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Question Number 174214 by Best1 last updated on 27/Jul/22

Answered by mr W last updated on 27/Jul/22

x^2 +y^2 =l^2   2x(dx/dt)+2y(dy/dt)=0  ⇒(dx/dt)=−(y/x)×(dy/dt)  with (dy/dt)=0.2 m/s and x=9m ⇒y=12m  ⇒(dx/dt)=−((12)/9)×0.2=−0.26 m/s

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={l}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}\frac{{dx}}{{dt}}+\mathrm{2}{y}\frac{{dy}}{{dt}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{dx}}{{dt}}=−\frac{{y}}{{x}}×\frac{{dy}}{{dt}} \\ $$$${with}\:\frac{{dy}}{{dt}}=\mathrm{0}.\mathrm{2}\:{m}/{s}\:{and}\:{x}=\mathrm{9}{m}\:\Rightarrow{y}=\mathrm{12}{m} \\ $$$$\Rightarrow\frac{{dx}}{{dt}}=−\frac{\mathrm{12}}{\mathrm{9}}×\mathrm{0}.\mathrm{2}=−\mathrm{0}.\mathrm{26}\:{m}/{s} \\ $$

Commented by Best1 last updated on 27/Jul/22

no answer from the choose?   A.(1/4)       B.(4/(15))         C.(1/3)     D.(3/4)

$${no}\:{answer}\:{from}\:{the}\:{choose}?\: \\ $$$${A}.\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:{B}.\frac{\mathrm{4}}{\mathrm{15}}\:\:\:\:\:\:\:\:\:{C}.\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:{D}.\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Commented by mr W last updated on 27/Jul/22

(4/(15))=0.26

$$\frac{\mathrm{4}}{\mathrm{15}}=\mathrm{0}.\mathrm{26} \\ $$

Commented by Tawa11 last updated on 27/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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