Can anyone help

A baseball diamond has the shape of a square with sides 90 feet long. A player 30 feet from third base is running at a speed of 28 feet per second. At what rate is the player’s distance s from the home plate?

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- January 6th 2007, 03:43 PMml_2007Related rates: urgent
Can anyone help

A baseball diamond has the shape of a square with sides 90 feet long. A player 30 feet from third base is running at a speed of 28 feet per second. At what rate is the player’s distance s from the home plate? - January 6th 2007, 09:25 PMticbol
The question is not very clear, but I assume that " A player 30 feet from third base is running ..." means the player is running from 2nd base to 3rd base.

With that assumption, imagine, or draw the figure on paper. It is a right triangle, with these:

---one leg = x ---------the portion of 2nd-to-3rd-base side, from the 3rd base.

---other leg = 90 ft ----the 3rd-to-home-base side.

---hypotenuse = s -----distance from runner to the home base.

By Pythagorean theorem,

s^2 = x^2 +90^2

Differentiate both sides with respect to time t,

2s(ds/dt) = 2x(dx/dt)

s(ds/dt) = x(dx/dt) ------------(1)

Now, in (1) we know that:

dx/dt = -28 ft/sec ----negative because x is getting shorter with time t.

x = 30 ft

We are looking for ds/dt.

But we do not know the s. So we find the s at that instant.

When x=30 ft,

s^2 = 30^2 +90^2

s^2 = 9000

s = 30sqrt(10)

So, plug those into (1),

[30sqr(10)](ds/dt) = 30(-28)

ds/dt = -28/sqrt(10)

ds/dt = -8.854 ft/sec ----negative, so s is getting shorter with time also.

Therefore, the distance of the player from the home base at that instant is decreasing at the rate of 8.854 ft/sec. ---------answer.