Respuesta :
Answer:
Step-by-step explanation:
To solve this problem, we can use trigonometry and related rates. Let's denote the angle of elevation as θ (theta), the distance between the airplane and the observer as x, and the altitude of the airplane as h.
Given that the airplane is flying directly overhead, forming a right triangle with the observer, we have:
\[ \tan(\theta) = \frac{h}{x} \]
Now, differentiate both sides of the equation with respect to time t:
\[ \sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{x}\frac{dh}{dt} - \frac{h}{x^2}\frac{dx}{dt} \]
Since we are given the altitude h (5 miles) and the speed of the plane (600 miles per hour), we can substitute these values and solve for \(\frac{d\theta}{dt}\) at different values of θ.
Let's go through the calculations for each part:
### Part (a) θ = 30°:
Given \( \tan(30°) = \sqrt{3}/3 \), we have:
\[ \sec^2(30°) = \frac{4}{3} \]
Substitute this into the equation:
\[ \frac{4}{3} \frac{d\theta}{dt} = \frac{1}{x}\frac{dh}{dt} - \frac{h}{x^2}\frac{dx}{dt} \]
Substitute \( h = 5 \) miles and \( x = h/\tan(30°) = 5/\frac{\sqrt{3}}{3} \):
\[ \frac{4}{3} \frac{d\theta}{dt} = \frac{1}{5/\frac{\sqrt{3}}{3}}\frac{dh}{dt} - \frac{5}{\left(\frac{\sqrt{3}}{3}\right)^2}\frac{dx}{dt} \]
Substitute \( \frac{dh}{dt} = 0 \) (since the altitude is constant):
\[ \frac{4}{3} \frac{d\theta}{dt} = -\frac{5}{\left(\frac{\sqrt{3}}{3}\right)^2}\frac{dx}{dt} \]
Substitute \( \frac{dx}{dt} = 600 \) miles per hour:
\[ \frac{4}{3} \frac{d\theta}{dt} = -\frac{5}{\left(\frac{\sqrt{3}}{3}\right)^2} \times 600 \]
Solve for \( \frac{d\theta}{dt} \):
\[ \frac{d\theta}{dt} = -\frac{5}{4} \times \frac{600}{\left(\frac{\sqrt{3}}{3}\right)^2} \]
### Part (b) θ = 60°:
Given \( \tan(60°) = \sqrt{3} \), we have:
\[ \sec^2(60°) = 4 \]
Substitute this into the equation:
\[ 4 \frac{d\theta}{dt} = \frac{1}{x}\frac{dh}{dt} - \frac{h}{x^2}\frac{dx}{dt} \]
Substitute \( h = 5 \) miles and \( x = h/\tan(60°) = 5/\sqrt{3} \):
\[ 4 \frac{d\theta}{dt} = \frac{1}{5/\sqrt{3}}\frac{dh}{dt} - \frac{5}{\left(\sqrt{3}\right)^2}\frac{dx}{dt} \]
Substitute \( \frac{dh}{dt} = 0 \) (since the altitude is constant):
\[ 4 \frac{d\theta}{dt} = -\frac{5}{\left(\sqrt{3}\right)^2}\frac{dx}{dt} \]
Substitute \( \frac{dx}{dt} = 600 \) miles per hour:
\[ 4 \frac{d\theta}{dt} = -\frac{5}{\left(\sqrt{3}\right)^2} \times 600 \]
Solve for \( \frac{d\theta}{dt} \):
\[ \frac{d\theta}{dt} = -\frac{5}{4} \times \frac{600}{\left(\sqrt{3}\right)^2} \]
### Part (c) θ = 75°:
Given \( \tan(75°) \), we have:
\[ \sec^2(75°) \]
Perform similar calculations as above to find \( \frac{d\theta}{dt} \) for this case.
Please note that the negative sign in the results indicates that the angle of elevation is decreasing.
