Question

The velocity of a particle moving in the x-y plane is given by $\dfrac{{dx}}{{dt}} = 8\pi sin2\pi t\;$ and $\dfrac{{dy}}{{dt}} = 5\pi cos2\pi t$ where,$t = 0$ $x = 8$ and$y = 0$, the path of the particle is
(A) A straight line
(B) An ellipse
(C) A circle
(D) A parabola

Answer 1

We proceed to solve this question by integrating $\dfrac{{dx}}{{dt}} = 8\pi sin2\pi t\;$ and $\dfrac{{dy}}{{dt}} = 5\pi cos2\pi t$ with the given limits to find $x$ and$y$. From the equation of $x$ we make $cos2\pi t$ the subject and in the equation of $y$ we make $sin2\pi t$ the subject. We add and square these two equations and compare it to see which equation of curve or straight line it fits in from the given options.

Complete step by step solution:
Integrating $\dfrac{{dx}}{{dt}} = 8\pi sin2\pi t\;$ to get $x$
$\int\limits_8^x {dx} = \int\limits_0^t {8\pi sin2\pi t}$
$\Rightarrow x - 8 = - \dfrac{{8\pi }}{{2\pi }}\left[ {cos2\pi } \right]_0^t$
$\Rightarrow x - 8 = 4(1 - cos2\pi t)$
$\Rightarrow cos2\pi t = \dfrac{{x - 12}}{4}$……….. (1)

Integrating $\dfrac{{dy}}{{dt}} = 5\pi cos2\pi t$ to get $y$

$\int\limits_0^y {dy} = 5\pi t\int\limits_0^t {cos2\pi t}$
$\Rightarrow y = \dfrac{5}{2}sin2\pi t$
$\Rightarrow sin2\pi t = \dfrac{y}{{(\dfrac{5}{2})}}$ .......(2)

From trigonometry
$co{s^2}\theta + si{n^2}\theta = 1$
$\therefore co{s^2}(2\pi t) + si{n^2}(2\pi t) = 1$
Squaring and adding equations (1) and (2) we get

${(\dfrac{{x - 12}}{4})^2} + (\dfrac{y}{{(\dfrac{5}{2})}}) = 1$
$\dfrac{{{{(x - 12)}^2}}}{{{4^2}}} + \dfrac{{{y^2}}}{{{{(\dfrac{5}{2})}^2}}} = 1$……… (3)
The equation of an ellipse is given as $\dfrac{{{{(x - h)}^2}}}{{{a^2}}} + \dfrac{{{{(y - k)}^2}}}{{{b^2}}} = 1$
We can see that equation 3 is in the form of the equation of ellipse.

Hence option (B) an ellipse is the correct answer.

Additional information The equation $\dfrac{{dx}}{{dt}} = 8\pi sin2\pi t\;$ and the equation $\dfrac{{dy}}{{dt}} = 5\pi cos2\pi t$ give us the velocity because the change in distance with respect to time is nothing but velocity. Here $y$ and $x$are distance travelled in $x$ and y direction.
In the equation of ellipse we notice that the numerator of equation are ${(x - h)^2}$ and ${(y - k)^2}$ this tells us that the centre of the ellipse is at $h,k$ If the is equation is just $\dfrac{{{{(x)}^2}}}{{{a^2}}} + \dfrac{{{{(y)}^2}}}{{{b^2}}} = 1$ then it means that the center of the ellipse is at the origin.

Note: To solve such questions one should know the equations of curves and straight lines. This is to compare the final equation with the equation of one of the curves. For this particular question, the equation of ellipse is required to be known.
Equation of straight line is $y = mx + c$
Equation of circle is ${x^2} + {y^2} = {r^2}$
Equation of parabola is $y = {x^2}$
Only the equation of an ellipse matches the answer hence it is the correct answer.