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Active Calculus - Multivariable

Activity 10.4.4.
In general, calculating \(\vN\) as a function of the parameter \(t\) ends up being very difficult because there are multiple compositions of functions involved which means there is a nesting of chain rules involved in the definition. In this activity, we will examine how to go through the direct calculation of \(\vN\) for a curve parameterized by \(\vr(t) =\langle t,t^2,t^3\rangle \text{.}\)
(a)
Calculate velocity \(\vv(t)\) and \(\text{speed}(t)\) for the parameterization \(\vr(t) =\langle t,t^2,t^3\rangle \text{.}\)
(b)
Now calculate \(\vT(t)\) for the path given by \(\vr(t) =\langle t,t^2,t^3\rangle \text{.}\)
(c)
Calculate the first component of \(\vT\, '=\frac{d\vT}{dt}\text{.}\)
Hint.
Remember that the derivative with respect to \(t\) is computed componentwise.
(d)
Calculate the second component of \(\vT\, '\text{.}\)
(e)
Calculate the third component of \(\vT\, '\text{.}\)
(f)
Put together your work for the three components of \(\vT\, '\) and compute \(\vecmag{\frac{d\vT}{dt}}\text{.}\)
(g)
In the previous part, you likely decided that trying to simplify your formulas for \(\frac{d\vT}{dt}\) and \(\vecmag{\frac{d\vT}{dt}}\) was more complicated than you had space for on your paper. Instead of doing more intricate algebraic computations, describe how you would calculate \(\vN\) if you had nice formulas for \(\frac{d\vT}{dt}\) and \(\vecmag{\frac{d\vT}{dt}}\text{.}\)