Problem 36 - 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 \(t\) . $$ \mathbf{r}(t)=e^{t} \mathbf{i}+e^{-t} \mathbf{j}, \quad t=0 $$

Step-by-Step Solution

Answer

The curve is a hyperbola. \(\mathbf{r}^{\prime}(t) = e^t \mathbf{i} - e^{-t} \mathbf{j}\). At \(t=0\), \(\mathbf{r}(0)\) and \(\mathbf{r}^{\prime}(0)\) can be sketched from \((1, 1)\).

Step 1: Sketch the Plane Curve

The vector equation given is \(\mathbf{r}(t) = e^t \mathbf{i} + e^{-t} \mathbf{j}\). This represents the curve in the \(xy\)-plane, with \(x = e^t\) and \(y = e^{-t}\). The equation is parameterized by \(t\). To sketch this, note that as \(t\) changes from negative to positive, \(x\) will increase exponentially, and \(y\) will decrease from a large positive value towards zero. This curve forms a hyperbola in the first quadrant of the Cartesian plane.

Step 2: Find the Derivative \( \mathbf{r}^{\prime}(t) \)

To find \(\mathbf{r}^{\prime}(t)\), differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\). Differentiating \(e^t\) gives \(e^t\), and differentiating \(e^{-t}\) gives \(-e^{-t}\). Therefore, \(\mathbf{r}^{\prime}(t) = e^t \mathbf{i} - e^{-t} \mathbf{j}\).

Step 3: Evaluate \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\) at \(t=0\)

Substitute \(t=0\) into \(\mathbf{r}(t)\): \(\mathbf{r}(0) = e^0 \mathbf{i} + e^{0} \mathbf{j} = \mathbf{i} + \mathbf{j}\). Substitute \(t=0\) into \(\mathbf{r}^{\prime}(t)\): \(\mathbf{r}^{\prime}(0) = e^0 \mathbf{i} - e^{0} \mathbf{j} = \mathbf{i} - \mathbf{j}\).

Step 4: Sketch the Position and Tangent Vectors

The position vector \(\mathbf{r}(0) = \mathbf{i} + \mathbf{j}\) points to the position \((1, 1)\) on the plane from the origin. The tangent vector \(\mathbf{r}^{\prime}(0) = \mathbf{i} - \mathbf{j}\) starts at \((1, 1)\) and points in the direction \((1, -1)\), indicating the direction of the curve at \(t=0\). Sketch these vectors to illustrate their orientation and positions on the curve.

Key Concepts

Plane Curve Sketching

When sketching the plane curve of a given vector equation, you start with the parameterized form. The vector equation \( \mathbf{r}(t) = e^t \mathbf{i} + e^{-t} \mathbf{j} \) describes a curve in the plane by breaking it down into components. Here, \( x = e^t \) and \( y = e^{-t} \).

This process involves understanding how each component changes as \( t \) varies. Consider the behavior of the exponential functions involved:

• \( x = e^t \) means that as \( t \) increases, \( x \) grows exponentially larger.
• \( y = e^{-t} \) implies that as \( t \) increases, \( y \) becomes smaller, moving towards zero.

The interplay of these changing values forms a hyperbola in the first quadrant. By plotting points for specific values of \( t \) and connecting them smoothly, you can visualize the full curve, showcasing how \( (x, y) \) coordinates move on the plane as \( t \) transitions from negative to positive.

Derivative of Vector Functions

Finding the derivative of vector functions, like \( \mathbf{r}(t) = e^t \mathbf{i} + e^{-t} \mathbf{j} \), requires differentiating each component with respect to \( t \). This process yields the velocity or tangent vector, which denotes the direction of the curve at any point.

To calculate \( \mathbf{r}^{\prime}(t) \):

• Differentiate \( e^t \): the derivative is \( e^t \), maintaining its form since \( e^x \) is its own derivative.
• Differentiate \( e^{-t} \): employing the chain rule, the derivative becomes \(-e^{-t} \).

Thus, the derivative is \( \mathbf{r}^{\prime}(t) = e^t \mathbf{i} - e^{-t} \mathbf{j} \).

By understanding these derivatives as velocities, you see how the curve dynamically moves through different points as \( t \) changes. These velocities convey not just magnitude but also direction, aiding in further analysis or graphical representation.

Position and Tangent Vectors

Position and tangent vectors provide insights into the specific orientation and direction of a curve at a certain point in time. Evaluating \( \mathbf{r}(t) \) and its derivative \( \mathbf{r}^{\prime}(t) \) at given values of \( t \) gives clear locations and directions.

For instance, when \( t = 0 \):

• Substitute into \( \mathbf{r}(t) \): \( \mathbf{r}(0) = \mathbf{i} + \mathbf{j} \), which translates to the point \((1, 1)\) on the plane.
• Substitute into \( \mathbf{r}^{\prime}(t) \): \( \mathbf{r}^{\prime}(0) = \mathbf{i} - \mathbf{j} \), indicating a vector pointing in the direction \((1, -1)\).

The position vector points to where the object is at \( t = 0 \), while the tangent vector indicates the object's immediate travel direction. Drawing these vectors on a plane not only grounds students in visually interpreting vector calculus problems but also enhances spatial understanding.