Problem 44 - 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

The product \(u \mathbf{r}\) is continuous if both are continuous, and differentiable with \(\frac{d}{d t}(u \mathbf{r}) = u \frac{d \mathbf{r}}{d t} + \mathbf{r} \frac{d u}{d t}\).

Step 1: Establish Continuity of Functions

First, let's understand what it means for a scalar function \(u(t)\) and a vector function \(\mathbf{r}(t)\) to be continuous. A function is continuous on an interval \([a, b]\) if it is continuous at every point \(t\) in the interval. This means there are no jumps, breaks, or undefined points in \(u(t)\) or \(\mathbf{r}(t)\) on \([a, b]\).

Step 2: Continuity of the Product

Consider the product \(w(t) = u(t) \mathbf{r}(t)\). Since \(\mathbf{r}(t)\) is a vector, each component of \(\mathbf{r}(t)\) is continuous whenever \(\mathbf{r}(t)\) is continuous. If both \(u(t)\) and \(\mathbf{r}(t)\) are continuous, then the product \(w(t)\), which is a vector with components \(u(t)r_1(t), u(t)r_2(t), ...\), is continuous because the product of continuous functions is continuous.

Step 3: Define Differentiability of Functions

Next, let's consider differentiability. A function is differentiable on \([a, b]\) if its derivative exists at every point in the interval. This implies \(u(t)\) has the derivative \(\frac{d u}{d t}\), and for \(\mathbf{r}(t)\), each component has a derivative giving \(\frac{d \mathbf{r}}{d t}\).

Step 4: Apply Product Rule to Differentiability

For the product \(w(t) = u(t) \mathbf{r}(t)\), we apply the product rule for derivatives. This gives:\[\frac{d}{d t}(u \mathbf{r}) = \frac{d u}{d t} \mathbf{r}(t) + u(t) \frac{d \mathbf{r}}{d t}\]This is analogous to the scalar product rule \(\frac{d}{d t}(uv) = \frac{du}{dt}v + u\frac{dv}{dt}\), extended here to vector functions by treating each component separately with the scalar product rule.

Step 5: Conclusion of Differentiability and Result

We conclude that the expression \(\frac{d}{d t}(u \mathbf{r}) = u \frac{d \mathbf{r}}{d t} + \mathbf{r} \frac{d u}{d t}\) holds, meaning that \(u \mathbf{r}\) is differentiable on \([a, b]\) as long as both functions \(u(t)\) and \(\mathbf{r}(t)\) are differentiable. This step finalizes our proof for differentiability and results in the expected derivative formula for the product of scalar and vector functions.

Key Concepts

Scalar Function

A scalar function is essentially a mathematical function that outputs a single real number for every point in its domain. It is generally denoted by variables, such as \( u(t) \) in our exercise. Scalar functions represent quantities that have only magnitude and no direction.

• Imagine temperature, which can be defined as a scalar function because at any point in space it only defines how hot or cold it is.
• A scalar function must be continuous, meaning it has no sudden jumps or gaps, for calculus operations like differentiation to be applicable seamlessly.

In the context of vector calculus, scalar functions can interact with vectors yielding new results, which is crucial for understanding the exercise involving the product of \( u(t) \) and a vector function \( \mathbf{r}(t) \).

Vector Function

A vector function outputs vectors, which have both magnitude and direction. For example, \( \mathbf{r}(t) \) is a vector function and generally outputs a vector in space as a function of a parameter \( t \).

• Each vector in \( \mathbf{r}(t) \) can be expressed in terms of its components, such as \( (r_1(t), r_2(t), r_3(t)) \).
• Vector functions are important in fields like physics and engineering as they can represent forces, velocities, and other quantities with directional components.

A vector function is considered continuous if each of its component functions are continuous. This is significant in mathematics because operations such as differentiation utilize component-wise calculations.

Continuity

Continuity is a vital trait of both scalar and vector functions used in calculus. A function is continuous if small changes in its input result in small changes in its output. This means the function does not have any abrupt changes or breaks.

• Both \( u(t) \) and \( \mathbf{r}(t) \) are said to be continuous over an interval \( [a, b] \) if they are continuous at every point in this range.
• The product of two continuous functions is continuous. Therefore, if \( u(t) \) and \( \mathbf{r}(t) \) are each continuous over the interval, their product combined as \( w(t) = u(t) \mathbf{r}(t) \) will also be continuous.

This property is essential in ensuring that operations like integration and differentiation don't encounter any discontinuity.

Differentiability

Differentiability is a property of a function that allows us to compute its derivative, which indicates the rate at which the function changes. A function is differentiable at a point if it has a defined tangent, or equivalently, if there exists a finite derivative at that point.

• For \( u(t) \), the scalar function is differentiable if \( \frac{d u}{d t} \) exists at each point in the interval.
• Similarly, \( \mathbf{r}(t) \) is differentiable if each of its components is differentiable at every point.

If both \( u(t) \) and \( \mathbf{r}(t) \) are differentiable, their product function \( w(t) = u(t) \mathbf{r}(t) \) is also differentiable, which ensures we can use calculus tools like the product rule to find the derivative.

Product Rule

The product rule is a differentiation technique used to find the derivative of a product of two functions. It is applicable when we have functions that are differentiable on a specific interval.

• For two scalar functions \( u(t) \) and \( v(t) \), the product rule states: \( \frac{d}{d t}(uv) = \frac{du}{dt}v + u\frac{dv}{dt} \).
• With vector functions, the concept is similar but applies component-wise. The derivative of the product \( w(t) = u(t) \mathbf{r}(t) \) becomes \( \frac{d}{d t}(u \mathbf{r}) = u \frac{d \mathbf{r}}{d t} + \mathbf{r} \frac{d u}{d t} \).

This ensures that each part of the function is accounted for, providing an accurate derivative. The product rule is crucial for accurately calculating derivatives of complex combinations of functions.