(1 point) an airplane, flying at 500 km/hr at a constant altitude of 5 km is approaching a camera mounted on the ground. let θθ be the angle of elevation above the ground at which the camera is pointed. when this angle is equal to π/4π/4, how many radians per hour is the rate of change for the camera's angle? give your answer in radians/hr.

[tex]\frac{d}{d\theta}\left (tan\theta \right ) =\frac{d}{d\theta}\left (\frac{h}{b} \right )\\\\sec^2\theta\frac{d\theta}{dt}=\frac{b\frac{dh}{dt}-h\frac{db}{dt}}{b^2}[/tex]

We have

h = 5 km

[tex]\frac{dh}{dt}=0\\\\\frac{db}{dt}=-500km/hr\\\\\theta =\frac{\pi }{4}\\\\tan\frac{\pi }{4}=\frac{h}{b}\Rightarrow 1=\frac{5}{b}\Rightarrow b=5km[/tex]

Substituting

[tex]sec^2\theta\frac{d\theta}{dt}=\frac{b\frac{dh}{dt}-h\frac{db}{dt}}{b^2}\\\\sec^2\frac{\pi }{4}\frac{d\theta}{dt}=\frac{5\times 0-5\times (-500)}{5^2}\\\\2\frac{d\theta}{dt}=\frac{1000}{25}\\\\2\frac{d\theta}{dt}=40\\\\\frac{d\theta}{dt}=20rad/hr[/tex]

The rate of change for the camera's angle = 20 rad/hr

xero099 xero099 · Answer 2 · 2021-11-14T04:30:04+01:00

Based on information from statement, we create a geometric diagram and derive a trigonometric formula for the angle of the camera:

[tex]\tan \theta = \frac{h}{x}[/tex] (1)

Where:

[tex]\theta[/tex] - Angle of elevation of the camera, in radians.
[tex]h[/tex] - Altitude of the airplane, in kilometers.
[tex]x[/tex] - Horizontal distance of the airplane with respect to camera, in kilometers.

Now we obtain an expression for the rate of change for the angle of the camera ([tex]\dot \theta[/tex]), in radians per hour, by differential calculus:

[tex]\sec^{2}\theta \cdot \dot {\theta} = -h\cdot x^{-2}\cdot \dot {x}[/tex]

[tex]\dot {\theta} = -h\cdot \cos^{2}\theta \cdot x^{-2}\cdot \dot {x}[/tex] (2)

If we know that [tex]h = 5\,km[/tex], [tex]\theta = \frac{\pi}{4}[/tex] and [tex]\dot x = -500\,\frac{km}{h}[/tex], then the rate of change of the angle of the camera is:

[tex]x = \frac{h}{\tan \theta}[/tex]

[tex]x = \frac{5\,km}{\tan \frac{\pi}{4} }[/tex]

[tex]x = 5\,km[/tex]

[tex]\dot \theta = -(5\,km)\cdot \left(\cos^{2} \frac{\pi}{4} \right)\cdot (5\,km)^{-2} \left(-500\,\frac{km}{h} \right)[/tex]

[tex]\dot \theta = 50\,\frac{rad}{h}[/tex]

The rate of change for the camera's angle is 50 radians per hour.

We kindly invite to check this question on rates of change: https://brainly.com/question/20414862

Note - The statement has typing errors. Correct form is presented below:

An airplane, flying at 500 kilometers per hour at a constant altitude of 5 kilometers is approaching a camera mounted on the ground. Let [tex]\theta[/tex] be the angle of elevation above the ground at which the camera is pointed. When this angle is equal to [tex]\frac{\pi}{4}[/tex] radians, how many radians per hour is the rate of change for the camera's angle? Give your answer in radians per hour.