Problem 26 - Advanced Engineering Mathematics

Question

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\). $$ \begin{aligned} &\mathbf{F}(x, y, z)=\left(2-e^{z}\right) \mathbf{i}+(2 y-1) \mathbf{j}+\left(2-x e^{2}\right) \mathbf{k} \\ &\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k},(-1,1,-1) \text { to }(2,4,8) \end{aligned} $$

Step-by-Step Solution

Answer

Use substitution to express \( \mathbf{F} \) and compute \( \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} \), then evaluate the integral numerically.

Step 1: Identify the vector field and parametric curve

The given vector field is \( \mathbf{F}(x, y, z) = (2-e^z) \mathbf{i} + (2y-1) \mathbf{j} + (2-xe^2) \mathbf{k} \). The parametric curve is \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \) with \( t \) ranging such that the curve passes from point \((-1,1,-1)\) to \((2,4,8)\).

Step 2: Determine the limits for parameter t

To find the values of \( t \), we compare the parametric point definitions with the given points: \((-1, 1, -1)\) corresponds to \( t = -1 \) and \((2, 4, 8)\) corresponds to \( t = 2 \). Hence, \( t \) ranges from \(-1\) to \(2\).

Step 3: Express F in terms of t

Substitute \( x = t \), \( y = t^2 \), and \( z = t^3 \) into \( \mathbf{F} \) to get: \( \mathbf{F}(t) = (2-e^{t^3}) \mathbf{i} + (2t^2-1) \mathbf{j} + (2-te^2) \mathbf{k} \).

Step 4: Compute the derivative of r(t)

The derivative \( \frac{d\mathbf{r}}{dt} \) is the velocity vector: \( \frac{d\mathbf{r}}{dt} = 1 \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).

Step 5: Compute the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \)

Calculate the dot product: \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = (2-e^{t^3}) \cdot 1 + (2t^2-1) \cdot 2t + (2-te^2) \cdot 3t^2 \).

Step 6: Simplify the expression

Simplify the expression: \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = 2 - e^{t^3} + 4t^3 - 2t + 6t^4 - 3t^3e^2 \).

Step 7: Integrate the dot product over t

Evaluate the integral \( \int_{-1}^{2} (2 - e^{t^3} + 4t^3 - 2t + 6t^4 - 3t^3e^2) \, dt \).

Step 8: Integrate term by term

Perform the integral term by term:- \( \int_{-1}^{2} 2 \, dt = 6 \)- \( \int_{-1}^{2} -e^{t^3} \, dt \) (requires substitution or numerical approximation)- \( \int_{-1}^{2} 4t^3 \, dt = \frac{31}{4} \)- \( \int_{-1}^{2} -2t \, dt = -6 \)- \( \int_{-1}^{2} 6t^4 \, dt = 93 \)- \( \int_{-1}^{2} -3t^3e^2 \, dt = -15.75e^2 \).

Step 9: Sum the results

Combine the results: \[ 6 + 0 + \frac{31}{4} - 6 + 93 - 15.75e^2 \] simplifying gives \[ \frac{103}{4} - 15.75e^2 \] (excluding the result of numeric approximation for the integral of the \(-e^{t^3}\) term).

Step 10: Numeric approximation of e^{t^3} term

The integral \( \int_{-1}^{2} -e^{t^3} \, dt \) is calculated numerically, contributing an additional constant. Given the complexity of \( e^{t^3} \), use numerical methods or software tools to approximate this value.

Key Concepts

Vector Field

A vector field is a function that assigns each point in space a vector. In this exercise, we have the vector field \( \mathbf{F}(x, y, z) = (2-e^z) \mathbf{i} + (2y-1) \mathbf{j} + (2-xe^2) \mathbf{k} \).
It describes how a vector (combination of magnitude and direction) changes as we move through the space defined by the coordinates \((x, y, z)\).

• Each component of the vector field corresponds to one of the coordinate directions \(\mathbf{i}, \mathbf{j},\mathbf{k}\).
• In our case, the vector components depend on the variables \(x, y, z\).

Vector fields find applications in physics, where they can represent force fields, velocity fields, or magnetic fields. Understanding these applications helps to visualize how a vector field operates in real-world scenarios, like how the wind blows across a plain or how an electric field exerts force on a charged particle.

Parametric Curve

A parametric curve provides a way of defining a curve by using parameters. In this instance, we use the variable \( t \) to express the position on a curve \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \).
This representation shows how each coordinate (x, y, z) changes as \( t \) changes.

• By plugging in different values of \( t \), you can find specific points on the curve.
• For instance, when \( t = -1 \), the point is \((-1, 1, -1)\); when \( t = 2 \), the point is \((2, 4, 8)\).

Parametric curves can be useful when analyzing the motion of objects that follow a path in space, like the route taken by a drone or the spiral of a corkscrew. By examining how \( t \) influences position vectors, we gain insight into the trajectory or path of an object.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers and returns a single number. It's a way to multiply two vectors to find a scalar.
In our exercise, we calculate the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \)
to understand how the vector field \( \mathbf{F} \) flows along the curve \( \mathbf{r}(t) \).

• The formula for the dot product is \( \mathbf{a} \cdot \mathbf{b} = a_xb_x + a_yb_y + a_zb_z \). This involves multiplying corresponding components and then summing those products.
• The dot product can tell us if vectors are perpendicular (result is 0), parallel (result is the product of their magnitudes), or something in-between.
• In work and energy terms, the dot product calculates how much of one vector influences or aligns with another, like force along a path.

The dot product enables us to analyze interactions between paths and fields, a concept valuable in physics where we often need to calculate work done by a force field.

Integration

Integration is a mathematical process of finding the integral of a function, which effectively accumulates quantities like area, volume, and other concepts over time or space. In this problem, we focus on integrating the expression from the dot product.
This operation helps determine how the vector field affects the entire path of the curve.

• Integration of the dot product \( \int_{-1}^{2} (2 - e^{t^3} + 4t^3 - 2t + 6t^4 - 3t^3 e^2) \, dt \) is crucial to obtaining the total effect over the curve.
• Breaking it down into parts, we integrate each term separately, like \( \int 2 \, dt \) or \( \int 6t^4 \, dt \), and sum these results.
• Integration can handle more complex functions using techniques like substitution or numerical methods, especially for terms like \(-e^{t^3}\).

Understanding integration is fundamental for problems involving continuous change, such as calculating the total accumulation of fluid or electrical charge along a curve. In practical terms, this results in knowing how much of a field influences overall motion, which is essential in engineering and physics applications.