Question: Dimensional analysis is a powerful tool for understanding the validity of formulas. For example, directly adding a velocity and an acceleration vector does not make much sense as the result will become dependent on the units.

a. Measuring lengths in meters and time in seconds, what are the units for position vectors, tangent vectors, acceleration vectors, arc length, curvature, and torsion?

b. Argue from the definition (involving limits) what the units for tangent vectors are.

c. For curvature, explain the units from the perspective of the definition and two of the equivalent computational formulas. Explain the 1 in the formula for the curvature formula involving plane curves of the form y = f (x).

d.(i) If the tangent vector of a curve at \({{r}_{m,s}}=\left\langle 1,2,3 \right\rangle \) is \(r{{'}_{m,s}}=\left\langle 4,5,6 \right\rangle \) in the meters–seconds units, what is the new tangent vector \(\left( r{{'}_{c,m}} \right)\) and where is it located \(\left( {{r}_{c,m}} \right)\) under a change of units to centimeters–minutes?

(ii) Compute \({{w}_{m,s}}={{r}_{m,s}}+r{{'}_{m,s}}\) in terms of the first system of units.

(iii) Compute \({{w}_{c,m}}={{r}_{c,m}}+r{{'}_{c,m}}\) in the second set of units.

(iv) Are \({{w}_{m,s}}\) and \({{w}_{c,m}}\) scalar multiples of each other? If so, what is the multiple?

(v) If we had a constant \(\alpha \) in front of r in the sum, what units should it have so that both vectors in the sum would have the same set of units?

(vi) Set \(\alpha \) equal to 1 for the unit system m, s. Now recompute the two vector sums: \({{w}_{m,s,\alpha }}={{\alpha }_{m,s}}{{r}_{m,s}}+r{{'}_{m,s}}\)

(vii) Are \({{w}_{m,s,\alpha }}\) and \({{w}_{c,m,\alpha }}\) scalar multiples of each other? If so, what is the multiple?

e. There is a question about changing units in the parametrization of the curve. Mathematically, we usually use arc length as the canonical parametrization. But in applications, time is time. To investigate the issue of units in curve parameterizations, it is important to understand how curves generally arise in practice. Consider the differential equation \(m\mathbf{r}''=-k\mathbf{r}\) where k and m are positive constants (representing a spring constant and a mass constant). The general solution, restricting our considerations to the plane, is \(\mathbf{r}\left( t \right)=\left\langle a\cos \left( gt+c \right),b\sin \left( ht+d \right) \right\rangle \). What units must k have? What units must the constants in the solution have (a, b, c, d, h, g)? Argue from dimensional considerations that \(g=h=\sqrt{\frac{k}{m}}\) is the simplest possibility for g and h. Confirm by differentiation that this makes sense. Finally, compute \(\mathbf{r}'\) and \(\mathbf{r}''\) and check that changing units from seconds to minutes yield equivalent answers.