### Instruction: A particle moves so that it is at $\left( 3 \sin \frac{t}{4}, 3 \cos \frac{t}{4} \right)$ at time $t.$  Find the speed of the particle, measured in unit of distance per unit of time.### Response: I recognize that the position of the particle is given by a parametric equation, where the x-coordinate is $3 \sin \frac{t}{4}$ and the y-coordinate is $3 \cos \frac{t}{4}.$
To find the speed of the particle, I need to find the magnitude of its velocity vector, which is the derivative of its position vector.
I can use the chain rule to find the derivatives of the x- and y-coordinates with respect to time.
For the x-coordinate, I get $\frac{dx}{dt} = 3 \cos \frac{t}{4} \cdot \frac{1}{4},$ by applying the chain rule to the sine function and the fraction.
For the y-coordinate, I get $\frac{dy}{dt} = -3 \sin \frac{t}{4} \cdot \frac{1}{4},$ by applying the chain rule to the cosine function and the fraction.
The velocity vector is then $\left( \frac{dx}{dt}, \frac{dy}{dt} \right) = \left( \frac{3}{4} \cos \frac{t}{4}, -\frac{3}{4} \sin \frac{t}{4} \right).$
The magnitude of the velocity vector is the square root of the sum of the squares of its components.
So, the speed of the particle is $\sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} = \sqrt{\left( \frac{3}{4} \cos \frac{t}{4} \right)^2 + \left( -\frac{3}{4} \sin \frac{t}{4} \right)^2}.$
I can simplify this expression by factoring out $\frac{9}{16}$ from both terms under the square root and using the identity $\cos^2 x + \sin^2 x = 1.$
I get $\sqrt{\frac{9}{16} \left( \cos^2 \frac{t}{4} + \sin^2 \frac{t}{4} \right)} = \sqrt{\frac{9}{16}} = \frac{3}{4}.$