A revolving searchlight in a lighthouse 2 km offshore is following a car traveling slowly along the shore. When the car is 1 km from the point on the shore that is closest to the lighthouse, the searchlight is Page 305turning at the rate of 0.25 rev/hr. How fast in kph is the car traveling at this moment?

when the car is x km from the specified point,

tanθ = x/2
so that means that
sec^2θ dθ/dt = 1/2 dx/dt
So now just find secθ when x=1, and plug in your numbers to find dx/dt

Well, well, well, looks like we have a revolving searchlight trying to keep up with a car. It's like a game of cat and mouse, or should I say lighthouse and car? Anyway, let's shed some light on this situation. Pun intended.

We know that the searchlight is turning at a rate of 0.25 rev/hr. Now, since we're dealing with kilometers and hours, we need to convert that to kph.

To do that, we simply need to multiply the rate of revolution (0.25 rev/hr) by the circumference of a circle with a radius of 2 kilometers (since the lighthouse is 2 km offshore). The circumference can be calculated by multiplying 2π (that's 2 times pi, not apple pie) by the radius, which gives us 4π km.

So, the car is 1 km away from the point on the shore closest to the lighthouse, and the searchlight is turning at a rate of 4π km/hr. At this very moment, the car's speed must match the rotational speed of the searchlight in order for it to always remain illuminated.

Hence, the car is traveling at a speed of 4π kph. And let's be honest, that's a lot of pie!

I hope that brightened your day!

To solve this problem, we can use the concept of related rates. Let's assume that the lighthouse is located at the origin (0,0) and the point on the shore closest to the lighthouse is P(1,0).

1. First, let's denote the position of the car at any given time t as C(x, y), where x is the distance of the car from P and y is the distance of the car from the lighthouse.
2. Since the car is moving along the shore, we know that y is constant and equal to 2 km (since the lighthouse is 2 km offshore).
3. We want to find how fast the car is traveling when it is 1 km from P, so we need to find dx/dt at x = 1.

Now, let's differentiate both sides of the equation x^2 + y^2 = d^2, where d is the distance between the car and the lighthouse (which remains constant at 2 km):

d/dt(x^2 + y^2) = d/dt(d^2)

2x(dx/dt) = 0

4. Simplifying the equation, we have:

2x(dx/dt) = 0

dx/dt = 0 / (2x)

dx/dt = 0

5. Since dx/dt = 0 when x = 1 (the car is 1 km from P), we need to find dy/dt (the rate at which the lighthouse is turning) when x = 1.

dy/dt = 0.25 rev/hr

6. To convert rev/hr to km/hr, we need to determine the circumference of the circle described by the searchlight.

The circumference of a circle is given by C = 2πr, where r is the radius.

Since the lighthouse is 2 km offshore, the radius is r = 2 km.

C = 2π(2) = 4π km = 4π * 1 km/hr = 4π km/hr

7. The car travels along a distance equal to the circumference of the circle in one hour, given that it is 1 km from P. Hence, the speed of the car is equal to the circumference of the circle when it is 1 km from P.

Speed of the car = 4π km/hr

Therefore, when the car is 1 km from the point on the shore that is closest to the lighthouse, the car is traveling at a speed of 4π km/hr.

To find the speed of the car at this moment, we can use the concept of related rates.

Let's denote the distance between the car and the lighthouse as "x" (in km) and the angle of rotation of the searchlight as "θ" (in radians). We are given that the searchlight is turning at a rate of 0.25 rev/hr.

First, let's find the relationship between 𝑥 and 𝜃. We know that the revolving searchlight is 2 km offshore, which forms a right triangle with 𝑥 as the adjacent side and the hypotenuse as 2 km. By using Pythagoras' theorem, we can express the relationship as:

𝑥^2 + 1^2 = 2^2
𝑥^2 + 1 = 4
𝑥^2 = 3

Differentiating both sides of this equation with respect to time (𝑡), we can find the rate of change of 𝑥 with respect to 𝑡:

2𝑥 * 𝑑𝑥/𝑑𝑡 = 0
𝑑𝑥/𝑑𝑡 = 0 / (2𝑥)
𝑑𝑥/𝑑𝑡 = 0

Since the distance between the car and the lighthouse is not changing, the rate of change of 𝑥 with respect to 𝑡 is 0 at this moment.

Next, we need to find the relationship between 𝜃 and 𝑡. We are given that the searchlight is turning at a rate of 0.25 rev/hr. One revolution is equivalent to 2π radians. Therefore, the rate of change of 𝜃 with respect to 𝑡 is:

𝑑𝜃/𝑑𝑡 = 0.25 * 2π

Finally, we can find the rate of change of 𝑥 with respect to 𝑡 (𝑑𝑥/𝑑𝑡) using the chain rule:

𝑑𝑥/𝑑𝑡 = 𝑑𝑥/𝑑𝜃 * 𝑑𝜃/𝑑𝑡

Since we already found 𝑑𝑥/𝑑𝑡 to be 0, the rate of change of 𝑥 with respect to 𝑡 is definitely 0 km/hr. Therefore, the car is not moving at this moment.

To summarize, the car is not moving (0 kph) at the moment when it is 1 km from the point on the shore closest to the lighthouse, and the searchlight is turning at a rate of 0.25 revolutions per hour.