Problem 11 - Thomas Calculus

Question

Evaluate \(\int _ { C } ( x y + y + z ) d s\) along the curve \(\mathbf { r } ( t ) = 2 \mathrm { ti } +\) \(t \mathbf { j } + ( 2 - 2 t ) \mathbf { k } , 0 \leq t \leq 1\)

Step-by-Step Solution

Answer

The value of the integral is 6.5.

Step 1: Parametrize the Curve

We are given the vector function \( \mathbf{r}(t) = 2ti + tj + (2 - 2t)k \). This represents a parameterized curve that spans \( 0 \leq t \leq 1 \). We need to express the components of this curve so \( x(t) = 2t, y(t) = t, z(t) = 2 - 2t \).

Step 2: Compute the Derivative of the Curve

Find \( \frac{d\mathbf{r}}{dt} \), the derivative of \( \mathbf{r}(t) \). We differentiate each component: \( \frac{dx}{dt} = 2 \), \( \frac{dy}{dt} = 1 \), and \( \frac{dz}{dt} = -2 \). Thus, \( \frac{d\mathbf{r}}{dt} = 2i + j - 2k \).

Step 3: Determine Magnitude of the Derivative

Calculate the magnitude \( \left| \frac{d\mathbf{r}}{dt} \right| \). We get \( \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).

Step 4: Substitute into Integral

Substitute \( x(t), y(t), z(t) \) and \( \left| \frac{d\mathbf{r}}{dt} \right| \) into the line integral expression: \[ \int_{0}^{1} ( (2t)(t) + t + (2 - 2t)) (3) dt \].

Step 5: Simplify the Expression

Simplify the integrand: \( 2t^2 + t + 2 - 2t \), which reduces to \( 2t^2 - t + 2 \).

Step 6: Execute the Integral

Evaluate the integral \[ \int_{0}^{1} 3(2t^2 - t + 2) dt = 3 \left( \int_{0}^{1} 2t^2 dt - \int_{0}^{1} t dt + \int_{0}^{1} 2 dt \right) \]. This is \[ 3 \left( \frac{2}{3}t^3 \bigg|_0^1 - \frac{1}{2}t^2 \bigg|_0^1 + 2t \bigg|_0^1 \right) = 3 \left( \frac{2}{3} - \frac{1}{2} + 2 \right) \].

Step 7: Compute the Final Result

Calculate \( 3 \left( \frac{2}{3} - \frac{1}{2} + 2 \right) = 3 \left( \frac{4}{6} - \frac{3}{6} + \frac{12}{6} \right) = 3 \left( \frac{13}{6} \right) = \frac{39}{6} = 6.5 \). Thus, the value of the integral is \( 6.5 \).

Key Concepts

Parametrization of Curves

When dealing with line integrals and curves, parametrization plays a crucial role. It helps to describe a curve using a parameter, usually denoted as \( t \). In our example, the curve is defined by the vector function \( \mathbf{r}(t) = 2ti + tj + (2 - 2t)k \). This essentially links each point on the curve to a unique value of \( t \).

The idea is simple: by letting \( t \) vary between certain values (here from 0 to 1), the vector function \( \mathbf{r}(t) \) traces out the path or curve in a three-dimensional space. This method allows us to transform problems involving curves into more manageable variable scenarios. It essentially provides a new perspective to break down and solve problems easily.

• \( x(t) = 2t \)
• \( y(t) = t \)
• \( z(t) = 2 - 2t \)

By expressing \( x, y, \) and \( z \) in terms of \( t \), each component of the space curve is effectively parameterized.

Vector Calculus

In vector calculus, we handle functions that have vectors as their outputs or inputs. When dealing with line integrals, derivatives of these vector functions become very crucial. They represent directions and magnitudes of the changes occurring on the curve.

For our specific example, \( \frac{d\mathbf{r}}{dt} \) was computed for the curve \( \mathbf{r}(t) = 2ti + tj + (2-2t)k \) to understand its rate of change.

• \( \frac{dx}{dt} = 2 \)
• \( \frac{dy}{dt} = 1 \)
• \( \frac{dz}{dt} = -2 \)

This results in \( \frac{d\mathbf{r}}{dt} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k} \), which indicates how the curve’s orientation and position changes as \( t \) increases. The understanding of vector calculus is vital in formulating these relationships and solving related problems.

Integral Calculus

Integral calculus is a branch of calculus focused on the idea of accumulation or areas under curves. In the context of line integrals, it is about adding up values along a specific curve rather than a straightforward area under a curve.

For the given function \( \int _ { C } ( x y + y + z ) d s \), where \( d s \) is the small linear segment, the integral represents the sum of contributions from each segment along the curve. Once the curve is parameterized, expressing the integral in terms of \( t \) simplifies it.

• Substitute expressions for \( x(t), y(t), \) and \( z(t) \)
• Include the magnitude of \( \frac{d\mathbf{r}}{dt} \)
• Integrate over the chosen interval

In our example, calculating \( \int_{0}^{1} 3(2t^2 - t + 2) dt \) resulted in a smooth process to find the accumulation along our curve.

Arc Length Differential

The arc length differential, often represented as \( ds \), is an essential part of line integrals. It helps in measuring how much of the curve is "covered" by the parameter \( t \).

For a parameterized curve \( \mathbf{r}(t) \), \( ds \) can be expressed as \( \left| \frac{d\mathbf{r}}{dt} \right| dt \), which provides a measure of the length of an infinitesimally small segment of the curve. Calculating the magnitude \( \left| \frac{d\mathbf{r}}{dt} \right| \) ensures that we accurately gauge the length of each small segment as \( t \) changes.

• Compute the derivative \( \frac{d\mathbf{r}}{dt} \)
• Find the magnitude: \( \sqrt{(2)^2 + (1)^2 + (-2)^2} = 3 \)
• Use this to weigh contributions along the curve

Understanding \( ds \) as a differential element makes calculating the total integral possible in terms of arc length.