Question

An airplane pilot wants to climb from an altitude of 6000 feet to 16,000 feet. If the plane climbs at an angle of 9 degrees, with a constant speed of 22,000 feet per minute, how long will it take to reach 16,000 feet?

Answer 1

So, the pilot goes from 6,000 to 16,000 or distance up of 10,000 feet
the plane is going at a speed of 22,000 per minute, namely, dx/dt = 22,000

so. .if you look at the picture below... we can simply use tangent
keeping in mind that, as the plane moves up and up, the angle is "constant", is never changing, whilst "y" and "x" and even "r" are changing, but we're not using "r" in this case.... we know what dx/dt is, we know the angle is a constant, this matters because the derivative of a constant is just 0

so, we really just need to find dy/dt then, because whilst dx/dt is telling us how fast is moving horizontally, dy/dt is telling us how long it took to cover those 10,000 feet

thus  $\bf tan(\theta)=\cfrac{y}{x}\implies tan(\theta)=yx^{-1}\implies tan(9^o)=yx^{-1}\\\\ -----------------------------\\\\ 0=\left( \cfrac{dy}{dt}\cdot \cfrac{1}{x} \right)+\left( y\cdot -\cfrac{1}{x^2}\cdot \cfrac{dx}{dt} \right)\implies 0=\cfrac{\frac{dy}{dt}}{x}-\cfrac{y\frac{dx}{dt}}{x^2}$

$\bf \\\\\\ \cfrac{y\frac{dx}{dt}}{x^2}=\cfrac{\frac{dy}{dt}}{x}\implies \cfrac{y\frac{dx}{dt}}{x}=\cfrac{dy}{dt} \\\\\\ \textit{so, what's "x" when y = 10000?} \\\\\\ tan(9^o)=\cfrac{10000}{x}\implies x=\cfrac{10000}{tan(9^o)}\implies x\approx 63175.5$

once you find what dy/dt is... you can simply do a quick conversion for the 10,000 feet, to see how many minutes it took