A baseball diamond is a square 90 ft on a side. A player runs from first base to second base at 10 ft/sec. At what rate is the player's distance from home base increasing when he is half way from first to second base?
it didnt work out to be correct.
I checked the math
it had to have the square root in it still
geeeshh, then put the square root back in
dd/dt = 100(4.5)/√10125 ft/sec
To find the rate at which the player's distance from the home base is increasing, you can use the concept of related rates.
First, let's set up a coordinate system for the baseball diamond, with the origin at the home base. Let's assume that the first base is located at (0, 0) and the second base is located at (90, 0).
Now, we need to determine the position of the player when they are halfway from first to second base. This point can be found by finding the midpoint of the line segment connecting first base and second base.
The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is given by the coordinates ((x1 + x2) / 2, (y1 + y2) / 2).
In this case, the coordinates of the midpoint would be ((0 + 90) / 2, (0 + 0) / 2), which simplifies to (45, 0).
Now, we need to find the player's distance from the home base when they are at the midpoint. This can be found using the distance formula:
Distance = √((x - x1)^2 + (y - y1)^2)
In this case, we can use the point (45, 0) as (x, y) and the home base (0, 0) as (x1, y1). Plugging these values into the formula, we get:
Distance = √((45 - 0)^2 + (0 - 0)^2) = √(2025) = 45 ft
Now, we can differentiate the distance equation with respect to time to find the rate of change:
d(Distance)/dt = d(45)/dt = 0 ft/s
Therefore, the player's distance from the home base is not changing when they are halfway from first to second base.
Let the time after he leaves 1st base be t seconds
Then the distance he has gone at that time be 10t ft from first, and we have a right-angled triangle with legs of 10t ft and 90 ft
if the distance from the runner to home is d ft,
d^2 = 90^2 + (10t)^2 = 8100 + 100t^2
2d dd/dt = 200t
dd/dt = 100t/d
when he is halfway, 10t = 45
t = 4.5 , and d^2 = 8100 + 100(4.5)^2 = 10125
d = √10125 = 100.623 ft
dd/dt = 100(4.5)/100.623 = appr 4.47 ft/s
check my arithmetic