Find the length of the curve given by \begin{cases} x(t) = e^{4t}\cos t,\\ y(t) = e^{4t}\sin t. \end{cases} for $0\leqslant t \leqslant 2$.
How do I approach this sort of question?
Find the length of the curve $x =e^{4t}\cos t , y =e^{4t}\sin t$ for $0\leqslant t \leqslant 2$
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multivariable-calculus
2 Answers
2
We know that the arc lengths of plane curves $x $ and $y $ when expressed in parametric form as a function of $t$ is given by: $$l = \int_{t_0}^{t_1} \sqrt {(\frac {dx}{dt})^2 + (\frac {dy}{dt})^2} \mathrm {d}t $$
Here $\frac {dx}{dt} = e^{4t}(-\sin t) + 4e^{4t}(\cos t) $ and $\frac {dy}{dt} = e^{4t}(\cos t) + 4e^{4t}(\sin t) $
Thus, $$l = \int_{0}^{2} e^{4t}\sqrt {17} \mathrm {d}t$$ $$ = \frac {\sqrt {17}}{4} (e^8-1) $$
Hope it helps.
1
Hint. Use : $$\ell=\int_0^2\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$