Angle of Elevation Related Rates help

[B3NR4Y]

I need help with this problem in my calculus book.
An airplane at an altitude of 10,000 feet is flying at a constant speed on a line which will take it directly over an observer on the ground. If, at a given instant, the observer notes that the angle of elevation of the airplane is 60 degrees and is increasing at a rate of one degree a second, find the speed of the airplane. (Hint: tangent of theta is equal to sine of theta over cosine of theta)

I've worked it and had tangent of theta equals "x" over 10,000. And took the derivative of each side to get secant squared of theta equals one over 10,000 times dx/dt. I solve for dx/dt with my information and get 40,000 feet per second. However the boom says 10,000*pi over 135.

I don't get where the pi comes from.

Sorry for no LaTeX, I'm on my phone.

 

[voko]

The full derivative of the angle term must also include the derivative of the angle, i.e., the angular speed. Whose value is given in degrees per second, but it must be converted to the radian measure.

 

[B3NR4Y]

Thank you. The reason I didn't include it was because it was 1 degree a second, but what you're saying is I should convert all angles to radian? I'll try that...

 

[B3NR4Y]

I still can't seem to get it. Can someone tell me how the book got that answer? I don't know why it would give me the hint it did if I don't really need it.

 

[voko]

What is the formula you got?

 

[B3NR4Y]

If you want to know what hint they gave: The tangent of theta being equal to the sine of theta over the cosine of theta.

If you want to know the answer I got: (40,000*pi)/180
I know it isn't right , so I didn't feel the need to simplify. The formula I get for the rate x changes with respect to time is:
dx/dt = pi/180 * sec^2 (pi/3)*10,000

 

[B3NR4Y]

If you want to know what hint they gave: The tangent of theta being equal to the sine of theta over the cosine of theta.

If you want to know the answer I got: (40,000*pi)/180
I know it isn't right , so I didn't feel the need to simplify. The formula I get for the rate x changes with respect to time is:
dx/dt = pi/180 * sec^2 (pi/3)*10,000

\begin{equation}
\frac{dx}{dt} = \frac{\pi sec^{2}(\frac{\pi}{3}) 10,000}{180}
\end{equation}

\begin{equation}
tan \theta =\frac{x}{10000}
\end{equation}
\begin{equation}
\frac{d}{dt} (tan \theta) = \frac{d}{dt} (\frac{x}{10000})
\end{equation}
\begin{equation}
sec^{2} \theta \frac{d\theta}{dt} = \frac{1}{10000} \frac{dx}{dt}
\end{equation}
\begin{equation}
10,000 (sec^{2} (\frac{\pi}{3}) \frac{\pi}{180}) = \frac{dx}{dt}
\end{equation}
\begin{equation}
10,000 (\frac{4\pi}{180})
\end{equation}
\begin{equation}
\frac{40,000\pi}{180}
\end{equation}
\begin{equation}
\frac{2,000\pi}{9}
\end{equation}

Where have I steered wrong

 

[B3NR4Y]

Oh, and the hint they gave was

\begin{equation}
tan\theta = \frac{sin\theta}{cos\theta}
\end{equation}

 

[voko]

$\tan \theta = \frac x {10000}$ is wrong. The correct formula is $\tan \theta = \frac {10000} x$.

 

[B3NR4Y]

So
\begin{equation}
sec^{2} \theta \frac{d\theta}{dt} = -\frac{10,000}{x^{2}} \frac{dx}{dt}
\end{equation}

 

[voko]

Yes. Eliminate x from the expression and then find v. Alternatively, you could use cot x = x/10000.

 

[B3NR4Y]

Thanks :D I got the correct answer. You are a calculus god, have the ceremonial loin cloth.