A plane flying with a constant speed of 14 km/min passes over a ground radar station at an altitude of 7 km and climbs at an angle of 45 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 2 minutes later?

I know you use law of cosines.

Write a formula for distance at any given time.

The derivate of this distance is the rate at which distance increases.

Draw the triangle. One leg vertical, 7km. The other two legs are from the top of the vertical @45 det to the plane (distance is 28km), the last leg is from the observer to the plane.

Figure by the law of cosines the distance from the observer to the plane.

Now, take derivative of the written law of cosines. You will have all you need. I will be happy to critique your work.

Well, I can definitely try to help you with that, but let's be honest, I'm more of a clown than a mathematician. So, brace yourself for some math humor!

Now, to get the formula for distance at any given time, let's call the distance from the radar station to the plane "d." We can use the Pythagorean theorem, but since we have an angle involved, let's make it the "Pythagorean Clown Theorem" just to keep things interesting.

According to the "Pythagorean Clown Theorem," we have the equation:

d² = 7² + (14t)² - 2 * 7 * 14t * cos(45),

where "t" represents time in minutes. Now, we want to find the rate at which distance is increasing, which is basically the derivative of "d" with respect to "t." So, let's put on our clown hats and differentiate this equation!

Taking the derivative of both sides with respect to "t," we get:

2d * d/dt = 0 + 2 * 14²t - 2 * 7 * 14 * cos(45) * 1.

Simplifying this a bit (hey, even clowns can simplify), we have:

2d * d/dt = 196t - 98√2.

Now, let's solve for d/dt, the rate at which distance is increasing:

d/dt = (196t - 98√2) / (2d).

But hold on a second, we want to find the rate at which distance is increasing 2 minutes later, right? So, let's plug in t = 2 into our equation:

d/dt = (196 * 2 - 98√2) / (2d)
= (392 - 98√2) / (2d).

And there you have it! The rate at which the distance from the plane to the radar station is increasing 2 minutes later is (392 - 98√2) / (2d) km/min. Just make sure to plug in the appropriate values for "d" to get the final answer.

I hope that brought a smile to your face, even if the math didn't! If you have any more questions (math-related or not), feel free to ask, and I'll do my best to entertain you.

To find the rate at which the distance from the plane to the radar station is increasing, we can start by considering the triangle formed by the plane, the radar station, and the distance between them.

Let's call the distance from the radar station to the plane "d" at any given time.

Using the given information, we have a vertical leg of the triangle with a length of 7 km, and the angle between the vertical leg and the ground is 45 degrees. The other leg of the triangle can be found by using basic trigonometry, since we know that the plane is flying with a constant speed of 14 km/min.

In 2 minutes, the plane will have covered a horizontal distance of (14 km/min) x (2 min) = 28 km.

Therefore, the length of the leg opposite to the 45-degree angle is 28 km.

Now, let's apply the Law of Cosines to find the distance "d" between the radar station and the plane at any given time. The Law of Cosines states:

c^2 = a^2 + b^2 - 2ab*cos(C)

In our triangle, "a" is the vertical leg (7 km), "b" is the leg opposite to the 45-degree angle (28 km), and "C" is the angle between the vertical leg and the leg opposite to the 45-degree angle (90 degrees).

Plugging these values into the Law of Cosines, we get:

d^2 = 7^2 + 28^2 - 2(7)(28)*cos(90)

Simplifying the expression, we have:

d^2 = 49 + 784 - 392*cos(90)

Since the cosine of 90 degrees is 0, the expression becomes:

d^2 = 49 + 784 - 0

d^2 = 833

Taking the square root of both sides, we can find the value of "d":

d = sqrt(833)

Now, we need to find the derivative of "d" with respect to time to determine the rate at which the distance is increasing after 2 minutes.

Differentiating both sides of the equation, we have:

d/dt(d) = d/dt(sqrt(833))

The derivative of sqrt(833) is 0 because it is a constant value.

Therefore, the rate at which the distance from the plane to the radar station is increasing 2 minutes later is 0 km/min.