Problem 24 - Calculus Early Transcendentals

Question

Find the curvature of \(\mathbf{r}(t)=\left\langle e^{t} \cos t, e^{t} \sin t, t\right\rangle\) at the point \((1,0,0) .\)

Step-by-Step Solution

Answer

The curvature at the point (1,0,0) is \( \frac{2\sqrt{2}}{3\sqrt{3}} \).

Step 1: Understand the curvature formula

The formula for the curvature \( \kappa \) of a vector function \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) is given by:\[\kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}\]Here, \( \mathbf{r}'(t) \) is the first derivative and \( \mathbf{r}''(t) \) is the second derivative of \( \mathbf{r}(t) \). We need to compute these derivatives to find the curvature.

Step 2: Compute the first derivative \( \mathbf{r}'(t) \)

Find the derivative of each component of \( \mathbf{r}(t) = \langle e^t \cos t, e^t \sin t, t \rangle \):- \( x'(t) = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t) \)- \( y'(t) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) \)- \( z'(t) = 1 \)Thus, \( \mathbf{r}'(t) = \langle e^t (\cos t - \sin t), e^t (\sin t + \cos t), 1 \rangle \).

Step 3: Compute the second derivative \( \mathbf{r}''(t) \)

Differentiate the first derivative \( \mathbf{r}'(t) \) again:- \( x''(t) = e^t (\cos t - \sin t) - e^t (\cos t + \sin t) = -2e^t \sin t \)- \( y''(t) = e^t (\sin t + \cos t) + e^t (\cos t - \sin t) = 2e^t \cos t \)- \( z''(t) = 0 \)The second derivative is \( \mathbf{r}''(t) = \langle -2e^t \sin t, 2e^t \cos t, 0 \rangle \).

Step 4: Compute the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \)

The cross product of \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \) is:\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \e^t (\cos t - \sin t) & e^t (\sin t + \cos t) & 1 \-2e^t \sin t & 2e^t \cos t & 0\end{vmatrix}\]This determinant evaluates to: \[ \mathbf{i} (0 - 2e^t \cos t) - \mathbf{j}(0 + 2e^t \sin t) + \mathbf{k}(2e^{2t} (\cos^2 t + \sin^2 t)) \]Simplifying, we obtain \( \langle -2e^t \cos t, -2e^t \sin t, 2e^{2t} \rangle \).

Step 5: Evaluate at the specific point \( (1,0,0) \)

The given point \( (1,0,0) \) corresponds to \( t = 0 \), as \( \mathbf{r}(0) = \langle e^0 \cos(0), e^0 \sin(0), 0 \rangle = \langle 1, 0, 0 \rangle \). At \( t = 0 \):- \( \mathbf{r}'(0) = \langle 1, 1, 1 \rangle \)- \( \mathbf{r}''(0) = \langle 0, 2, 0 \rangle \)- \( \mathbf{r}'(0) \times \mathbf{r}''(0) = \langle -2, 0, -2 \rangle \) - The magnitude is \( \| \mathbf{r}'(0) \times \mathbf{r}''(0) \| = \sqrt{(-2)^2 + 0^2 + (-2)^2} = 2\sqrt{2} \)- \( \| \mathbf{r}'(0) \| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \)

Step 6: Calculate the curvature \( \kappa \)

Plugging into the curvature formula gives:\[\kappa = \frac{\| \mathbf{r}'(0) \times \mathbf{r}''(0) \|}{\| \mathbf{r}'(0) \|^3} = \frac{2\sqrt{2}}{(\sqrt{3})^3} = \frac{2\sqrt{2}}{3\sqrt{3}}\]Upon simplifying the expression using \( \sqrt{3^3} = \sqrt{27} = 3\sqrt{3} \), we further simplify it to get the final curvature value.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that involves differentiating and integrating vector fields, often in two or three-dimensional Euclidean spaces. It extends the concepts of calculus to multi-dimensional spaces using vectors, which can represent quantities that have both magnitude and direction. In the context of the given problem, you are working with a vector function, \( \mathbf{r}(t) \), that describes a path in three-dimensional space. This function is defined by components incorporating exponential, trigonometric, and linear functions.

Key operations in vector calculus include:

• Gradient: A vector operator that represents the rate and direction of change in a scalar field.
• Divergence: A scalar representation of the rate of expansion of a vector field in a given region.
• Curl: A vector field that represents the rotational motion or circulation at every point in a vector field.

Vector calculus is essential for describing physical phenomena, such as electromagnetic fields and fluid dynamics, providing a framework to model the movement of objects in space.

Cross Product

The cross product is a binary operation on two vectors in three-dimensional space, producing a third vector that is perpendicular to both input vectors. This operation is essential when determining the curvature of a vector path, as seen in the exercise. The cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is given by:\[\mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle\]This product yields the normal vector, which can help in determining the nature of the curvature of a path.

The magnitude of the cross product can be interpreted as the area of the parallelogram that the vectors span. In curvature calculation, the magnitude of \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) aids in deriving the curvature \( \kappa \) which indicates how sharply a curve bends at a given point.

Derivative

The derivative measures how a function changes as its input changes. For a vector function, taking the derivative means finding the rate of change of each component of the vector. In the problem, both the first and second derivatives of \( \mathbf{r}(t) = \langle e^t \cos t, e^t \sin t, t \rangle \) are necessary.

First Derivative: The first derivative, \( \mathbf{r}'(t) \), indicates the velocity or tangent to the curve at each point. For the vector function, this involves differentiating each component with respect to \( t \):

• \( x'(t) = e^t (\cos t - \sin t) \)
• \( y'(t) = e^t (\sin t + \cos t) \)
• \( z'(t) = 1 \)

Second Derivative: The second derivative, \( \mathbf{r}''(t) \), reflects the change in the velocity, or the acceleration along the curve:

• \( x''(t) = -2e^t \sin t \)
• \( y''(t) = 2e^t \cos t \)
• \( z''(t) = 0 \)

Derivatives provide critical insights into the nature of the trajectory and are crucial for applications in physics, engineering, and beyond.

Curvature Formula

Curvature measures how quickly a curve changes direction at a given point. The formula for the curvature \( \kappa \) of a space curve given by the vector function \( \mathbf{r}(t) \) is:\[\kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}\]This formula entails the cross product of the first and second derivatives, and its magnitude provides a measure of the "bending" at any point along the curve.

To find \( \kappa \) at a specific point, such as \( (1, 0, 0) \), you evaluate \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \) at \( t = 0 \). The calculation involves:

• Finding \( \| \mathbf{r}'(0) \times \mathbf{r}''(0) \| \).
• Calculating \( \| \mathbf{r}'(0) \|^3 \).
• Dividing these outcomes to obtain \( \kappa \).

This approach reveals how much the curve deviates from being straight and is an important concept in fields such as mechanical design, where understanding curvature helps in designing smooth paths and trajectories.