Calculus - angle of elevation

im preety sure there is no answer to this :(

Hi LambLamb,

I am assuming that you are on the ground and that the plane is flying in your compas direction but that it ia maintainng a height of 5km

\(\frac{d\theta}{dt}=\frac{d\theta}{dx}\times \frac{dx}{dt}\\ \text{now }\frac{dx}{dt} \text{ is the velocity of the plane so it equals 600km/hour so}\\ \frac{d\theta}{dt}=\frac{d\theta}{dx}\times -600\\~\\ \)

At any point in time,  

\(sin\theta=\frac{height\; of \;plane}{x}\\ sin\theta=\frac{5}{x}\\ x=\frac{5}{sin\theta}\qquad \text{except when }sin\theta =0\\ x=5*({sin\theta})^{-1}\\ \frac{dx}{d\theta}=-5(sin\theta)^{-2}*cos\theta\\ \frac{dx}{d\theta}=\frac{-5cos\theta}{(sin\theta)^2}\\ \frac{d\theta}{dx}=\frac{-(sin\theta)^2}{5cos\theta}\\\)

SO

\(\frac{d\theta}{dt}=\frac{d\theta}{dx}*-600\\ \frac{d\theta}{dt}=\frac{-(sin\theta)^2}{5cos\theta}*-600\\ When\;\theta=30^\circ\\ \frac{d\theta}{dt}=[-(sin30^\circ)^2 \div 5cos30^\circ]*-600\\ \frac{d\theta}{dt}=[\frac{-1}{4} \div \frac{5\sqrt3}{2}]*-600\\ \frac{d\theta}{dt}=[\frac{-1}{4} \times \frac{2}{5\sqrt3}]*-600\\ \frac{d\theta}{dt}=[ \frac{-1}{10\sqrt3}]*-600\\ \frac{d\theta}{dt}= \frac{60}{\sqrt3}\;degrees/ hour\\ \frac{d\theta}{dt}=\frac{1}{\sqrt3}\;degrees/ minute\\\)

1 /sqrt(3) = 0.5773502691896258

When the angle ofelevation is 30 degrees it (the angle) is increasing at the rate of approx 0.577 degrees per minute 

Thank you so much, Melody. This answer is perfect. I forgot to mention the altitude remains constant and dx/dt is in fact -600km/h, but you safely assumed those.