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Question Number 103400 by abony1303 last updated on 14/Jul/20

$$\mathrm{A}{\mathrm{particle}}^{\prime }\mathrm{s}\mathrm{trajectory}\mathrm{is}\mathrm{described}\mathrm{by}$$$$\mathrm{x}={\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}}\mathrm{y}=2\mathrm{t}$$$$\mathrm{Find}\mathrm{the}\mathrm{distance}\mathrm{that}\mathrm{the}\mathrm{particle}$$$$\mathrm{traveled}\mathrm{for}0⩽\mathrm{t}⩽2$$

Commented by abony1303 last updated on 14/Jul/20

$$\mathrm{Pls}\mathrm{help}$$

Answered by mr W last updated on 14/Jul/20

$$dx=(e^t-e^{-t})dt$$$$dy=2dt$$$$ds=\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}=\sqrt{{(e^t-e^{-t})}^2+4}dt$$$$=(e^t+e^{-t})dt$$$$s\left(t\right)={\int }_0^{t}ds={\int }_0^t(e^t+e^{-t})dt={[e^t-e^{-t}]}_0^t$$$$=e^t-e^{-t}$$$$\Rightarrow s\left(2\right)=e^2-\frac{1}{e^2}$$

Commented by abony1303 last updated on 14/Jul/20

$$\mathrm{thank}\mathrm{you}\mathrm{ser}.\mathrm{Can}\mathrm{you}\mathrm{pls}\mathrm{explain}\mathrm{why}$$$$\mathrm{you}\mathrm{differentiated}\mathrm{x}\mathrm{and}\mathrm{y},\mathrm{and}\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}$$$$\mathrm{understand}\mathrm{the}\mathrm{formula}\mathrm{in}3\mathrm{rd}\mathrm{raw}.$$

Commented by mr W last updated on 14/Jul/20

$$doyouknowthis:$$$$ds=\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}=\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$$

Commented by mr W last updated on 14/Jul/20

$$ifx=f\left(t\right),y=g\left(t\right),$$$$intimedttheobjectcoversa$$$$distanceinx-directiondx=f^{\prime }\left(t\right)dt$$$$andiny-directiondy=g^{\prime }\left(t\right)dt.the$$$$totaldistanceittravelsinthistime$$$$isds=\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}=\sqrt{{\left(f^{\prime }\left(t\right)\right)}^2+{\left(g^{\prime }\left(t\right)\right)}^2}dt$$

Answered by OlafThorendsen last updated on 14/Jul/20

$$0⩽t⩽2\iff 0⩽y⩽4$$$$x=e^t+e^{-t}=2\mathrm{ch}t=2\mathrm{ch}\frac{y}{2}$$$$\frac{dx}{dy}=2\times \frac{1}{2}\mathrm{sh}\frac{y}{2}=\mathrm{sh}\frac{y}{2}$$$$\mathrm{d}={\int }_0^4\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}dy$$$$\mathrm{d}={\int }_0^4\sqrt{1+{\mathrm{sh}}^2\frac{y}{2}}dy$$$$\mathrm{d}={\int }_0^4\mathrm{ch}\frac{y}{2}dy$$$$\mathrm{d}={\left[2\mathrm{sh}\frac{y}{2}\right]}_0^4=2\mathrm{sh}2=e^2-\frac{1}{e^2}$$

Answered by Dwaipayan Shikari last updated on 14/Jul/20

$$\frac{dx}{dt}=e^t-e^{-t}$$$$\frac{dy}{dt}=2$$$$Resultant=\sqrt{△x^2+△y^2}=\sqrt{{(e^t-e^{-t})}^2+4}=e^t+e^{-t}$$$$Timeisbounded(0,2)$$$${\int }_0^2{e}^t-e^{-t}dt=e^2-\frac{1}{e^2}\left(Resultanttracetory\right)$$

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