am i correct? calculus

check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer.  That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

[tex]\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y \\\\\\ \stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt} \\\\\\ \boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\ -------------------------------\\\\ [/tex]

[tex]\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\ -------------------------------\\\\ \cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15} \\\\\\ \cfrac{d\theta }{dt}=\cfrac{11}{75}[/tex]