Related Rates Problem?

Question

A baseball diamond is a square with side 90ft. A batter hits the ball and runs towards first base with a speed of 24ft/s. At what rate is his distance from second base decreasing when he is halfway (45ft) into first base? At what rate is his distance from 3rd base increasing at the same moment?

Answer 1

--The time at 45ft--
The person is moving at 24ft/s, and travels 45ft.
45ft ÷ 24ft/s = 1.875s


--For second base--

Let the distance from second base be represented by the function:

d2(t) = [90^2 + (90 - 24t)^2]^0.5
= [8100 + (90 - 24t)^2]^0.5

d2 ' (t) = {.5[8100 + (90 - 24t)^2]^-0.5}{2(90 - 24t)(24)}
= {.5[8100 + (90 - 24(1.875))^2]^-0.5}{2(90 - 24(1.875))(24)}
= {.5[(8100 + 45^2)^.5]}{-48 * 45}
= (-2160) / 2(8100 + 2025)^.5
= (-2160) / (201.25)
= -10.73ft/s

therefore, distance from the second base is decreasing by 10.73ft/s

--For third base--

Let the distance from third base be represented by the function:

d3(t) = [90^2 + (24t)^2]^0.5
= [8100 + 576t^2]^0.5

d2 ' (t) = {.5[8100 + 576t^2]^-0.5} * {2(576t)}
= 1152t / 2*[8100 + 576t^2]^0.5
= 1152t / [16200 + 1152t^2]^0.5
= 1152(1.875) / [16200 + 1152(1.875)^2]^0.5
= 2160 / 201.25
= 10.73ft.s

therefore, distance from the third base is increasing by 10.73ft/s

--Check--
The person's position is equally between second and third base, and he is moving parallel to the second-third base line. Therefore, it makes sense that at this moment, the person is moving towards second base at the same rate he is moving away from third.

Answer 2

First thing to do is rotate how you usually look at a baseball diamond. Place homeplate on the origin, the first baseline on the x-axis, and the third baseline on the y-axis. So, 3rd base is at (0,90), 2nd base is at (90,90), at the location of the runner is (x,0).
Since the runner has a speed of 24ft/s, then dx/dt = 24.

Think of the runner at some arbitrary point on the first base line. Call his distance to 3rd base a, and his distance to 2nd base b. So, x^2 + 90^2 = a^2 and (90-x)^2 + 90^2 = b^2.
Differentiate both sides of both equations with respect to t.
2x*dx/dt = 2a*da/dt and -2(90-x)*dx/dt = 2b*db/dt
Solving the first equation for da/dt gives you:
(x/a)*dx/dt = da/dt. When x = 45, a = sqrt(45^2+90^2) = 45sqrt(5), and so x/a = 45/(45sqrt(5)) = 1/sqrt(5) = (1/5)sqrt(5). da/dt = (1/5)sqrt(5)*24 = 10.73ft/s
solving the second equation for db/dt gives you:
(-(90-x)/b)*dx/dt = db/dt. But since we're going for the point halfway between home and first a=b and x = 90-x. So the rate at which is distance from 2nd is decreasing is also about 10.73ft/s, i.e. db/bt = -(1/5)sqrt(5)*24 = -10.73ft/s.

Answer 3

It helps to draw a picture of the proceedings. When he hits the ball, he is at a 45 degree corner of a 45/45/90 triangle, which is a diagonal of the square. As he runs, he is still at the corner of a right triangle (the right angle is at first base). The distance to second base from first stays constant (90 ft). However, the distance to first base (h1) is decreasing at 24 ft/sec. Thus, the distance to second base is a hypotenuse (H), with one of its legs decreasing. So we can write
H^2= 90^2 + (h1)^2 or H= sqrt(8100+h1^2)=function(h1)
and dH/dt = d/dt(function(h1)= d/d(h1)(function)*d(h1)/dt.
So you find the derivitive of the term on the right, which will include both h1=45 ft and d(h1)/dt=-24. You solve.

The second part involves another right triangle. At the moment the ball is hit, the batter is 90 ft from third base. As he run up the first base line, he is at the apex of an angle of a right triangle (the 90 degree angle is at home plate) which has a side along the first base line INCREASING at 24 m/sec. Again, you are solving for the hypotenuse.

If you are having difficulties visualizing this, draw a baseball diamond. With third-home as one leg and first-second as the other, draw an "M" with apexes at second and third and the point touching the home-first base line at its center. Now visualize the point being pushed up towards first. The lower part of the M is getting bigger and the upper part is getting shorter.