can anyone help me with this question?
A baseball diamond is a square whose sides are 90 ft long. Suppose that a player running from second base to third base has a speed of 30 ft/sec at the instant when he is 20 ft from the third base. At what rate is the player’s distance from home plate changing at that instant.
let the distance from 3rd base be x. Then the distance z from home is
z^2 = 90^2 + x^2
so, to find dz/dt,
2z dz/dt = 2x dx/dt
So, find z when x=20, then plug in dx/dt and x.
Certainly! This question can be solved by using the concept of related rates. We will first find an expression for the player's distance from home plate, and then differentiate it with respect to time.
Let's assume that home plate is at the origin of a Cartesian coordinate system, with the x-axis extending along the base paths. In this case, second base will be located at (0, 90) and third base will be located at (90, 0), forming a square.
The player's position can be represented by a point on a line connecting second and third bases. Let's call this point P, with coordinates (x, y). Since the player is always on this line, the slope of the line connecting P and second base is the same as the slope of the line connecting second and third bases.
The slope of this line is given by:
m = (0 - 90) / (90 - 0) = -1
Now let's find an equation for the line connecting second and third bases. Using the slope-intercept form, we have:
y = mx + b
Substituting the slope and the coordinates of second base:
90 = -1 * 0 + b
b = 90
Therefore, the equation of the line connecting second and third bases is:
y = -x + 90
Now, let's find an equation for the line connecting P and second base. The slope of this line can be found using the given information. At the instant when the player is 20 ft from third base, the x-coordinate is 70 (90 - 20) since we are moving along the x-axis. We are also given the speed of the player, which is 30 ft/sec.
Using the slope formula, we have:
slope = Δy / Δx
slope = (y - 90) / (x - 0)
Differentiating both sides with respect to time (t), we have:
(dy / dt) = [(dx / dt) * (x - 0) - (y - 90) * (0 - 0)] / (x - 0)^2
Since the player is running along the line connecting P and second base, the speed in the y-direction (dy / dt) is zero. We're interested in finding the speed at the instant when the player is 20 ft from third base, so we substitute the given values:
0 = [(dx / dt) * (70 - 0) - (y - 90) * (0 - 0)] / (70 - 0)^2
Simplifying, we have:
0 = (dx / dt) * (70 - 0) / 70^2
Solving for (dx / dt), we find:
(dx / dt) = 0
Therefore, the rate at which the player's distance from home plate is changing at that instant is 0 ft/sec.
In summary, the player's distance from home plate is not changing at the instant when he is 20 ft from third base.