Problem 40 - 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 \([a, b] .\) 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

\( u(t)\mathbf{r}(t) \) is continuous and differentiable on \([a, b]\). Its derivative is \( u \frac{d\mathbf{r}}{dt} + \mathbf{r} \frac{du}{dt} \).

Step 1: Understand the Problem

We need to prove two different parts for a function composed of a scalar and vector products: a) The continuity of the product if both components are continuous, and b) The differentiability using the product rule for derivatives.

Step 2: Define Continuity for Function

A scalar function \( u(t) \) is continuous on \([a, b]\) if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( | t - c | < \delta \) for any \( c \) in \([a, b]\) then \( | u(t) - u(c) | < \epsilon \). Similarly, a vector function \( \mathbf{r}(t) \) is continuous if each component is continuous.

Step 3: Show Continuity of the Product

Given that both \( u(t) \) and \( \mathbf{r}(t) = [r_1(t), r_2(t), r_3(t)] \) are continuous on \([a, b]\), their product \( u(t) \mathbf{r}(t) \) is continuous. This is because the product of continuous functions is continuous. Each output dimension \( u(t)r_i(t) \) is also continuous for \( i = 1, 2, 3 \).

Step 4: Define Differentiability

A function is differentiable if its derivative exists at each point. For a scalar function \( u(t) \), differentiability implies that the derivative \( \frac{du}{dt} \) exists. For a vector function \( \mathbf{r}(t) \), each component function differentiates individually.

Step 5: Apply Product Rule for Differentiation

If both \( u(t) \) and \( \mathbf{r}(t) \) are differentiable, we use the product rule on \( u(t) \mathbf{r}(t) \). The derivative of the product \( \frac{d}{dt}(u \mathbf{r}) \) is \( u \frac{d\mathbf{r}}{dt} + \mathbf{r} \frac{du}{dt} \). This applies individually to each vector component due to the definitions of scalar and vector differentiation products.

Step 6: Conclusion of Differentiability

Thus, \( u \mathbf{r} \) is differentiable on \([a, b]\), and the expression for its derivative has been derived as \( u \frac{d\mathbf{r}}{dt} + \mathbf{r} \frac{du}{dt} \).

Key Concepts

Continuity of Scalar and Vector Functions

When we talk about continuity in mathematics, it refers to how a function behaves without any interruptions along its domain. For a scalar function \( u(t) \), this means for every small distance \( \epsilon \) away from a point, there exists a small step \( \delta \) such that stepping within \( \delta \) keeps the change in \( u(t) \) less than \( \epsilon \). Similarly, a vector function \( \mathbf{r}(t) \) is continuous if each of its components (like \( r_1(t), r_2(t), r_3(t) \)) follow the same criteria of continuity.

Therefore, if both \( u(t) \) and \( \mathbf{r}(t) \) are continuous over the interval \([a, b]\), then their product \( u(t) \mathbf{r}(t) \) is also continuous over that same interval. This conclusion arises from the fact that the product of continuous functions remains continuous. In more simple terms, smooth functions staying smooth even when multiplied together.

Differentiability in Scalar and Vector Contexts

Differentiability is a measure of how well a function can be approximated by a linear function at small scales. For a scalar function \( u(t) \), differentiation involves computing its rate of change, \( \frac{du}{dt} \). A vector function \( \mathbf{r}(t) \) is differentiable if you can differentiate each of its components individually (for example, calculating \( \frac{dr_1}{dt}, \frac{dr_2}{dt},\) etc.).

If both a scalar function \( u(t) \) and a vector function \( \mathbf{r}(t) \) are differentiable, then their product \( u \mathbf{r} \) is differentiable as well. This is deduced from the product rule of differentiation. Understanding that \( u \mathbf{r} \) behaves so smoothly, we conclude it is approachable by a tangent plane at any point.

Product Rule for Differentiation

The product rule is a vital tool in calculus. It allows us to differentiate products of two functions efficiently. For any functions \( u(t) \) and \( \mathbf{r}(t) \) that are differentiable, the rule states that the derivative of their product \( \frac{d}{dt}(u \mathbf{r}) \) is given by: \[ \frac{d}{dt}(u \mathbf{r}) = u \frac{d\mathbf{r}}{dt} + \mathbf{r} \frac{du}{dt} \] This formula arises from distributing the derivative across the product. The derivative of a product includes both possible "waves" of change from \( u(t) \) and \( \mathbf{r}(t) \). This is like saying, the wave produced by \( u(t) \) pushing on the functions combined with the wave from \( \mathbf{r}(t) \) being pushed, cover all paths of how our function changes.

The product rule is important as it gives a straightforward way to handle computationally complex derivative expressions involving products of functions.