Question

Relationship between \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) 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: Given the parametric equation for a circle of radius \(a\), \(\mathbf{r}(t) = \langle a \cos t, a \sin t \rangle\), prove that the velocity vector (the derivative of the parametric equation) is orthogonal to the position vector at all points on the circle. Answer: The velocity vector, \(\mathbf{r}^{\prime}(t) = \langle -a \sin t, a \cos t \rangle\), is orthogonal to the position vector, \(\mathbf{r}(t) = \langle a \cos t, a \sin t \rangle\), at all points on the circle because the dot product of these two vectors is always equal to zero: \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t) = -a^{2} \cos t \sin t + a^{2} \cos t \sin t = 0\).

Step-by-step solution

Compute the derivative of the given parametric equation for the circle.

First, we need to compute the derivative of the given parametric equation: \(\mathbf{r}(t) = \langle a \cos t, a \sin t \rangle\). The derivative of a parametric equation is computed by taking the derivative of each component of the equation with respect to the parameter \(t\): \(\mathbf{r}^{\prime}(t) = \langle -a \sin t, a \cos t \rangle\)

Calculate the dot product of \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\).

Next, we need to compute the dot product of \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\) to show that they are orthogonal for all values of \(t\). The dot product is calculated as follows: \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t) = (a \cos t)(-a \sin t) + (a \sin t)(a \cos t)\)

Simplify the dot product and check if it is equal to zero.

Now, we need to simplify the expression for the dot product and check if it is equal to zero: \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t) = -a^{2} \cos t \sin t + a^{2} \cos t \sin t = 0\) Since the dot product of \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\) is equal to zero for all values of \(t\), we can conclude that \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal at all points on the circle.

Key concepts

Derivative of Parametric Equations

Understanding the derivative of parametric equations is essential in calculus, particularly when dealing with the motion of an object along a path. In parametric equations, each coordinate of a point is expressed as a function of a third variable, usually time, denoted by 't'.

Let's take the parametric equation \(\mathbf{r}(t) = \langle a \cos t, a \sin t \rangle\) from the exercise, which represents a circle with radius 'a'. To find the derivative, we differentiate each component with respect to 't'. Hence, \(\mathbf{r}^{\prime}(t)\) or the velocity vector, is \(\langle -a\sin t, a\cos t \rangle\), representing the instantaneous rate of change of \(\mathbf{r}\) with respect to 't'.

Why is this important? The derivative provides information on the tangent to the path at any point, and for a circle, it gives the direction of motion at any instant. It's also crucial for finding whether vectors are orthogonal at specified points, an integral part of understanding the geometry of curves in a parametric form.

Dot Product

Moving onto the concept of the dot product, also known as the scalar product, it's a way of multiplying two vectors to yield a scalar quantity. This operation is invaluable when we wish to find the angle between vectors or to determine if vectors are orthogonal.

As seen in the exercise solution, the dot product of vectors \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\) is computed by multiplying corresponding components of the vectors and then adding them together. In this case, \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t) = (a \cos t)(-a \sin t) + (a \sin t)(a \cos t)\). When simplified, we see that the dot product equals zero, which has a particular significance.

Why does it equate to zero? A zero dot product indicates that vectors are orthogonal, meaning they are at a 90-degree angle to each other. This reveals not just their geometric relationship but also can signify perpendicular motion, especially relevant in physics and engineering problems.

Orthogonal Vectors

Lastly, the concept of orthogonal vectors plays a pivotal role in both geometry and algebra. Vectors are said to be orthogonal if their dot product equals zero, which implies that they are perpendicular to each other.

In the context of the exercise, the vectors \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) being orthogonal mean that at every point on the circle, the radius vector and the velocity vector are perpendicular. This aligns perfectly with the geometric properties of a circle, where the radius is always perpendicular to the circle's tangent at any given point.

Understanding this concept assists in visualizing complex shapes and their properties, aids in solving problems regarding forces in physics, and helps in understanding gradient vectors in higher dimensions in calculus. It's a fundamental concept that helps bridge the abstract mathematical theories with tangible, real-world applications.