A calculus problem by Pallav Doshi

$\begin{aligned} y^2 &= x^2 + 1 \\ \implies (\frac{1250}{1000})^2 &= x^2 + 1 \\ \implies x &= \frac{3}{4} \\ \end{aligned}$

Differentiating $y^2 = x^2 + 1$ wrt to time $t$

$\begin{aligned} 2y\frac{dy}{dt} &= 2x\frac{dx}{dt} + 0 \\ \implies \frac{dy}{dt} &= \frac{x}{y} \cdot \frac{dx}{dt} \\ \implies v_y &= \frac{\frac{3}{4}}{\frac{5}{4}} \cdot 800 \\ \implies \boxed{v_y = 480} \end{aligned}$

draw a line between the aircraft and the observer when it is 1250 meters away from him, and a vertical line between the aircraft and the ground. Now a triangle is formed. By applying Pythagoras theorem to this triangle,

dis^{2}=x^{2}+(1)^{2}

Where x is the horizontal distance between the observer and the aircraft.

'dis' is the distance variable.

Differentiate the above equation, and substitute the known values and get d(dis)/dt....