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Question Number 207901 by Tawa11 last updated on 29/May/24

Answered by mr W last updated on 30/May/24

v=(dr/dt)=e^t (cos t−sin t) i+e^t (sin t+ cos t) j a=(dv/dt)=−2e^t sin t i+2e^t cos t j ∣a∣=(√((−2e^t cos t)^2 +(2e^t sin t)^2 ))=2e^t

$$\boldsymbol{{v}}=\frac{{d}\boldsymbol{{r}}}{{dt}}={e}^{{t}} \left(\mathrm{cos}\:{t}−\mathrm{sin}\:{t}\right)\:\boldsymbol{{i}}+{e}^{{t}} \left(\mathrm{sin}\:{t}+\:\mathrm{cos}\:{t}\right)\:\boldsymbol{{j}} \\ $$$$\boldsymbol{{a}}=\frac{{d}\boldsymbol{{v}}}{{dt}}=−\mathrm{2}{e}^{{t}} \:\mathrm{sin}\:{t}\:\boldsymbol{{i}}+\mathrm{2}{e}^{{t}} \:\mathrm{cos}\:{t}\:\boldsymbol{{j}} \\ $$$$\mid\boldsymbol{{a}}\mid=\sqrt{\left(−\mathrm{2}{e}^{{t}} \:\mathrm{cos}\:{t}\right)^{\mathrm{2}} +\left(\mathrm{2}{e}^{{t}} \:\mathrm{sin}\:{t}\right)^{\mathrm{2}} }=\mathrm{2}{e}^{{t}} \\ $$

Commented by Tawa11 last updated on 30/May/24

Thanks sir. I appreciate

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate} \\ $$

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