Mathematics

Question

Find the distance traveled by a particle with position (x, y) as t varies in the given time interval. x = 5 sin2 t, y = 5 cos2 t, 0 ≤ t ≤ 3π

2 Answer

  • [tex]\begin{cases}x(t)=5\sin2t\\y(t)=5\cos2t\end{cases}\implies\begin{cases}\frac{\mathrm dx}{\mathrm dt}=10\cos2t\\\frac{\mathrm dy}{\mathrm dt}=-10\sin2t\end{cases}[/tex]

    The distance traveled by the particle is given by the definite integral

    [tex]\displaystyle\int_C\mathrm dS=\int_0^{3\pi}\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]

    where [tex]C[/tex] is the path of the particle. The distance is then

    [tex]\displaystyle\int_0^{3\pi}\sqrt{100\cos^22t+100\sin^22t}\,\mathrm dt=10\int_0^{3\pi}\mathrm dt=30\pi[/tex]
  • The distance travelled by the particle is [tex]30\pi[/tex]

    The distance travelled by the particle is the same as the arc length as [tex]t[/tex] varies within the interval [tex]0\le t\le3\pi[/tex]. It is given by

    [tex]d=\int^{3\pi}_{0}{\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}}[/tex]

    since the trajectory of the particle is

    [tex]x = 5sin2 t, y = 5cos2 t[/tex]

    we can substitute, and simply to get the distance

    [tex]d=\int^{3\pi}_{0}{\sqrt{(10cos2t)^2+(-10sin2t})^2}dt}\\=\int^{3\pi}_{0}{\sqrt{100(cos^2(2t)+sin^2(2t)})}dt}\\=10\int^{3\pi}_{0}{dt}\\=10\times3\pi-10\times0=30\pi[/tex]

    Learn more about arc length here: https://brainly.com/question/16229252

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