Question Number 174214 by Best1 last updated on 27/Jul/22
Answered by mr W last updated on 27/Jul/22
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={l}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}\frac{{dx}}{{dt}}+\mathrm{2}{y}\frac{{dy}}{{dt}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{dx}}{{dt}}=−\frac{{y}}{{x}}×\frac{{dy}}{{dt}} \\ $$$${with}\:\frac{{dy}}{{dt}}=\mathrm{0}.\mathrm{2}\:{m}/{s}\:{and}\:{x}=\mathrm{9}{m}\:\Rightarrow{y}=\mathrm{12}{m} \\ $$$$\Rightarrow\frac{{dx}}{{dt}}=−\frac{\mathrm{12}}{\mathrm{9}}×\mathrm{0}.\mathrm{2}=−\mathrm{0}.\mathrm{26}\:{m}/{s} \\ $$
Commented by Best1 last updated on 27/Jul/22
$${no}\:{answer}\:{from}\:{the}\:{choose}?\: \\ $$$${A}.\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:{B}.\frac{\mathrm{4}}{\mathrm{15}}\:\:\:\:\:\:\:\:\:{C}.\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:{D}.\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 27/Jul/22
$$\frac{\mathrm{4}}{\mathrm{15}}=\mathrm{0}.\mathrm{26} \\ $$
Commented by Tawa11 last updated on 27/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$