"Consider the path of a particle in a conservative force field represented by the vector-valued function $r(t) = \langle 4(\sin t - t \cos t), 4(\sin t + t \sin t), (\frac{3}{2})t^2 \rangle$."
"A) Find the arc length function $s$."
"D) Show that $|r'(t)| = 1$."
To do this, I took the first derviative of the function. Then, I set up the square root of the sum of each dervived component squared. However, I could not get this to simplify down to $1$, as I assume I should be able to by the instructions in part D. My prof suggests reparameterizing the original function to make the problem simpler. Any thoughts on the new parameter?
1 Answer
We have: $$s(t)=\int_0^t |\mathbf r'(\tau)|d\tau$$ Therefore: $$s'(t)=|\mathbf r'(t)|$$
The chain rule and the inverse derivative rule tell us that: $$\mathbf r'(s)=\frac{d\mathbf r}{ds} = \frac{d\mathbf r}{dt} \frac{dt}{ds} = \frac{\mathbf r'(t)}{s'(t)} $$
So that: $$\mathbf r'(s)=\frac{\mathbf r'(t)}{|\mathbf r'(t)|} \Rightarrow |\mathbf r'(s)|=\frac{|\mathbf r'(t)|}{|\mathbf r'(t)|}=1$$
