1. A baseball diamond is a square with a distance of 90 feet between the bases. A runner is running from first base to second at the constant rate of 20 feet per second. How fast is the distance between the runner and the catcher changing when the runner is halfway from first to second?

2. Water is dripping slowly onto a metal countertop, creating a circular wet spot whose area is increasing at the constant rate of ÆÄÀÌ/2 square inches per second. How fast is its radius increasin g when the radius is 4 inches?

3. An observer stands 25 feet from the base of a 50 foot flagpole and watches a flag being lowered at a rate of 5ft/sec. Determine the rate at which the angle of elevation from the observer to the flag is changing at the instant that the flag is 25 feet.

Let the distance from home to the runner be d.

Let the distance from first base to the runner be x

d^2 = 90^2 + x^2
2d dd = 2x dx

when x = 45, d = 45 sqrt(5) = 100.62

201.24 dd = 90 (20)
dd = 1800/201.24 = 8.94 ft/s

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can't read 2nd problem - font garbling

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h/25 = tan(theta)
dh/25 = sec^2(theta) * dtheta

-5/25 = 2 * dtheta
-.1 = dtheta

For 2, I got 0.625 in/sec. dA/dt is pi/2.

1. To find the rate at which the distance between the runner and the catcher is changing, we can use the Pythagorean theorem. Let's call the distance between the runner and the catcher "x" and the distance the runner has traveled from first base to second base "d". Since the runner is halfway from first to second, d will be 45 feet.

We can set up a right triangle where the hypotenuse is the distance between the runner and the catcher (x) and one of the legs is the distance from the runner to first base (d). The other leg represents the distance from the catcher to first base, which is constant at 90 feet.

Using the Pythagorean theorem, the equation becomes x^2 = d^2 + 90^2. Differentiating both sides with respect to time, we get 2x(dx/dt) = 2d(dd/dt). We want to find dx/dt, which represents the rate at which the distance between the runner and the catcher is changing.

Plugging in the known values, we have x = sqrt(45^2 + 90^2) = sqrt(2025 + 8100) = sqrt(10125). Then, we can substitute this value into the equation, which becomes sqrt(10125)(dx/dt) = 45(20).

Simplifying further, dx/dt = (45 * 20) / sqrt(10125).

2. Let's assume the radius of the wet spot is "r" and the area is "A". We are given that the area is increasing at the constant rate of ÆÄÀÌ/2 square inches per second.

The formula for the area of a circle is A = πr^2. Differentiating both sides with respect to time, we get dA/dt = 2πr(dr/dt). We want to find dr/dt, which represents the rate at which the radius is increasing.

Plugging in the known value for dA/dt (ÆÄÀÌ/2) and the given radius of 4 inches, we have ÆÄÀÌ/2 = 2π(4)(dr/dt).

Simplifying further, dr/dt = ÆÄÀÌ / (16π) inches per second.

3. To find the rate at which the angle of elevation is changing, we can use trigonometry. Let's call the distance from the observer to the flag "x" and the height of the flagpole "h". We are given that the flag is being lowered at a rate of 5 feet per second.

Using the tangent function, tan(θ) = h / x, where θ represents the angle of elevation. Differentiating both sides with respect to time, we get sec^2(θ) dθ/dt = (dh/dt) / x.

We want to find dθ/dt, which represents the rate at which the angle of elevation is changing. Given that the flag is 25 feet, we can calculate the distance from the observer to the flag using the Pythagorean theorem as x = sqrt(25^2 + 50^2) = sqrt(625 + 2500) = sqrt(3125) = 5√5.

Plugging in the known values, we have sec^2(θ) dθ/dt = (5 / x)(-5).

Using trigonometric identities, sec^2(θ) = 1 + tan^2(θ), we can substitute it into the equation and get (1 + tan^2(θ)) dθ/dt = (5 / x)(-5).

Plugging in the known value for x (5√5), we have (1 + tan^2(θ)) dθ/dt = (-5 / 5√5)(-5).

Simplifying further, (1 + tan^2(θ)) dθ/dt = 1 / √5.

Finally, dθ/dt = (1 / √5(1 + tan^2(θ))).