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 is his distance from third base increasing at the same moment?
a.) Well, let's get the party started with some math! We can begin by finding the distance between second base and the halfway point to first base. Since the baseball diamond is a square, the distance between second base and the halfway point to first base is (90/2) ft = 45 ft.
Now, to find the rate at which his distance from second base is decreasing, we need to find his rate of change at that particular moment. This can be done by taking the derivative of the distance function with respect to time. Since he is moving towards first base, the distance function can be expressed as d = √(x^2 + y^2), where x represents the distance from second base and y represents the distance from first base.
Taking the derivative, we get d' = (x/x') + (y/y'), where x' and y' are the rates of change of x and y, respectively. Since he is halfway to first base, we know that x = 45 ft, so we can solve for x'.
Plugging in the given values, we have d' = (45/x') + (y/y').
Now, we also know that he is running towards first base with a speed of 24 ft/s. Therefore, his rate of change of y is -24 ft/s (since he's getting further away from third base). Plugging this value into the equation, we have d' = (45/x') - (y/24).
Since he is halfway to first base, we can also say that y = (90 - 45) ft = 45 ft. Plugging this value into the equation, we have d' = (45/x') - (45/24).
Simplifying the equation, we get d' = 45(1/x' - 1/24).
Now, we can solve for x' by substituting the given value for d' (24 ft/s). Plugging this value into the equation, we have 24 = 45(1/x' - 1/24).
Solving the equation, we find that x' = 45/43 ft/s (approximately).
Therefore, the rate at which his distance from second base is decreasing when he is halfway to first base is approximately 45/43 ft/s, or you could say that it's running away at a pretty "fast" pace!
b.) Alright, let's move on to the next base on this amusement park ride of a question! We want to find the rate at which his distance from third base is increasing at the same moment.
To do this, we can use a similar approach as in part a. We know that the distance between third base and the halfway point to first base is also 45 ft. So the distance function becomes d = √((90 - x)^2 + y^2), where x represents the distance from first base and y represents the distance from third base.
Taking the derivative, we get d' = (-x/x') + (y/y').
Since he is halfway to first base, we know that x = 45 ft. Plugging this value into the equation, we have d' = (-45/x') + (y/y').
Since he is running towards first base with a speed of 24 ft/s, we can say that x' = 24 ft/s. Plugging this value into the equation, we have d' = (-45/24) + (y/y').
Now, we need to find the rate of change of y. Since he is running towards first base, the rate of change of y is also -24 ft/s. Plugging this value into the equation, we have d' = (-45/24) + (y/-24).
Since he is halfway to first base, we can also say that y = 45 ft. Plugging this value into the equation, we have d' = (-45/24) + (-45/-24).
Simplifying the equation, we get d' = (-45/24) + (45/24).
Now, we can solve for d' by adding the fractions together. The denominator stays the same (24), so we just add the numerators. We get d' = 0 ft/s.
Therefore, at the same moment, his distance from third base is not increasing or decreasing. It's just chillin' and stayin' the same.
To solve this problem, we can use the concept of similar triangles. Let's break it down step by step:
a) To find the rate at which his distance from second base is decreasing, we can use the similar triangles formed by the baseball diamond.
In the given scenario, the runner is halfway to first base, which means he is at a distance of 45 ft from the starting point (home plate). We need to find how fast he is moving towards second base at this point.
Let's define:
x = distance of the runner from second base
y = distance of the runner from home plate (starting point)
Since the baseball diamond is a square, the distance between home plate and second base is also 90 ft (as stated in the question).
We have two similar triangles:
1) The runner's position triangle: x/y
2) The baseball diamond triangle: (90-x)/90
The ratio of these two triangles will remain constant:
x/y = (90-x)/90
Now, let's differentiate both sides of the equation with respect to time (t):
d(x)/dt / d(y)/dt = -d(90-x)/dt / d(90)/dt
Given that the speed of the runner (dy/dt) is 24 ft/s, we can substitute it into the equation:
dx/dt / 24 = -(-dx/dt) / 0
Simplifying, we get:
dx/dt = -((dx/dt) * 24) / 0
Since dx/dt cannot be zero, we divide by dx/dt on both sides:
1 = -24 / 0
The expression on the right is undefined, indicating that the rate at which the distance is decreasing is infinite. Therefore, at the halfway point to first base, the runner is approaching second base at an infinitely fast rate.
b) To find the rate at which his distance from third base is increasing, we can use the same concept.
Similarly, we have two similar triangles:
1) The runner's position triangle: (90-x)/y
2) The baseball diamond triangle: x/90
The ratio of these two triangles will remain constant:
(90-x)/y = x/90
Now, let's differentiate both sides of the equation with respect to time (t):
d((90-x)/y) / dt = d(x/90) / dt
Given that the speed of the runner (dy/dt) is 24 ft/s, we can substitute it into the equation:
(-dx/dt)/y + (90-x)(-dy/dt)/y^2 = dx/dt/90
Since we are interested in the rate at which his distance from third base is increasing, we want to find dx/dt. Rearranging the equation, we get:
dx/dt = (90-x)(-dy/dt)/y^2 + dx/dt/90
Substituting the values we have:
dx/dt = (90-45)(-24)/(45^2) + dx/dt/90
Simplifying:
dx/dt = (-45)(-24)/(45^2) + dx/dt/90
dx/dt = 1080/2025 + dx/dt/90
dx/dt = 0.533 + dx/dt/90
Therefore, the rate at which his distance from third base is increasing at the halfway point to first base is approximately 0.533 ft/s.
To answer these questions, we can use the concept of related rates. Related rates involve finding the rate at which one quantity is changing with respect to another related quantity. In this case, we want to find the rates of change of the distances from the batter to second base (part a) and third base (part b) as the batter moves towards first base.
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 find this rate, we need to determine how the distance from the batter to second base (let's call it x) is related to the distance from the batter to first base (90 ft - x). We can use the Pythagorean theorem to establish this relationship:
x^2 + (90 ft - x)^2 = (90 ft)^2
Simplifying this equation, we get:
2x^2 - 180x + 8100 = 8100
Now, we can differentiate both sides of the equation with respect to time (t), to find the rate at which x is changing:
d/dt[2x^2 - 180x + 8100] = d/dt[8100]
4x(dx/dt) - 180(dx/dt) = 0
Simplifying further, we get:
4x(dx/dt) - 180(dx/dt) = 0
(dx/dt)(4x - 180) = 0
Since the batter is moving towards first base (which means x is decreasing), we can ignore the solution x = 0. Therefore, we have:
4x - 180 = 0
4x = 180
x = 45 ft
Now, we can substitute this value back into the equation to find the value of dx/dt at this moment:
dx/dt = (dx/dt) @ x = 45 ft
Next, we can use the given information that the batter is running towards first base with a speed of 24 ft/s. This means that dx/dt = -24 ft/s (negative sign since x is decreasing). Substituting this value into the equation, we have:
-24 ft/s = (dx/dt) @ x = 45 ft
Therefore, the rate at which his distance from second base is decreasing when he is halfway to first base is -24 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?
Similar to part a, we need to determine how the distance from the batter to third base (let's call it y) is related to the distance from the batter to first base (90 ft - y). Again, we can use the Pythagorean theorem:
y^2 + (90 ft - y)^2 = (90 ft)^2
Simplifying this equation, we get:
2y^2 - 180y + 8100 = 8100
Differentiating both sides of the equation with respect to time (t) to find the rate of change of y:
d/dt[2y^2 - 180y + 8100] = d/dt[8100]
4y(dy/dt) - 180(dy/dt) = 0
Simplifying further:
4y(dy/dt) - 180(dy/dt) = 0
(dy/dt)(4y - 180) = 0
Since the batter is moving towards first base (which means y is also decreasing), we can ignore the solution y = 0. Therefore, we have:
4y - 180 = 0
4y = 180
y = 45 ft
Now, substituting this value back into the equation:
dy/dt = (dy/dt) @ y = 45 ft
Using the information given that the batter is running towards first base with a speed of 24 ft/s, we have:
dy/dt = 24 ft/s
Therefore, the rate at which his distance from third base is increasing at the same moment is 24 ft/s.
In summary:
a.) The rate at which his distance from second base is decreasing when he is halfway to first base is -24 ft/s.
b.) The rate at which his distance from third base is increasing at the same moment is 24 ft/s.