Question-162673

Question Number 162673 by mkam last updated on 31/Dec/21

Answered by mr W last updated on 31/Dec/21

(i)×2−(ii): −3x=3t+2 ⇒x=−t−(2/3) −1+t+(2/3)+(dy/dt)=2t+1 (dy/dt)=t+(4/3) ⇒y=(t^2 /2)+((4t)/3)+C

$$\left({i}\right)×\mathrm{2}−\left({ii}\right): \\ $$$$−\mathrm{3}{x}=\mathrm{3}{t}+\mathrm{2}\:\Rightarrow{x}=−{t}−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$−\mathrm{1}+{t}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{{dy}}{{dt}}=\mathrm{2}{t}+\mathrm{1} \\ $$$$\frac{{dy}}{{dt}}={t}+\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow{y}=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{4}{t}}{\mathrm{3}}+{C} \\ $$

Commented by mkam last updated on 31/Dec/21

put sir how x = − (2/3) and (dx/dt) = 0 ? can you clearly that please ?

$${put}\:{sir}\:{how}\:{x}\:=\:−\:\frac{\mathrm{2}}{\mathrm{3}}\:{and}\:\frac{{dx}}{{dt}}\:=\:\mathrm{0}\:?\:{can}\:{you}\:{clearly}\:{that}\:{please}\:? \\ $$

Commented by mkam last updated on 31/Dec/21