A particle moves with its position given by $x=\cos \left( 4t \right)$ and$y=\sin \left( t \right)$, where positions are given in feet from the origin and time t is in seconds. What is the speed of the particle?

Hint: We are given the x and y component of a particle and asked to find the speed of that particle. We know that the first derivative of position is always its velocity. So we will first differentiate the position with respect to time. We also know that the magnitude of velocity is speed. Hence, after finding the velocity we will find its magnitude, that is the particle’s speed.

Complete step by step solution:
Given that a particle moves with its position as $x=\cos \left( 4t \right)$ and$y=\sin \left( t \right)$,
We can find the velocity by adding up the components, which we find by taking the first derivative of the x and y functions:
$\dfrac{dx}{dt}=-4\sin \left( 4t \right)$
$\dfrac{dy}{dt}=\cos \left( t \right)$
We know that velocity is a vector with components as derived above.
 Speed is the magnitude of this vector, which can be found via Pythagorean theorem (magnitude of velocity):
$s=\sqrt{{{\left( -4\sin \left( 4t \right) \right)}^{2}}+{{\cos }^{2}}\left( t \right)}$
$s=\sqrt{16{{\sin }^{2}}\left( 4t \right)+{{\cos }^{2}}\left( t \right)}$
Therefore, $s=\sqrt{16{{\sin }^{2}}\left( 4t \right)+{{\cos }^{2}}\left( t \right)}$ is the speed of the given particle moving with its position as $x=\cos \left( 4t \right)$ and $y=\sin \left( t \right)$.

Note:
One must not get confused between speed and velocity, if in the problem we were asked to find velocity of the particle we would only find the first derivative of the position of the particle but here we are asked to find the speed of the particle so we first found the velocity by differentiating the position one time and then found its magnitude in order to find the speed.