Question

Consider the circle \(\mathbf{r}(t)=\langle a \cos t, a \sin t\rangle,\) for \(0 \leq t \leq 2 \pi\) where \(a\) is a positive real number. Compute \(\mathbf{r}^{\prime}\) and show that it is orthogonal to \(\mathbf{r}\) for all \(t\).

Short answer

Question: Show that the derivative of the position vector \(\mathbf{r}(t) = \langle a \cos t, a \sin t \rangle\) is orthogonal to \(\mathbf{r}(t)\) for all \(t\). Answer: By computing the derivative of the position vector \(\mathbf{r}(t)\) as \(\mathbf{r'}(t) = \langle -a \sin t, a \cos t \rangle\), calculating their dot product, and showing that it equals zero, we can conclude that the two vectors are orthogonal for all \(t\).

Step-by-step solution

Computing the derivative of \(\mathbf{r}(t)\)

To find the derivative of the position vector \(\mathbf{r}(t)\), we need to take the derivative of each of its components with respect to \(t\). The given position vector is \(\mathbf{r}(t) = \langle a \cos t, a \sin t \rangle\). So, differentiating each component, we have: $$ \mathbf{r'}(t) = \langle -a \sin t, a \cos t \rangle $$

Calculate the dot product of \(\mathbf{r}(t)\) and its derivative

Now, we must find the dot product of \(\mathbf{r}(t)\) and its derivative \(\mathbf{r'}(t)\). The dot product of two vectors \(\mathbf{u} = \langle u_1, u_2 \rangle\) and \(\mathbf{v} = \langle v_1, v_2 \rangle\) is defined as \(\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2\). So we have: $$ \mathbf{r}(t) \cdot \mathbf{r'}(t) = (a \cos t)(-a \sin t) + (a \sin t)(a \cos t) $$

Prove that the dot product is zero

Now, we need to simplify the dot product expression to prove that it equals zero. $$ \mathbf{r}(t) \cdot \mathbf{r'}(t) = -a^2 \cos t \sin t + a^2 \sin t \cos t = 0 $$ The dot product between the position vector and its derivative is zero, which confirms that the two vectors are orthogonal for all \(t\).

Key concepts

Parametric Equations

Parametric equations allow us to describe a geometric object, such as a curve, by defining its coordinates in terms of one or more parameters. In the given exercise, we're looking at the circle described by the vector function \( \mathbf{r}(t) = \langle a \cos t, a \sin t \rangle \). The parameter \( t \) represents the angle in the circle, ranging from \( 0 \) to \( 2\pi \), often referred to as the parameter interval.

• \( a \) is a positive real number, representing the radius of the circle.
• The function \( \langle a \cos t, a \sin t \rangle \) generates points on the circle in the plane, where \( a \cos t \) and \( a \sin t \) are the \( x \)- and \( y \)-coordinates, respectively.

This representation allows us to work with curves using simple trigonometric functions, providing a clear mathematical way to handle circular paths without the need for complex algebraic equations. Parametric equations are particularly useful in calculus-related problems such as this, where they provide a straightforward way to compute derivatives and understand the movement along the curve.

Dot Product

In vector calculus, the dot product is an algebraic operation that takes two equal-length sequences of numbers, or two vectors, and outputs a single number. The dot product is vital in this exercise to examine the relationship between \( \mathbf{r}(t) \) and its derivative \( \mathbf{r'}(t) \). The dot product of vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \) is calculated as:
\[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 \]
In the exercise, we used the dot product to determine if the vectors are orthogonal.

• For the vectors \( \mathbf{r}(t) = \langle a \cos t, a \sin t \rangle \) and \( \mathbf{r'}(t) = \langle -a \sin t, a \cos t \rangle \), we calculated:
• \( \mathbf{r}(t) \cdot \mathbf{r'}(t) = (a \cos t)(-a \sin t ) + (a \sin t)(a \cos t) \)
• The final value is 0, as the terms cancel each other out, confirming orthogonality.

Understanding the dot product helps in analyzing angles between vectors and checking for perpendicularity or orthogonality. It has wide applications not only in mathematics but also in physics and engineering.

Orthogonality

Orthogonality is a concept that means two vectors are perpendicular to each other. This is shown mathematically when their dot product is zero.

• If \( \mathbf{u} \cdot \mathbf{v} = 0 \), then \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal.

In this exercise, after calculating the derivative of the position vector, we checked the dot product of \( \mathbf{r}(t) \) and \( \mathbf{r'}(t) \). It simplifies to zero, indicating that the velocity (derivative) vector is orthogonal to the position vector at every point on the curve.

Importance of Orthogonality

• Orthogonal vectors have no projection on each other. They indicate completely separate directions.
• In the context of motion along a curve, this signifies that the direction of movement is at right angles to the curve at that point.

Recognizing orthogonality is crucial because it implies certain physical properties about the system, such as balance and symmetry.

Differentiation

Differentiation is the process of finding a derivative, which is a fundamental operation in calculus. It helps us understand how a function changes as its input changes. In this problem, differentiation of the parametric equations is crucial to find the tangent (or velocity) vector to the path described. The calculation of \( \mathbf{r}'(t) \) reveals the rate of change of \( \mathbf{r}(t) \) with respect to \( t \).

For the circle, we differentiated:

• The \( x \)-component: \( -a \sin t \)
• The \( y \)-component: \( a \cos t \)

This derivative is crucial because it provides the direction of movement along the curve.

Applications of Differentiation

• Helps in finding tangents and normals to curves, often used in physics to find velocity and acceleration.
• Allows for optimization in calculus, where we seek maximum or minimum values.

Differentiation provides insights not just into changes and motion, but also into how systems behave over time. Understanding it is key to solving complex calculus problems, like the given exercise.