Problem 55 - Thomas Calculus

Question

Products of scalar and vector functions Suppose that the scalar function \(u(t)\) and the vector function \(\mathbf{r}(t)\) are both defined for \(a \leq t \leq b\) . a. Show that \(u \mathbf{r}\) is continuous on \([a, b]\) if \(u\) and \(\mathbf{r}\) are continuous on \([a, b] .\) If \(u\) and \(\mathbf{r}\) are both differentiable on \([a, b],\) show that \(u \mathbf{r}\) is differentiable on \([a, b]\) and that $$ \frac{d}{d t}(u \mathbf{r})=u \frac{d \mathbf{r}}{d t}+\mathbf{r} \frac{d u}{d t} $$

Step-by-Step Solution

Answer

If both \(u(t)\) and \(\mathbf{r}(t)\) are continuous on \([a, b]\), then \(u\mathbf{r}\) is continuous. If differentiable, then \(\frac{d}{dt}(u\mathbf{r}) = u \frac{d\mathbf{r}}{dt} + \mathbf{r} \frac{du}{dt}\).

Step 1: Check Continuity Criteria

To show that the product of a scalar function and a vector function, \(u(t)\mathbf{r}(t)\), is continuous on \([a, b]\), first recall that a function is continuous on an interval if it does not have any discontinuities in that interval. Because both \(u(t)\) and \(\mathbf{r}(t)\) are continuous on \([a, b]\), the product \(u(t)\mathbf{r}(t)\) is also continuous. This is due to the property that the product of continuous functions is continuous.

Step 2: Condition for Differentiability

To demonstrate differentiability, we need to prove that the product \(u(t)\mathbf{r}(t)\) is differentiable on \([a, b]\) provided that both \(u(t)\) and \(\mathbf{r}(t)\) are differentiable. Since the derivative of a product is given by the product rule, in general, a product of differentiable functions is differentiable. Here, both \(u\) and \(\mathbf{r}\) are differentiable on \([a, b]\), so their product \(u\mathbf{r}\) is differentiable as well.

Step 3: Apply the Product Rule for Differentiation

Using the product rule for differentiation, we can write the derivative of \(u(t)\mathbf{r}(t)\) as the sum of two components: the derivative of \(u(t)\) times \(\mathbf{r}(t)\), plus \(u(t)\) times the derivative of \(\mathbf{r}(t)\). This is expressed mathematically as: \[ \frac{d}{dt}(u(t)\mathbf{r}(t)) = u(t) \frac{d \mathbf{r}(t)}{dt} + \mathbf{r}(t) \frac{d u(t)}{dt} \]This illustrates that the product rule applicable to scalar and vector function products results in two terms, confirming that \(u\mathbf{r}\) is differentiable and providing its derivative.

Key Concepts

Scalar Functions

A scalar function is a mathematical expression that assigns a single real number to each point in its domain. It results in outputs that are independent of direction, effectively mapping from inputs like time or space to a scalar value. Scalar functions are used to represent quantities such as temperature or pressure in physics, where only the magnitude matters.Scalar functions can be analyzed for various properties such as continuity and differentiability. A scalar function is said to be continuous over an interval if there are no breaks, jumps, or holes within that interval. This means that as you approach any point within the interval, the function’s value gets arbitrarily close to the function's value at that point.In the context of our exercise, the scalar function is denoted as \( u(t) \), which is assumed to be continuous and differentiable on the interval \([a, b]\). Understanding the behavior of scalar functions is crucial when dealing with more complex expressions involving products with vector functions.

Vector Functions

Vector functions are functions that result in vectors rather than single scalar quantities. These vectors often vary with respect to different values of a single variable such as time. For instance, a vector function might describe the position of an object in space at any given time.For our example, the vector function \( \mathbf{r}(t) \) is a crucial element. It captures quantities that have both magnitude and direction, such as velocity or force in physics. A vector function over an interval, just like a scalar one, can also be checked for continuity and differentiability.To ensure continuity of a vector function within an interval \([a, b]\), there should be no abrupt changes in the vector’s direction or magnitude. Differentiability requires that the derivative of the vector, capturing the rate of change, exists at each point along the interval. These properties ensure the vector function works seamlessly in mathematical operations and real-world applications.

Product Rule for Differentiation

The product rule is a fundamental theorem in calculus that allows us to find the derivative of the product of two functions. This is crucial when working with both scalar and vector functions, as shown in the exercise.When you have a product of a scalar function \( u(t) \) and a vector function \( \mathbf{r}(t) \), the derivative is computed by the formula:\[ \frac{d}{dt}(u(t)\mathbf{r}(t)) = u(t) \frac{d \mathbf{r}(t)}{dt} + \mathbf{r}(t) \frac{d u(t)}{dt} \]This rule states that the derivative of the product involves the derivative of each function, one at a time, while the other is kept constant. Each component is crucial as it shows that both the rate of change of the scalar function and the vector's rate of change are considered.Understanding this rule helps in correctly differentiating products of functions, which is necessary for exploring dynamic systems in physics and engineering.

Continuity

Continuity is a key concept that helps ensure a function behaves well over its domain. For a function to be continuous over an interval \([a, b]\), it must not exhibit any jumps, breaks, or infinite behaviors.For the function \( u(t)\mathbf{r}(t) \), it is continuous if both \( u(t) \) and \( \mathbf{r}(t) \) are continuous over the interval. In simpler terms, as you move through values of \( t \) within \([a, b]\), the product should not suddenly change its value, ensuring a smooth transition.This uninterrupted behavior is essential in both real-world applications and theoretical mathematics because it allows for predictable and reliable interactions when modeling physical systems.

Differentiability

Differentiability is the property that indicates whether a function has a well-defined derivative at all points in its domain. For a product of a scalar and vector function like \( u(t) \mathbf{r}(t) \), differentiability is crucial as it implies the function's derivative is correctly defined everywhere on the interval \([a, b]\).If both the scalar function \( u(t) \) and the vector function \( \mathbf{r}(t) \) are differentiable in the interval, their product is differentiable. This ensures that the changes in the function can be tracked accurately and the product rule can be applied effectively.Differentiability is a cornerstone for calculus, as it facilitates understanding the rate of change and enhancing the analysis of behavior in scientific and engineering problems.