Problem 35 - Essential Calculus Early Transcendentals

Question

\(33-38=\) (a) Sketch the plane curve with the given vector equation. (b) Find \(\mathbf{r}^{\prime}(t) .\) (c) Sketch the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{r}^{\prime}(t)\) for the given value of \(t\) . $$ \mathbf{r}(t)=\sin t \mathbf{i}+2 \cos t \mathbf{j}, \quad t=\pi / 4 $$

Step-by-Step Solution

Answer

The position vector is \( \mathbf{r}\left(\pi/4\right) = \frac{\sqrt{2}}{2} \, \mathbf{i} + \sqrt{2} \, \mathbf{j} \) and the tangent vector is \( \mathbf{r}'\left(\pi/4\right) = \frac{\sqrt{2}}{2} \, \mathbf{i} - \sqrt{2} \, \mathbf{j} \). The curve is an ellipse centered at the origin.

Step 1: Understanding the Vector Equation

The provided vector equation is \( \mathbf{r}(t) = \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \). This equation represents a curve in the plane as a function of \( t \), and consists of an \( i \) component (\( x \)-axis) and a \( j \) component (\( y \)-axis). The coordinates of any point on this curve are \((\sin t, 2\cos t)\).

Step 2: Sketching the Plane Curve

To sketch the plane curve, observe that the equation \( (x, y) = (\sin t, 2 \cos t) \) resembles a parametric equation for an ellipse. For any value of \( t \), \( x = \sin t \) and \( y = 2\cos t \), meaning the curve is an ellipse centered at the origin with a semi-major axis along the \( y \)-axis twice as long as that along the \( x \)-axis.

Step 3: Calculating the Derivative

To find \( \mathbf{r}'(t) \), differentiate \( \mathbf{r}(t) = \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \):\[ \mathbf{r}'(t) = \cos t \, \mathbf{i} - 2 \sin t \, \mathbf{j}. \]

Step 4: Substituting the Given Value of t

Substitute \( t = \frac{\pi}{4} \) into \( \mathbf{r}(t) \) and \( \mathbf{r}'(t) \):- For \( \mathbf{r}\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) \, \mathbf{i} + 2\cos\left(\frac{\pi}{4}\right) \, \mathbf{j} \), this yields \( \frac{\sqrt{2}}{2} \, \mathbf{i} + \sqrt{2} \, \mathbf{j} \).- For \( \mathbf{r}'\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) \, \mathbf{i} - 2\sin\left(\frac{\pi}{4}\right) \, \mathbf{j} \), this yields \( \frac{\sqrt{2}}{2} \, \mathbf{i} - \sqrt{2} \, \mathbf{j} \).

Step 5: Sketching the Vectors

Using the calculated vectors from Step 4:- Plot the position vector \( \mathbf{r}\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \, \mathbf{i} + \sqrt{2} \, \mathbf{j} \) starting from the origin.- Plot the tangent vector \( \mathbf{r}'\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \, \mathbf{i} - \sqrt{2} \, \mathbf{j} \) starting at the point found by \( \mathbf{r}(t) \). This vector should point in the direction corresponding to \( t = \frac{\pi}{4} \).

Key Concepts

Parametric Equations

Parametric equations are an elegant way to define a curve by expressing the coordinates of points on the curve as functions of a variable called a parameter, typically denoted as \( t \). These equations allow us to describe complex shapes and motions in a plane or space. Unlike the Cartesian form, which might directly express \( y \) as a function of \( x \), parametric equations treat both \( x \) and \( y \) as separate functions of \( t \):\( x = f(t) \) and \( y = g(t) \).

For instance, the parametric equations for an ellipse are \( x = a \cos t \) and \( y = b \sin t \), where \( a \) and \( b \) determine the lengths of the semi-major and semi-minor axes, respectively. This differs from the Cartesian equation of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The benefit of parametric equations is that they are particularly useful in describing motion, where \( t \) can represent time. In our example, \( \mathbf{r}(t) = \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \), the curve forms an ellipse where \( \sin t \) gives the \( x \) component and \( 2 \cos t \) provides the \( y \) component.

Let's summarize some advantages:

• Easier to describe curves that are difficult to express in Cartesian coordinates.
• Useful in physics and engineering for describing trajectories.
• Enhanced flexibility in mathematical modeling.
• Allows for parameterization of surfaces in higher dimensions.

Vector Functions

Vector functions are pivotal in vector calculus as they extend the concept of functions to dimensions higher than just numbers. As vectors have magnitude and direction, a vector function assigns a vector to each value of the parameter, \( t \). The vector function \( \mathbf{r}(t) \) can be expressed as a combination of basis vectors, such as \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), in three-dimensional space.

In our exercise, the vector function \( \mathbf{r}(t) = \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \) maps the parameter \( t \) to a vector in the plane. Each component \( \sin t \) and \( 2 \cos t \) are the scalar functions multiplied by the unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), respectively. The beauty of vector functions lies in how they can neatly represent paths traced out by moving points in space.

Key Points About Vector Functions:

• Essential for modeling dynamics, like the position of a moving particle over time.
• Enables us to express and manipulate shapes and motions in multiple dimensions.
• Facilitate the transition from two-dimensional to three-dimensional analysis.
• Provide a framework for computing derivatives and integrals in vector calculus.

Derivatives in Calculus

Derivatives are a core concept in calculus, representing the rate of change of a function. In the context of vector calculus, derivatives also apply to vector functions, highlighting changes in both magnitude and direction.

When differentiating a vector function like \( \mathbf{r}(t) = \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \), we do it component-wise to get \( \mathbf{r}'(t) = \cos t \, \mathbf{i} - 2 \sin t \, \mathbf{j} \). This derivative, called the tangent vector, points in the direction of motion at any given \( t \).

The process of taking derivatives showcases the dynamic aspect of vector functions, enabling exploration of:

• Velocity: The first derivative \( \mathbf{r}'(t) \) tells us how position changes.
• Acceleration: The second derivative \( \mathbf{r}''(t) \) examines changes in velocity.
• Tangential and normal components: Useful in calculating curvatures and centripetal forces.

Calculating these derivatives allows us to understand the geometrical properties of curves, such as slope, curvature, and inflection points, in ways that are analogous to scalar functions in regular calculus.