The original equation of motion of $A$ with respect to time $t$ was this in vector cartesian or parametric form (I do not know which one it is). I am assuming Cartesian:
$$r(t)= 24\sin(2πt)i + (-24\cos(2πt)+35)j.$$
I was asked to find the velocity (rate of change of motion with respect to time), so I took the first derivative of $r(t)$ which is this:
$$r'(t)= 48π\cos(2πt)i + 48π\sin(2πt)j.$$
After this I was asked to find the speed of $A$ at time $t$. I assume this is the magnitude of the first derivative. Since the speed is magnitude quantity of velocity. The question now arises is how.
I know to find the magnitude of the vector you use take the whole square root of the number behind i squared and the number behind j squared. The Pythagorean theorem basically. But I cannot do this with this one because I have to deal with the t inside the trigonometric function, I was told to turn this into a parametric equation but I do not know how.
I am literally stuck, my professor told me, it is already in parametric form, but on the internet the parametric form starts with $x=\text{something}$ and $y=\text{something}$ in curly brackets. I just want to find the magnitude of the velocity function, and nothing is helping me.
Thanks a lot.
Your description of the magnitude of a vector is correct. The magnitude of $r'(t)$ as you have it written is $$\sqrt{(48\pi\cos(2\pi t))^2+(48\pi\sin(2\pi t))^2}$$ Now what do we know about $\cos^2(\theta)+\sin^2(\theta)$?
Note: just because a vector depends on a parameter, say $t$, this does not change how one would compute the magnitude. We just have to do a little more work in the algebra with a variable. However, in this case, the classic trig-identity which I allude to pops out a nice answer.