Question

A rocket is launched vertically and is tracked by a radar station located on the ground 5 km from the launch pad. Suppose that the elevation angle x of the line of sight to the rocket is increasing at 3 degrees per second when x______.

a) 5 km
b) 3 degrees
c) 1 km
d) 0.1 degrees

Answer 1

The elevation angle x of the line of sight to the rocket is increasing at 3 degrees per second when x is at d) 0.1 degrees.

Step-by-step explanation:

The relationship between the distance from the radar station (d), the elevation angle (x), and the rate of change of x with respect to time
`(\( (dx)/(dt) \))` in a right-angled triangle can be described by the tangent function:

`\[ \tan(x) = (h)/(d) \]`

where h is the height of the rocket. We are given that
`\( (dx)/(dt) = 3 \)` degrees per second. To find x when
`\( (dx)/(dt) = 3 \)` we can use the arctangent function:

`\[ x = \tan^(-1)\left((h)/(d)\right) \]`

Now, we want to find x when
`\( (dx)/(dt) = 3 \)`. If x is increasing, it means the rocket is moving away from the radar station. At this point, x is relatively small, so we can make the approximation
`\( \tan(x) \approx x \)` for small x. Therefore:

`\[ 3 \approx (h)/(d) \]`

Given that d = 5 km, we can solve for h:

`\[ h \approx 3 * 5 = 15 \]`

Now, we substitute h = 15 and d = 5 into the tangent function to find x:

`\[ x = \tan^(-1)\left((15)/(5)\right) = \tan^(-1)(3) \]`

This is approximately 0.1 degrees. Hence, the correct answer is d) 0.1 degrees.