Question

Relationship between \(\mathbf{r}\) and \(\mathbf{r}^{\prime}.\) Consider the ellipse \(\mathbf{r}(t)=\langle 2 \cos t, 8 \sin t, 0\rangle,\) for \(0 \leq t \leq 2 \pi\) Find all points on the ellipse at which \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal.

Short answer

Question: Identify the points on the ellipse parametrized by the position vector $\mathbf{r}(t) = \langle 2\cos{t}, 8\sin{t}, 0 \rangle$ at which the position vector is orthogonal to its derivative. Answer: The points are $\langle 2, 0, 0 \rangle$, $\langle 0, 8, 0 \rangle$, $\langle -2, 0, 0 \rangle$, and $\langle 0, -8, 0 \rangle$.

Step-by-step solution

1. Find the derivative of the position vector

The derivative of the position vector \(\mathbf{r}(t)\) with respect to \(t\) can be found component-wise: $$ \mathbf{r}^{\prime}(t) = \frac{d\mathbf{r}}{dt} = \left\langle \frac{d}{dt}(2\cos{t}), \frac{d}{dt}(8\sin{t}), \frac{d}{dt}(0) \right\rangle. $$ Now differentiate each component: $$ \mathbf{r}^{\prime}(t) = \left\langle -2\sin{t}, 8\cos{t}, 0 \right\rangle. $$

2. Orthogonality condition using the dot product

Two vectors are orthogonal if their dot product is zero. Therefore, we can use this condition to find the values of \(t\) when \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\) are orthogonal: $$ \mathbf{r}(t)\cdot\mathbf{r}^{\prime}(t) = \langle 2\cos{t}, 8\sin{t}, 0 \rangle \cdot \langle -2\sin{t}, 8\cos{t}, 0 \rangle = 0. $$

3. Solve for the parameter \(t\)

Compute the dot product and solve for \(t\): $$ -4\cos{t}\sin{t} + 64\sin{t}\cos{t} = 0. $$ Move the terms to the left side and factor out \(\sin{t}\cos{t}\): $$ 60\sin{t}\cos{t} = 0. $$ To find when this product is zero, one or both of \(\sin{t}\) and \(\cos{t}\) must be zero, which occurs at the following values of \(t\): $$ t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}. $$

4. Find the points on the ellipse

Now that we have the values of \(t\) where the position vector and its derivative are orthogonal, plug them back into the position vector \(\mathbf{r}(t)\) to find the corresponding points on the ellipse: $$ \mathbf{r}(0) = \langle 2\cos{0}, 8\sin{0}, 0 \rangle = \langle 2, 0, 0 \rangle, $$ $$ \mathbf{r}\left(\frac{\pi}{2}\right) = \langle 2\cos{\frac{\pi}{2}}, 8\sin{\frac{\pi}{2}}, 0 \rangle = \langle 0, 8, 0 \rangle, $$ $$ \mathbf{r}(\pi) = \langle 2\cos{\pi}, 8\sin{\pi}, 0 \rangle = \langle -2, 0, 0 \rangle, $$ $$ \mathbf{r}\left(\frac{3\pi}{2}\right) = \langle 2\cos{\frac{3\pi}{2}}, 8\sin{\frac{3\pi}{2}}, 0 \rangle = \langle 0, -8, 0 \rangle. $$ Thus, the points on the ellipse at which the position vector \(\mathbf{r}\) and its derivative \(\mathbf{r}^{\prime}\) are orthogonal are \(\langle 2, 0, 0 \rangle\), \(\langle 0, 8, 0 \rangle\), \(\langle -2, 0, 0 \rangle\), and \(\langle 0, -8, 0 \rangle\).

Key concepts

Vector Differentiation

Understanding vector differentiation is crucial when studying dynamics or physics. Essentially, it involves finding the rate at which a vector-valued function changes. In the given exercise, we differentiate the position vector \( \mathbf{r}(t) \), a function that assigns a vector to each value of \( t \). This differentiation is performed component-wise, meaning we find the derivatives of each component of \( \mathbf{r}(t) \) with respect to \( t \).

For example, given \( \mathbf{r}(t) = \langle 2 \cos t, 8 \sin t, 0\rangle \), the derivative, \( \mathbf{r}^\prime(t) \), is \( \left\langle -2\sin{t}, 8\cos{t}, 0 \right\rangle \). This rate of change is crucial as it leads us to understand the motion along the curve that \( \mathbf{r}(t) \) describes – in this case, an ellipse.

Dot Product

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. It's defined by multiplying corresponding entries and then summing those products. The key property of the dot product in relation to orthogonality is that when two vectors are orthogonal (perpendicular), their dot product is zero.

In our exercise, we use the dot product to find when the position vector \( \mathbf{r}(t) \) and its derivative are orthogonal. Mathematically, this is presented as \( \mathbf{r}(t) \cdot \mathbf{r}^\prime(t) = 0 \). When the dot product of these two vectors is zero, they are perpendicular to each other at that specific point on the ellipse.

Parametric Equations of an Ellipse

The parametric equations of an ellipse provide a way to describe the locus of all points on an ellipse as a function of a single parameter, commonly \( t \), which is often associated with time in physical problems. In the given exercise, the parametric equations are \( \mathbf{r}(t) = \langle 2 \cos t, 8 \sin t, 0\rangle \), with \( t \) ranging from \( 0 \) to \( 2\pi \).

These equations associate each value of \( t \) with a unique point on the ellipse, providing a smooth curve that represents the ellipse when \( t \) is varied continuously. Understanding these equations is fundamental for analyzing movement along elliptical paths or optimizing various physical and engineering problems.

Trigonometric Functions

Trigonometric functions such as sine and cosine are at the heart of the parametric equations that describe the ellipse in our exercise. They allow the description of complex shapes and motion using simple, periodic functions. For example, in the context of an ellipse, cosine typically controls the horizontal component, and sine the vertical component, or vice versa depending on the parametrization.

The significance of these functions becomes apparent when solving the orthogonality condition: \( 60\sin{t}\cos{t} = 0 \). Here, we need to understand the basic property that these functions are zero at multiples of \( \pi/2 \), allowing us to find the values of \( t \) where the vectors are orthogonal. In simpler terms, trigonometric functions help to relate angles with side ratios in right-angled triangles, providing a bridge between geometric figures and algebraic equations.