Question

A plane flying horizontally at an altitude of 2 mi and a speed of 540 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station. (Round your answer to the nearest whole number.)

Answer 1

Answer:

The plane's distance from the radar station will increase about 8 miles per minute when it is 5 miles away from it.

Step-by-step explanation:

When the plane passes over the radar station, the current distance is the altitude h = 2. Then it moves b horizontally so that the distance to the station is 5. We can form a rectangle triangle using b, h and the hypotenuse 5. Therefore, b should satisfy

h²+b² = 5², since h = 2, h² = 4, as a result

b² = 25-4 = 21, thus

b = √21.

Since it moved √21 mi, then the time passed is √21/540 = 0.008466 hours, which is 0.51 minutes. Note that in 1 minute, the plane makes 540/60 = 9 miles.

The distance between the plane and the radar station after x minutes from the moment that the plane passes over it is given by the function

$f(x) = \sqrt{((9x)^2 + 2^2)} = \sqrt{(81x^2+4)}$

We have to compute the derivate of f in x = 0.51. The derivate of f is given by

$f'(x) = \frac{162x}{2\sqrt{81x^2+4}}$

also,

$f'(0.51) = \frac{162*0.51}{2\sqrt{81*0.51^2+4}} = 8.2486$

The plane's distance from the station will increase about 8 miles per minute.