Question

Evaluate each line integral using the given curve \(C\). \(\int x^{2} d x+d y+y d z ; \mathbf{C}\) is the curve \(\mathbf{r}(t)=\left\langle t, 2 t, t^{2}\right\rangle,\) for \(0 \leq t \leq 3\)

Short answer

Question: Evaluate the line integral \(\int (x^2 dx + dy + ydz)\) along the curve defined by the parameterization \(\mathbf{r}(t) = \left\langle t, 2t, t^2 \right\rangle\), where \(0 \leq t \leq 3\). Answer: The value of the line integral is 51.

Step-by-step solution

Parameterize the curve

Given the curve with parameterization \(\mathbf{r}(t) = \left\langle t, 2t, t^2 \right\rangle\). We have: - x = t - y = 2t - z = t^2 for \(0 \leq t \leq 3\).

Compute the differential elements dx, dy, and dz

To compute the differential elements, we take the derivatives of x(t), y(t), and z(t) with respect to t: - \(\frac{d}{dt}x(t) = 1\), so dx = dt. - \(\frac{d}{dt}y(t) = 2\), so dy = 2dt. - \(\frac{d}{dt}z(t) = 2t\), so dz = 2tdt.

Substitute the expressions for the differentials and variables into the line integral

Now substitute the expressions found in the previous steps into the given line integral function: \(\int (x^2 dx + dy + ydz)\). This gives us: \(\int_0^3 (t^2(1)dt + (2)dt + (2t)(2t)dt) = \int_0^3 (t^2 dt + 2dt + 4t^2 dt)\).

Perform the integration

Now, integrate the expression with respect to t: \(\int_0^3 (t^2 dt + 2dt + 4t^2 dt) = \left[\frac{1}{3}t^3 + 2t + \frac{4}{3}t^3\right]_0^3\) Evaluate at the limits: \(\left[\frac{1}{3}t^3 + 2t + \frac{4}{3}t^3\right]_0^3 = \left(\frac{1}{3}(3)^3 + 2(3) + \frac{4}{3}(3)^3\right) - \left(\frac{1}{3}(0)^3 + 2(0) + \frac{4}{3}(0)^3\right)\) Calculate the final result: \(= \left(\frac{1}{3}(27) + 6 + \frac{4}{3}(27)\right) - (0) = 9 + 6 + 36 = \boxed{51}\).

Key concepts

Parameterization

Parameterization is a crucial step in solving line integrals along curves. It involves expressing a curve or path in terms of a single variable, usually denoted by parameter \(t\). In this exercise, the given curve \( \mathbf{C} \) is represented by the vector function \( \mathbf{r}(t) = \langle t, 2t, t^2 \rangle \). This decomposition shows how each component, \(x(t)\), \(y(t)\), and \(z(t)\), varies as \(t\) changes.
\(x = t\), \(y = 2t\), and \(z = t^2\) make it easier to compute other required quantities. The chosen interval for \(t\) (from 0 to 3 in this problem) determines the section of the curve that is relevant for the integration.
By parameterizing a curve, we transform it into a form suitable for calculus operations, allowing us to apply integration techniques more easily.

Differential Elements

Differential elements are small changes in a variable, and they are essential in line integrals. The original components of \(x\), \(y\), and \(z\) depend on \(t\), so we derive their changes by differentiating with respect to \(t\). This gives us the differential elements needed for the integral:

• \( dx = dt \) arises from \( \frac{d}{dt}x(t) = 1 \)
• \( dy = 2dt \) because \( \frac{d}{dt}y(t) = 2 \)
• \( dz = 2tdt \) from \( \frac{d}{dt}z(t) = 2t \)

This differential process breaks down the changes along the curve into manageable components. Each differential represents a tiny segment of movement along the path in its respective direction. Understanding these elements is crucial for setting up and solving the line integral.

Integration

Integration is the mathematical process of summing up differential elements to find the total effect along a curve or within a domain. For line integrals, this involves substituting expressions for differentials and the corresponding parameterized variables into the integral.
In the solved exercise, the line integral \[ \int (x^2 dx + dy + ydz) \]becomes\[ \int_0^3 (t^2 dt + 2dt + 4t^2 dt) \]
after substituting the parameterized variables and their differentials. Performing integration over this expression with respect to \(t\) involves integrating each term separately, combining them, and then evaluating the result within the given bounds (from 0 to 3). This process enables us to calculate the accumulated quantity, which, in this case, results in the final answer of 51.

Vector Calculus

Vector calculus is the branch of mathematics that deals with vector fields and operations on vectors. It provides tools to deal with physical phenomena like flow and movement, common in physics and engineering. Line integrals are a core part of vector calculus, focusing on integrating vector fields along curves.
In this exercise, the line integral is evaluated over a parameterized curve, with each part of the equation (\(x^2 dx\), \(dy\), and \(ydz\)) involving vector elements. Concepts of vector calculus allow us to interpret and solve these integrals, understanding how different components of a vector field contribute along a curve. This holistic approach ties together multidimensional calculus operations within continuous fields, offering insights into diverse applications like electromagnetism and fluid dynamics.