = 5/2 * (12 + 50t)/[?(136 + 12t + 25t²)] = 5(12 + 50t)/[?(136 + 12t + 25t²)] whilst t = 0 (at that given element) dCS/dt = 60/?(136) feet/s i e they're shifting removed from one yet another at a velocity of 60/?(136) feet/s (~5.14 feet/s)