Question

The velocity of a particle moving in the x-y plane is given by $\frac{dx}{dt}=8\pi sin2\pi tand\frac{dy}{dt}=5\pi cos2\pi t.Att=0,x=8andy=0$, the path of the particle is:

• A. A straight line
• B. An ellipse
• C. A circle
• D. A parabola

Answer 1

The correct option is B An ellipse

The y-x graph gives the trajectory of the particle.
$\frac{dx}{dt}=8\pi sin2\pi t{\int }_8^{x}dx={\int }_0^{t}8\pi sin2\pi tdtx-8=-\frac{8\pi }{2\pi }[cos2\pi t{]}_0^{t}x-8=4(1-cos2\pi t)cos2\pi t=-\frac{x-12}{4}\frac{dy}{dt}=5\pi cos2\pi tdt{\int }_0^{y}dy=5\pi {\int }_0^{t}cos2\pi tdty=\frac{5}{2}sin2\pi t\Rightarrow sin2\pi t=\frac{y}{\frac{5}{2}}{\left(\frac{x-12}{4}\right)}^2+\frac{y^2}{(\frac{5}{2}{)}^2}=1\frac{(x-12{)}^2}{(4{)}^2}+\frac{y^2}{(\frac{5}{2}{)}^2}=1$
This is the equation of an ellipse.