Answer:
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.
Explanation:
Geometrically speaking, the distance between the rocket and the observer ([tex]r[/tex]), measured in kilometers, can be represented by a right triangle:
[tex]r = \sqrt{x^{2}+y^{2}}[/tex] (1)
Where:
[tex]x[/tex] - Horizontal distance between the rocket and the observer, measured in kilometers.
[tex]y[/tex] - Vertical distance between the rocket and the observer, measured in kilometers.
The angle of elevation of the rocket ([tex]\theta[/tex]), measured in sexagesimal degrees, is defined by the following trigonometric relation:
[tex]\tan \theta = \frac{y}{x}[/tex] (2)
If we know that [tex]x = 5\,km[/tex], then the expression is:
[tex]\tan \theta = \frac{y}{5}[/tex]
And the rate of change of this angle is determined by derivatives:
[tex]\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y[/tex]
[tex]\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}[/tex]
[tex]\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}[/tex]
[tex]\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}[/tex]
Where:
[tex]\dot \theta[/tex] - Rate of change of the angle of elevation, measured in sexagesimal degrees.
[tex]\dot y[/tex] - Vertical speed of the rocket, measured in kilometers per hour.
If we know that [tex]y = 4\,km[/tex] and [tex]\dot y = 400\,\frac{km}{h}[/tex], then the rate of change of the angle of elevation is:
[tex]\dot \theta = 48.780\,\frac{\circ}{s}[/tex]
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.
