UPSC NDA Math Model Paper 3

Let OC be the wall as shown in the figure below. At a certain instant t, let AB be the position of the ladder such that OA = x meter and OB = y meter. Given: dx/dt = 2 m/sec From the figure given above we can see that: In right angle triangle Δ AOB, $\Rightarrow x^{2}+y^{2}=5^{2}=25$ .......-(By pythagoras theorem) Now by differentiating the above equation with respect t we get $\Rightarrow 2 x \frac{\dot{d x}}{d t}+2 y \frac{\dot{d y}}{d t}=0$ Now by substituting dx/dt = 2 m/sec in the above equation we get, $\Rightarrow 2 x \dot{2}+2 y \frac{\dot{d y}}{d t}=0$ $\Rightarrow \frac{dy}{dt}=-\frac{2x}{y}$ Now when $x=4 \Rightarrow y=\sqrt{5^{2}-4^{2}}=3$ By substituting the value of $x$ and $y$ in the equation $\Rightarrow \frac{dy}{dt}=-\frac{2x}{y}$ we get $\Rightarrow \frac{dy}{dt}=-\frac{2 \times 4}{3}=-\frac{8}{3} \text{m}/\text{sec}$ Hence, the required rate of decrease in the height of the ladder on the wall is: 8/3 m/sec