Question

A rod 10 ft. moves with its ends A and B on two perpendicular lines OX and OY respectively. If A moves at the rate of 2 ft/sec when it is 8 ft. from O, find at what rate, end B is moving.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{x}\mathrm{be}\mathrm{the}\mathrm{distance}\mathrm{covered}\mathrm{by}\mathrm{end}\mathrm{A}\mathrm{along}\mathrm{OX}\mathrm{and}\mathrm{y}\mathrm{by}\mathrm{end}\mathrm{B}\mathrm{along}\mathrm{OY}\mathrm{at}\mathrm{any}\mathrm{instant}\mathrm{t}. \\ \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{rod}=10\mathrm{ft} \\ \mathrm{In}∆\mathrm{AOB}, \\ {\mathrm{OA}}^2+{\mathrm{OB}}^2={\mathrm{AB}}^2\mathbf{}\left[\mathbf{Pythagoras}\mathbf{}\mathbf{theorem}\right] \\ \Rightarrow {\left(10\right)}^2={\mathrm{y}}^2+{\mathrm{x}}^2 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=100....\left(1\right) \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{t},\mathrm{we}\mathrm{get} \\ 2\mathrm{x}\frac{\mathrm{dx}}{\mathrm{dt}}+2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dt}}=0 \\ \Rightarrow \mathrm{x}\frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dt}}=0 \\ \Rightarrow \mathrm{x}\times 2+\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dt}}=0\left[\mathbf{as}\mathbf{,}\mathbf{}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathbf{ft}\mathbf{/}\mathbf{s}\right] \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-\frac{2\mathrm{x}}{\mathrm{y}} \\ \mathrm{When}\mathrm{x}=8,\mathrm{then}\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}\mathrm{y}=6. \\ \mathrm{Now},\frac{\mathrm{dy}}{\mathrm{dt}}=-2\times \frac{8}{6}=-\frac{8}{3}\mathrm{ft}/\mathrm{s} \\ \mathrm{So},\mathrm{end}\mathrm{B}\mathrm{is}\mathrm{moving}\mathrm{with}\mathrm{the}\mathrm{speed}\mathrm{of}-\frac{8}{3}\mathrm{ft}/\mathrm{s}. \end{gathered}$$