I am given the position vector $r(t) = (5cos3t) i + (6t) j + (5sin3t) k$. I am asked to find the component of the acceleration in the direction tangent to the curve and hence deduce the direction of acceleration relative to the curve.
In the earlier parts of the question I was asked to find the $(i) velocity$, $(ii) speed$, $(iii) distance\quad travelled$, $(iv) acceleration$, $(v) magnitude \quad of \quad acceleration$.
I am not sure what the question is asking, is it asking me to find the tangential component of acceleration because I tried it but it doesn't make sense. How do I go about this question? Thanks everyone for your help.
First, find the acceleration:
$$\vec a (t)=\ddot {\vec r}(t)=-45\cos (3t) \vec i -45 \sin (3t) \vec k$$
Next, find the direction that is tangent to the parametric curve at time $t$:
$$\vec \tau (t)= \dot {\vec r}(t)=-15\sin (3t) \vec i +6 \vec j + 15 \cos (3t) \vec k$$
Normalize to get $$\frac{\vec \tau (t)}{|\vec \tau (t)|}=\frac{1}{\sqrt{261}}\bigl(-15\sin (3t) \vec i +6 \vec j + 15 \cos (3t) \vec k\bigr)$$
To get the magnitude of the component of acceleration in the direction of the tangent, compute the dot product of the two (this essentially gives the projection of one vector on to another):
$$\vec a (t) \cdot \frac{\vec \tau (t)}{|\vec \tau (t)|}=\frac{1}{\sqrt{261}}\bigl(675\cos (t)\sin (t)+0-675\cos (t)\sin (t)\bigr)=0$$
This shows that the acceleration has no component in the tangent direction. Indeed,
Here, I have plotted $\vec r (t)$ in green, the position at a particular time ($t=1/4$) in white, its acceleration in blue and the tangent (which is also its velocity) in red.