Problem 21 - Advanced Engineering Mathematics

Question

Graph the curve \(C\) that is described by \(\mathbf{r}\) and graph \(\mathbf{r}^{\prime}\) at the indicated value of \(\boldsymbol{t}\). $$ \mathbf{r}(t)=2 \cos t \mathbf{i}+6 \sin t \mathbf{j} ; t=\pi / 6 $$

Step-by-Step Solution

Answer

The curve is an ellipse, and the tangent at \( t = \pi/6 \) is \(-\mathbf{i} + 3\sqrt{3} \mathbf{j} \).

Step 1: Parametric Equations

The vector function \( \mathbf{r}(t) = 2 \cos t \mathbf{i} + 6 \sin t \mathbf{j} \) can be written with parametric equations for \( x \) and \( y \). From \( \mathbf{r}(t) \), we have \( x = 2 \cos t \) and \( y = 6 \sin t \). These equations describe an ellipse.

Step 2: Sketching the Parametric Curve

Recognizing that \( x = 2 \cos t \) and \( y = 6 \sin t \) outlines an ellipse, we know that its major axis is along the \( y \)-axis with length 12, and the minor axis is along the \( x \)-axis with length 4. Sketch the ellipse centered at the origin.

Step 3: Finding the Derivative of \( \mathbf{r}(t) \)

Differentiate \( \mathbf{r}(t) \) with respect to \( t \). The derivative \( \mathbf{r}'(t) = \frac{d}{dt}(2 \cos t \mathbf{i} + 6 \sin t \mathbf{j}) = -2 \sin t \mathbf{i} + 6 \cos t \mathbf{j} \).

Step 4: Evaluating \( \mathbf{r}'(t) \) at \( t = \pi/6 \)

Substitute \( t = \pi/6 \) into \( \mathbf{r}'(t) \): \( \mathbf{r}'(\pi/6) = -2 \sin(\pi/6) \mathbf{i} + 6 \cos(\pi/6) \mathbf{j} = -2(1/2) \mathbf{i} + 6(\sqrt{3}/2) \mathbf{j} = -\mathbf{i} + 3\sqrt{3} \mathbf{j} \). This vector is the slope of the tangent at \( t = \pi/6 \).

Step 5: Plot the Tangent Vector

At the point \((\sqrt{3}, 3)\), which corresponds to \( t = \pi/6 \), draw the vector \(-\mathbf{i} + 3\sqrt{3} \mathbf{j}\), starting at this point and indicating its direction.

Key Concepts

Ellipse

An ellipse is a smooth, closed curve that resembles a stretched circle. It's one of the conic sections that can be formed by slicing a cone with a plane. In the exercise, the ellipse is represented using parametric equations:

• The equation for the x-coordinate is: \( x = 2 \cos t \)
• The equation for the y-coordinate is: \( y = 6 \sin t \)

These equations are used to outline the shape of the ellipse on the Cartesian plane.
The coefficients of the trigonometric functions indicate the semi-axis lengths:

• The semi-major axis along the y-axis has a length of 6.
• The semi-minor axis along the x-axis has a length of 2.

The ellipse is centered at the origin because values of \( t \) resulting in zero coordinate shifts do not change the origin location. The primary distinguishing feature of an ellipse is that the sum of the distances from any point on the ellipse to two focus points is constant.

Vector Calculus

Vector calculus is a branch of mathematics focused on vector fields and differential operators. In the context of the problem, the function \( \mathbf{r}(t) = 2 \cos t \mathbf{i} + 6 \sin t \mathbf{j} \) is a vector equation representing a parametric curve.
This type of equation allows us to describe geometric shapes, like the ellipse, by defining a position vector \( \mathbf{r}(t) \) that changes over time with respect to the parameter \( t \).

• The vector \( \mathbf{r}(t) \) is composed of its components along the \( x \)-axis and \( y \)-axis.
• In this case, \( \mathbf{i} \) and \( \mathbf{j} \) represent unit vectors in the direction of the x-axis and y-axis, respectively.

Such vector descriptions are instrumental for performing operations like rotations, translations, and finding intersections in space. Through vector calculus, one can also calculate derivatives, as shown in the solution, to find slopes of tangents and velocities along curves.

Derivative

The derivative of a function provides the rate at which a quantity changes with respect to another quantity, most often time. When dealing with parametric equations and vector functions, differentiating helps us understand how certain points on a curve behave as the parameter \( t \) changes.
In this exercise, we differentiated the vector function:

• The original function is \( \mathbf{r}(t) = 2 \cos t \mathbf{i} + 6 \sin t \mathbf{j} \).
• The derivative or tangent vector, \( \mathbf{r}'(t) \), is given by \( \frac{d}{dt}(2 \cos t \mathbf{i} + 6 \sin t \mathbf{j}) = -2 \sin t \mathbf{i} + 6 \cos t \mathbf{j} \).

This derivative tells us the direction and speed of the point on the ellipse as \( t \) changes.
Evaluating \( \mathbf{r}'(t) \) at a specific \( t \) value, like \( \pi/6 \), allows us to find the exact slope of the tangent line at that point on the curve. The vector \( -\mathbf{i} + 3\sqrt{3} \mathbf{j} \) obtained indicates how the shape is stretching or moving at that instant, providing insights into the curve's local behavior.