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: Show that the given parameterized circle \(\mathbf{r}(t) = \langle a \cos t, a \sin t \rangle\) is orthogonal to its derivative \(\mathbf{r}^\prime(t)\) for all \(t\). Answer: The parameterized circle \(\mathbf{r}(t)\) and its derivative \(\mathbf{r}^\prime(t)\) are orthogonal because their dot product is always 0, as shown in the step-by-step solution above.

Step-by-step solution

Compute \(\mathbf{r}^\prime(t)\)

First, we need to find the derivative of the parameterized circle with respect to \(t\). To do this, we differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\): $$\mathbf{r}^\prime(t) = \frac{d}{dt} \langle a \cos t, a \sin t \rangle = \langle -a \sin t, a \cos t \rangle.$$

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

Next, we need to find the dot product of \(\mathbf{r}(t)\) and its derivative, \(\mathbf{r}^\prime(t)\). The dot product is given by: $$\mathbf{r}(t) \cdot \mathbf{r}^\prime(t) = \langle a \cos t, a \sin t \rangle \cdot \langle -a \sin t, a \cos t \rangle = -a^2 \cos t \sin t + a^2 \sin t \cos t.$$

Show the dot product is 0 for all \(t\)

To show that the two functions are orthogonal, we need to prove that their dot product is 0 for all values of \(t\). From step 2, we have: $$\mathbf{r}(t) \cdot \mathbf{r}^\prime(t) = -a^2 \cos t \sin t + a^2 \sin t \cos t = 0$$ Since it is always 0, this means that \(\mathbf{r}(t)\) is orthogonal to \(\mathbf{r}^\prime(t)\) for all \(t\).

Key concepts

Parametric Equations

When delving into the world of vector calculus, parametric equations are a fundamental concept to grasp. These equations express a set of quantities as explicit functions of one or more independent variables, known as parameters. For instance, consider the problem where the path of a circle in a plane is described by \(\mathbf{r}(t)=\langle a \cos t, a \sin t\rangle\), with \(t\) being the parameter and \(a\) being a fixed positive real number. Here, the position of a point on the circle at any given time \(t\) is provided by the functions \(a \cos t\) and \(a \sin t\).

In practice, parametric equations are incredibly useful for mapping out the trajectory of moving objects, tracing paths in space, and solving complex geometric problems. They give us the ability to describe motion and change in a clear and concise manner. As seen in the exercise, this circle’s location at any time is neatly encapsulated by the cosine and sine functions, giving us a dynamic view of its geometry.

Dot Product

The dot product, also known as the scalar product, measures the extent to which two vectors in the same space are aligned. Mathematically, for two vectors \(\mathbf{u}\) and \(\mathbf{v}\), the dot product is defined as \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \ldots + u_nv_n\), where \(u_1, u_2, \ldots, u_n\) and \(v_1, v_2, \ldots, v_n\) are the components of \(\mathbf{u}\) and \(\mathbf{v}\) respectively. The result is a single number.

In our exercise, we used the dot product to investigate the relationship between a vector and its derivative. Specifically, we have vector \(\mathbf{r}(t)\) and its derivative \(\mathbf{r}^\prime(t)\), calculated as \(\langle -a \sin t, a \cos t \rangle\). The dot product \(\mathbf{r}(t) \cdot \mathbf{r}^\prime(t)\) elucidates whether the original vector and its rate of change are aligned or, as we aim to show, orthogonal (perpendicular to each other). It’s an essential tool for understanding vector dynamics and geometry.

Orthogonal Vectors

Orthogonal vectors are two vectors that meet at a right angle, thereby being perpendicular to each other. In terms of the dot product, two vectors are orthogonal if and only if their dot product is zero. This scenario signifies that they have no component of one in the direction of the other.

In context with our circle problem, showing orthogonality between the position vector \(\mathbf{r}(t)\) and its derivative \(\mathbf{r}^\prime(t)\) indicates that the tangent (represented by \(\mathbf{r}^\prime(t)\)) to the circle at any point is perpendicular to the radius (represented by \(\mathbf{r}(t)\)) at that point. This holds true for all \(t\), a property intrinsic to circles in Euclidean geometry. Understanding orthogonality is key in many areas of mathematics and physics, as it often comes into play in topics ranging from geometry and trigonometry to signal processing and linear algebra.