Chapter 14: Problem 42
Prove: If \(\mathbf{r}(t)\) is a smoothly parametrized function, then the angles between \(\mathbf{r}^{\prime}(t)\) and the vectors \(\mathbf{i}, \mathbf{j},\) and \(\mathbf{k}\) are continuous functions of \(t\)
Short Answer
Expert verified
The angles between \(\mathbf{r}'(t)\) and the vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are continuous functions of \(t\).
Step by step solution
01
Define the Dot Product and Angle Relationship
The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is related to the angle \(\theta\) between them by the formula: \(\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos(\theta)\). This implies that the cosine of the angle \(\theta\) is given by: \(\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\).
02
Express Derivative and Unit Vectors in Dot Product
For a vector function \(\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}\), the derivative is \(\mathbf{r}'(t) = x'(t)\mathbf{i} + y'(t)\mathbf{j} + z'(t)\mathbf{k}\). The unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are constants. Apply the dot product: \(\mathbf{r}'(t) \cdot \mathbf{i} = x'(t)\), \(\mathbf{r}'(t) \cdot \mathbf{j} = y'(t)\), \(\mathbf{r}'(t) \cdot \mathbf{k} = z'(t)\).
03
Compute Magnitudes and cosines
The magnitude of \(\mathbf{r}'(t)\) is \(\sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}\). Compute cosines for each angle: \(\cos(\theta_\mathbf{i}) = \frac{x'(t)}{\|\mathbf{r}'(t)\|}\), \(\cos(\theta_\mathbf{j}) = \frac{y'(t)}{\|\mathbf{r}'(t)\|}\), and \(\cos(\theta_\mathbf{k}) = \frac{z'(t)}{\|\mathbf{r}'(t)\|}\).
04
Verify Continuity of Cosine Functions
Since \(x'(t)\), \(y'(t)\), \(z'(t)\), and consequently \(\|\mathbf{r}'(t)\|\) are continuous functions (as derivatives of smooth functions), each of these cosine functions is continuous. The continuity of \(\cos(\theta)\) implies the continuity of each angle \(\theta_\mathbf{i}\), \(\theta_\mathbf{j}\), and \(\theta_\mathbf{k}\).
05
Conclusion: Angles are Continuous
Since the cosine functions that determine the angles between \(\mathbf{r}'(t)\) and \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are continuous, the angles themselves are continuous. Thus, the angles between \(\mathbf{r}'(t)\) and the vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are continuous functions of \(t\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Smoothly Parametrized Functions
A smoothly parametrized function is essentially a way to represent a curve or path in a space where its evolution follows a smooth and continuous trajectory. Imagine a smooth roller coaster track; it doesn’t have sharp corners or breaks. Similarly, in mathematics, a smoothly parametrized function has derivatives that exist and change continuously.
- This means no sudden changes or discontinuities in direction.
- Smooth functions are well-behaved and predictable over their domain.
Dot Product
The dot product is a key concept in understanding the relationship between two vectors. It is a scalar quantity that reflects how much one vector extends in the direction of another.
- The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta) \).
- Here, \( \theta \) represents the angle between \( \mathbf{a} \) and \( \mathbf{b} \).
Angle Between Vectors
Understanding the angle between vectors involves looking at how two vectors are positioned relative to each other. Using the dot product, it's easy to quantify this relationship.
- The angle \( \theta \) between two vectors is determined using the formula \( \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \).
- This helps in determining how aligned two vectors are within a space.
Derivatives of Vector Functions
Derivatives in calculus help us understand how a function changes. When dealing with vector functions, the derivative provides a vector representing the rate of change with respect to some parameter.
- For a vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \), the derivative is \( \mathbf{r}'(t) = x'(t) \mathbf{i} + y'(t) \mathbf{j} + z'(t) \mathbf{k} \).
- The derivative acts as a tangent vector to the path traced out by the vector function.