(Another related rate question)

A plane (alt. 4000ft) is flying west at 700ft/sec. A searchlight under its path, tracks it. How fast if the light pivoting when the plane is 1000ft east?

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Kind of similar to the lamp post question I posted earlier. I know dp/dt, 700ft/sec. We're looking for the rate of the light, dl/dt. 4000ft is a constant. I also think p would be -1000ft, but I'm not quite sure.

Again, I'm not quite sure about the equation. I do know, based on my picture, that there will be trig functions involved, based on the angle of the searchlight moving to track the plane. Help again?

Thanks!

To solve this problem, we can use similar trigonometric functions like in the lamp post question. Let's denote the distance of the searchlight from the point directly under the plane as "d" (in feet). We are given that the plane is flying west at a constant speed of 700 ft/sec, so the rate of change of the plane's x-coordinate is -700 feet per second.

To find the rate of change of the light pivoting, or dl/dt, we need to find a relationship between d and the x-coordinate of the plane.

Since the searchlight is tracking the plane, it forms a right triangle with the vertical altitude of the plane and the horizontal distance between the point directly under the plane and the light. Let's call the x-coordinate of the plane "p" (in feet). From the information given, when the plane is 1000 feet east, p = -1000 ft.

Using the Pythagorean theorem, we can relate p, d, and the altitude of the plane (4000 ft):

p^2 + d^2 = 4000^2

Differentiating both sides of the equation with respect to time t:

(2p)(dp/dt) + (2d)(dd/dt) = 0

Since we're looking for dl/dt, we need to express dl/dt in terms of p, d, and dp/dt. To do this, we'll need to involve trigonometric ratios.

We can write that dl/dt is related to dp/dt and the angle θ between the altitude and the line connecting the light to the plane by:

tan(θ) = d / p

Differentiating both sides of the equation with respect to time t:

sec^2(θ) (dθ/dt) = (dp/dt)(p) / (d^2 + p^2)

Rearranging the equation to solve for dθ/dt (the rate at which the angle θ changes):

dθ/dt = [(dp/dt)(p) / (d^2 + p^2)] * (1 / sec^2(θ))

Now we can substitute known values into the equation. We have dp/dt = -700 ft/sec, p = -1000 ft, and the known constant altitude of the plane (4000 ft). We are given that the plane is 1000 ft east, so d = 0.

Plugging in these values:

dθ/dt = [(dp/dt)(p) / (d^2 + p^2)] * (1 / sec^2(θ))
= [(-700)(-1000) / (0^2 + (-1000)^2)] * (1 / sec^2(θ))
= 700000 / 1000000 * (1 / sec^2(θ))
= 0.7 * (1 / sec^2(θ))

Now, we need to find sec^2(θ). In the right triangle formed, sec(θ) is the ratio of the hypotenuse (4000 ft) to the altitude (4000 ft). So, sec^2(θ) = (4000/4000)^2 = 1.

Plugging this value back into the equation:

dθ/dt = 0.7 * (1 / sec^2(θ))
= 0.7 * (1/1)
= 0.7

Thus, the rate at which the light pivots when the plane is 1000 ft east is 0.7 radians per second.