Calculus of Vector-Valued Functions, Worked Problems

Mark Vuletic

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Source of problems: Rogawski J. 2008. Calculus: Early Transcendentals. New York: W. H. Freeman. pp. 753-755.

Problems worked on this page: 23, 24

 

Problem 13.2.23:

Let \( \vec{r_1}(t) = (t^2, t^3, 4t) \), \(\vec{r_2}(t) = (t^{-1}, 1+t, 2) \), and \( F(t) = \vec{r_1}(t) \cdot \vec{r_2}(t) \). (a) Calculate \( F'(t) \) using the Product Rule. (b) Expand the product \( \vec{r_1}(t) \cdot \vec{r_2}(t) \) and differentiate. Compare with part (a).

Solution:

(a) Using the Product Rule, we have:

\( F'(t) = \frac{d}{dt}\vec{r_1}(t) \cdot \vec{r_2}(t) + \vec{r_1}(t) \cdot \frac{d}{dt}\vec{r_2}(t) \)

\( = \frac{d}{dt}(t^2, t^3, 4t) \cdot (t^{-1}, 1+t, 2) + (t^2, t^3, 4t) \cdot \frac{d}{dt}(t^{-1}, 1+t, 2) \)

\( = (2t, 3t^2, 4) \cdot (t^{-1}, 1+t, 2) + (t^2, t^3, 4t) \cdot (-t^{-2}, 1, 0) \)

\( = 2 + 3t^2 + 3t^3 + 8 -1 + t^3 + 0 \)

\( = 4t^3 + 3t^2 + 9 \)

(b) Expanding, then differentiating, we have:

\( F'(t) = \frac{d}{dt} \left[ \vec{r_1}(t) \cdot \vec{r_2}(t) \right] \)

\( = \frac{d}{dt} \left[ (t^2, t^3, 4t) \cdot (t^{-1}, 1+t, 2) \right] \)

\( = \frac{d}{dt} (t + t^3 + t^4 + 8t) \)

\( = 4t^3 + 3t^2 + 9 \)

The two results are the same.

 

Problem 13.2.24:

Let \( \vec{r_1}(t) = (t^2, t^3, 4t) \), \(\vec{r_2}(t) = (t^{-1}, 1+t, 2) \), and \( G(t) = \vec{r_1}(t) \times \vec{r_2}(t) \). (a) Calculate \( G'(t) \) using the Product Rule. (b) Expand the product \( \vec{r_1}(t) \times \vec{r_2}(t) \) and differentiate. Compare with part (a).

Solution:

(a) Using the Product Rule, we have:

\( G'(t) = \frac{d}{dt}\vec{r_1}(t) \times \vec{r_2}(t) + \vec{r_1}(t) \times \frac{d}{dt}\vec{r_2}(t) \)

\( = \frac{d}{dt}(t^2, t^3, 4t) \times (t^{-1}, 1+t, 2) + (t^2, t^3, 4t) \times \frac{d}{dt}(t^{-1}, 1+t, 2) \)

\( = (2t, 3t^2, 4) \times (t^{-1}, 1+t, 2) + (t^2, t^3, 4t) \times (-t^{-2}, 1, 0) \)

\( = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 2t & 3t^2 & 4\\ t^{-1} & 1+t & 2 \end{vmatrix} + \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ t^2 & t^3 & 4t\\ -t^{-2} & 1 & 0 \end{vmatrix} \)

\( = (6t^2 -4 -4t, 4t^{-1} -4t, 2t+2t^2 - 3t) + (-4t, -4t^{-1}, t^2 + t) \)

\( = (6t^2 - 8t - 4, -4t, 3t^2 )\)

(b) Expanding, then differentiating, we have:

\( G'(t) = \frac{d}{dt} \left[ \vec{r_1}(t) \times \vec{r_2}(t) \right] \)

\( = \frac{d}{dt} \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ t^2 & t^3 & 4t\\ t^{-1} & 1+t & 2 \end{vmatrix} \)

\( = \frac{d}{dt} (2t^3 -4t -4t^2, 4 -2t^2, t^2 + t^3 - t^2) \)

\( = ( 6t^2 - 8t - 4, -4t, 3t^2 ) \)

The two results, again, are the same.

Last updated: 15 Nov 2014

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