Problem 47 - Advanced Engineering Mathematics

Question

Suppose \(\mathbf{r}\) is a differentiable vector function for which \(\|\mathbf{r}(t)\|=c\) for all \(t\). Show that the tangent vector \(\mathbf{r}^{\prime}(t)\) is perpendicular to the position vector \(\mathbf{r}(t)\) for all \(t\).

Step-by-Step Solution

Answer

The tangent vector \( \mathbf{r}^{\prime}(t) \) is perpendicular to the position vector \( \mathbf{r}(t) \) for all \( t \).

Step 1: Understand the given condition

The problem states that the magnitude of the vector function \( \mathbf{r}(t) \) is constant and equal to \( c \) for all \( t \). This means \( \| \mathbf{r}(t) \| = c \) where \( c \) is a constant value.

Step 2: Write the expression for magnitude

The magnitude of a vector \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) is \( \| \mathbf{r}(t) \| = \sqrt{x(t)^2 + y(t)^2 + z(t)^2} \). Since \( \| \mathbf{r}(t) \| = c \), square both sides to simplify: \( x(t)^2 + y(t)^2 + z(t)^2 = c^2 \).

Step 3: Differentiate the magnitude expression

Take the derivative of both sides with respect to \( t \). The derivative of \( x(t)^2 + y(t)^2 + z(t)^2 \) with respect to \( t \) is \( 2x(t)x'(t) + 2y(t)y'(t) + 2z(t)z'(t) \). For a constant \( c^2 \), the derivative with respect to \( t \) is 0. Thus, \( 2x(t)x'(t) + 2y(t)y'(t) + 2z(t)z'(t) = 0 \).

Step 4: Express the derivative in vector form

The expression \( 2x(t)x'(t) + 2y(t)y'(t) + 2z(t)z'(t) = 0 \) can be written as \( 2 \cdot \mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t) = 0 \) using the dot product in vector form. This states that \( \mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t) = 0 \).

Step 5: Conclude the relationship between vectors

Since the dot product of \( \mathbf{r}(t) \) and \( \mathbf{r}^{\prime}(t) \) equals 0, the vectors are perpendicular. This is because the dot product of two vectors is zero if and only if the vectors are orthogonal to each other.

Key Concepts

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. It is one of the basic operations that can be performed on vectors. The dot product is computed by multiplying corresponding entries and then summing those products. For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is given by:

• \( a_1b_1 + a_2b_2 + a_3b_3 \)

The dot product has several important characteristics:

• If the dot product of two vectors is zero (\( \mathbf{a} \cdot \mathbf{b} = 0 \)), the vectors are orthogonal (perpendicular).
• It is used to calculate the angle between two vectors.
• The dot product can be used to determine the orthogonal projection of one vector onto another.

Understanding the dot product is crucial in many fields, including physics, computer graphics, and engineering, as it relates to projections, work done by a force, and more.
In our exercise, we used the dot product to prove that the tangent vector \( \mathbf{r}'(t) \) is orthogonal to \( \mathbf{r}(t) \). This happened by showing that their dot product equals zero.

Differentiable Vector Functions

A differentiable vector function is a function that maps a real parameter, typically time \( t \), to a vector space, and it can be differentiated with respect to that parameter. In simpler terms, it is a vector defined as a function of one or more variables where derivatives exist. If \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) is a vector function, the derivative \( \mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle \) represents the rate of change of the vector function.

• The components of \( \mathbf{r}'(t) \) are simply the derivatives of the components of \( \mathbf{r}(t) \).
• The derivative vector \( \mathbf{r}'(t) \) can be viewed as a tangent to the path described by \( \mathbf{r}(t) \).

In the given exercise, the vector function \( \mathbf{r}(t) \) was differentiable, which allowed us to find \( \mathbf{r}'(t) \). The differentiability was essential in proving the orthogonality condition by using the properties of derivatives and the dot product.

Orthogonal Vectors

Orthogonal vectors are vectors that meet at right angles, meaning the angle between them is 90 degrees. In terms of vector mathematics, when two vectors are orthogonal, their dot product is zero. This is quite a powerful concept because it provides a straightforward way to verify orthogonality.

• If \( \mathbf{a} \cdot \mathbf{b} = 0 \), then \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal.
• Orthogonality leads to simplifications in computations such as projections and vector decompositions.

Why are orthogonal vectors important? They provide the basis for constructing coordinate systems, such as the Cartesian plane, where axes are perpendicular (orthogonal).
In the given exercise, we established that the tangent vector \( \mathbf{r}'(t) \), which is the derivative of the position vector \( \mathbf{r}(t) \), is always orthogonal to \( \mathbf{r}(t) \). We concluded this by proving their dot product is zero, confirming that the two vectors have a 90-degree angle between them.