Exercise 5.3 — Vector-Valued Functions: Velocity, Acceleration & Speed

All questions in this exercise are listed below. Click on a question to view its solution.

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Key Concepts

This exercise focuses on the following concepts:

• Vector-valued position functions $\mathbf{r}(t)$

• Domain and range of vector functions (intersection of component domains)

• Velocity as the derivative of position: $\mathbf{v}(t)={\mathbf{r}}^{\prime }(t)$

• Acceleration as the derivative of velocity: $\mathbf{a}(t)={\mathbf{v}}^{\prime }(t)={\mathbf{r}}^{\prime \prime }(t)$

• Speed as a scalar-valued magnitude function $|\mathbf{v}(t)|$

• Differentiation of vector functions (component-wise)

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Important Formulas

Kinematic Vectors

$\mathbf{v}(t)={\mathbf{r}}^{\prime }(t)=\frac{d\mathbf{r}}{dt}$

$\mathbf{a}(t)={\mathbf{v}}^{\prime }(t)={\mathbf{r}}^{\prime \prime }(t)=\frac{d^2\mathbf{r}}{d{t}^2}$

Speed (Scalar Magnitude of Velocity)

$|\mathbf{v}(t)|=\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2+[z^{\prime }(t){]}^2}$

(For 2D motion, omit the $z$-component.)

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Worked Example

Let $\mathbf{r}(t)=t^3\hat{i}+e^{-2t}\hat{j}+\sin 2t\hat{k}$.

Step 1 — Velocity vector: $\mathbf{v}(t)={\mathbf{r}}^{\prime }(t)=3t^2\hat{i}-2e^{-2t}\hat{j}+2\cos 2t\hat{k}$

Step 2 — Acceleration vector: $\mathbf{a}(t)={\mathbf{v}}^{\prime }(t)=6t\hat{i}+4e^{-2t}\hat{j}-4\sin 2t\hat{k}$

Step 3 — Speed at any time $t$: $|\mathbf{v}(t)|=\sqrt{(3t^2{)}^2+(-2e^{-2t}{)}^2+(2\cos 2t{)}^2}=\sqrt{9t^4+4e^{-4t}+4{\cos }^{2}2t}$

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Summary

This exercise covers the analysis of particle motion using vector-valued functions:

1. Domain of $\mathbf{r}(t)$ = intersection of domains of all component functions.
2. Velocity: differentiate each component of $\mathbf{r}(t)$ once.
3. Acceleration: differentiate each component of $\mathbf{r}(t)$ twice.
4. Speed: take the magnitude of the velocity vector using the Pythagorean formula.