A baseball diamond is a square with side 90 ft. a batter hits the ball and runs toward first base with a speed of 24 ft/s.
(a) At what rate is his distance from second base decreasing when he is halfway to first base?
(b) At what rate is his distance from third base increasing at the same moment?
draw a triangle vertexs:runner, secondbase, firstbase
a) x= distance to second base frm runner
90 ft = distance second to first
l= distance frm runner to first base
so l^2+90^2=x^2
and given dl/dt=24ft/sec
2l dl/dt+0=2xdx/dt
solve for dx/dt when l=45
you will have to calculalte x.
b. draw the traingle from second to runner to third to second. Set it up the same, same methodology
To solve this problem, we can use the concept of related rates. We need to find the rates at which the distances from the batter to second base and third base are changing at a specific moment.
Let's start with part (a):
(a) At what rate is his distance from second base decreasing when he is halfway to first base?
To solve this, we need to find a relationship between the distances involved. We can use the Pythagorean theorem as follows:
Let x be the distance from the batter to second base, and y be the distance from the batter to first base. We can write the Pythagorean equation:
x^2 + y^2 = 90^2
Next, we differentiate both sides of the equation with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0
Since the batter is halfway towards the first base, the distance from the batter to first base is y = 45 ft. We need to find dx/dt, the rate at which distance from the batter to second base is changing when y = 45 ft.
Plugging in the values we have:
2x(dx/dt) + 2(45)(dy/dt) = 0
Since we are given the speed of the batter towards the first base, dy/dt = 24 ft/s.
Simplifying the equation:
2x(dx/dt) + 90(24) = 0
Dividing by 2:
x(dx/dt) = -90(24)
Now, we can determine dx/dt by dividing both sides by x:
(dx/dt) = (-90(24)) / x
To find x, we can substitute it back into the original equation:
x^2 + (45^2) = 90^2
x^2 + 2025 = 8100
x^2 = 8100 - 2025
x^2 = 6075
x = sqrt(6075) ≈ 77.97 ft
Now, we can find dx/dt by plugging the value of x into the derived equation:
(dx/dt) = (-90(24)) / 77.97
(dx/dt) ≈ -27.58 ft/s
So, the rate at which the distance from the batter to second base is decreasing when he is halfway to first base is approximately -27.58 ft/s.
Now, let's move on to part (b):
(b) At what rate is his distance from third base increasing at the same moment?
To solve this, we can use a similar approach. Let z be the distance from the batter to third base.
Using the Pythagorean theorem, we have:
y^2 + z^2 = 90^2
Differentiating both sides of the equation with respect to time:
2y(dy/dt) + 2z(dz/dt) = 0
Since y = 45 ft, which is halfway to first base, we can substitute this value into the equation:
2(45)(24) + 2z(dz/dt) = 0
Simplifying:
90(24) + 2z(dz/dt) = 0
Dividing by 2:
45(24) + z(dz/dt) = 0
Substituting z = sqrt(90^2 - 45^2) ≈ 81.59 ft:
45(24) + 81.59(dz/dt) = 0
dz/dt = -45(24) / 81.59
dz/dt ≈ -13.32 ft/s
So, the rate at which the distance from the batter to third base is increasing at the same moment is approximately -13.32 ft/s.