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.

Problem 13.2.19:

Evaluate $$\frac{d}{dt}(\vec{r_1}(t) \cdot \vec{r_2}(t))$$, where $$\vec{r_1} = (8t, 4, -t^3), \vec{r_2} = (0, e^t, -6)$$.

Solution:

Using the appropriate product rule in the first step, we have:

$$\frac{d}{dt}(\vec{r_1}(t) \cdot \vec{r_2}(t)) = \frac{d}{dt}\vec{r_1}(t) \cdot \vec{r_2}(t) + \vec{r_1}(t) \cdot \frac{d}{dt}\vec{r_2}(t)$$

$$= \frac{d}{dt}(8t, 4, -t^3) \cdot (0, e^t, -6) + (8t, 4, -t^3) \cdot \frac{d}{dt}(0, e^t, -6)$$

$$= (8, 0, -3t^2) \cdot (0, e^t, -6) + (8t, 4, -t^3) \cdot (0, e^t, 0)$$

$$= (8 \cdot 0) + (0 \cdot e^t) + (-3t^2 \cdot -6) + (8t \cdot 0) + (4 \cdot e^t) + (-t^3 \cdot 0)$$

$$= 18t^2 + 4e^t$$

Problem 13.2.20:

Evaluate $$\frac{d}{dt}(t^4\vec{r_1}(t))$$, where $$\vec{r_1}$$ and $$\vec{r_2}$$ are as given in 13.2.19.

Using the appropriate product rule in the first step, we have:

Solution:

$$\frac{d}{dt}(t^4\vec{r_1}(t)) = \left( \frac{d}{dt}t^4 \right) \vec{r_1}(t) + t^4 \left( \frac{d}{dt}\vec{r_1}(t) \right)$$

$$= 4t^3 (8t, 4, -t^3) + t^4 (8, 0, -3t^2)$$

$$= (32t^4 + 8t^4, 16t^3 + 0, -4t^6 -3t^6)$$

$$= (40t^4, 16t^3, -7t^6)$$

Problem 13.2.21:

Evaluate $$\frac{d}{dt}(\vec{r_1}(t) \times \vec{r_2}(t))$$, where $$\vec{r_1}$$ and $$\vec{r_2}$$ are as given in 13.2.19

Solution:

Using the appropriate product rule in the first step, we have:

$$\frac{d}{dt}(\vec{r_1}(t) \times \vec{r_2}(t)) = \frac{d}{dt}\vec{r_1}(t) \times \vec{r_2}(t) + \vec{r_1}(t) \times \frac{d}{dt}\vec{r_2}(t)$$

$$= \frac{d}{dt}(8t, 4, -t^3) \times (0, e^t, -6) + (8t, 4, -t^3) \times \frac{d}{dt}(0, e^t, -6)$$

$$= (8, 0, -3t^2) \times (0, e^t, -6) + (8t, 4, -t^3) \times (0, e^t, 0)$$

$$= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 8 & 0 & -3t^2\\ 0 & e^t & -6 \end{vmatrix} + \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 8t & 4 & -t^3\\ 0 & e^t & 0 \end{vmatrix}$$

$$= (3t^2e^t, 48, 8e^t) + (t^3e^t, 0, 8te^t)$$

$$= (t^2e^t(3+t), 48, 8e^t(1+t) )$$

Problem 13.2.22:

Evaluate $$\frac{d}{dt}(\vec{r_1}(t) \cdot \vec{r_3}(t)) \bigg|_{t=5}$$, where $$\vec{r_1}$$ is as given in 3.2.19, $$\vec{r_3}(5) = (3, 1, 2)$$, and $$\vec{r_3}'(5) = (-1, 2, 7)$$

Solution:

Using the appropriate product rule in the first step, we have:

$$\frac{d}{dt}(\vec{r_1}(t) \cdot \vec{r_3}(t)) = \frac{d}{dt}\vec{r_1}(t) \cdot \vec{r_3}(t) + \vec{r_1}(t) \cdot \frac{d}{dt}\vec{r_3}(t)$$

$$= \frac{d}{dt}(8t, 4, -t^3) \cdot \vec{r_3}(t) + (8t, 4, -t^3) \cdot \frac{d}{dt}\vec{r_3}$$

$$= (8, 0, -3t^2) \cdot \vec{r_3}(t) + (8t, 4, -t^3) \cdot \frac{d}{dt}\vec{r_3}$$

To evaluate at $$t = 5$$ we plug the value in for $$t$$ where it occurs and plug in the given values of $$\vec{r_3}(5)$$ and $$\vec{r_3}'(5)$$:

$$= (8, 0, -75) \cdot (3, 1, 2) + (40, 4, -125) \cdot (-1, 2, 7)$$

$$= 24 + 0 - 150 - 40 + 8 - 875$$

$$= -1033$$

Last updated: 15 Nov 2014

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