Problem 13 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 20 , evaluate the line integral over the given curve. \(\int_{c}(x+y) d x+(y+z) d y+(x+z) d z ; C:\) the line segment from the origin to the point \((1,2,4)\)

Step-by-Step Solution

Answer

The value of the line integral is 16.5

Step 1: Parameterize the Curve

The given curve is a line segment from the origin to the point \(1,2,4\). For a line segment, we can use a parameter \(t\) such that \(0 \leq t \leq 1\). The parameterization of the curve \(C\) can be written as: \(\mathbf{r}(t) = (t, 2t, 4t)\).

Step 2: Compute the Derivatives

Next, find the derivatives of the parameterized functions with respect to \(t\): \[\frac{dx}{dt} = 1, \ \frac{dy}{dt} = 2, \ \frac{dz}{dt} = 4\].

Step 3: Substitute and Integrate

Substitute \(x = t, \ y = 2t, \ z = 4t\) and their corresponding derivatives into the line integral: \ \ \ \[ \int_{C} (x+y) \frac{dx}{dt} dt + (y+z) \frac{dy}{dt} dt + (x+z) \frac{dz}{dt} dt \] becomes \[ \int_{0}^{1} \left( t + 2t \right) 1 dt + \left( 2t + 4t \right) 2 dt + \left( t + 4t \right) 4 dt \].

Step 4: Simplify the Integral

Simplify the integrals: \[ \int_{0}^{1} 3t dt + \int_{0}^{1} 12t dt + \int_{0}^{1} 20t dt \] giving a total integrates to: \[ \int_{0}^{1} 3t dt + \int_{0}^{1} 12t dt + \int_{0}^{1} 20t dt \] becomes \frac{3}{2} + 6 + 10 = 16.5\.

Key Concepts

Parameterize the Curve

One essential step in evaluating a line integral is to parameterize the curve. This process converts a geometric path into a form that can be easily manipulated through calculus.
For a line segment connecting two points, we use a parameter, usually denoted by \(t\), which takes values in an interval like \([0, 1]\). In our exercise, the line segment starts at the origin \((0,0,0) \) and ends at the point \((1,2,4) \).
To parameterize this segment, we express each coordinate as a linear function of \(t\):
\ \(x(t) = t\)\ \(y(t) = 2t\)\ \(z(t) = 4t\).
Here, \( t \) varies from 0 to 1, sweeping the path of our curve. Using such a parameterization simplifies the subsequent steps of the integration process and ensures that we correctly account for how each coordinate changes along the curve.

Derivatives with Respect to Parameter

Once the curve is parameterized, the next crucial step is to find the derivatives of these parameter functions with respect to \(t\). These derivatives help us determine how fast the coordinates change along the curve.
In our problem, the parameterizations are: \ \(x(t) = t\)\ \(y(t) = 2t\)\ \(z(t) = 4t\).
Taking the derivatives with respect to \(t\), we get:
\ \(\frac{dx}{dt} = 1\)\ \(\frac{dy}{dt} = 2\)\ \(\frac{dz}{dt} = 4\).
These derivatives will be used as part of the integrand in the line integral. Each derivative tells us the rate of change for each coordinate which will be multiplied by the corresponding differential \(dt\) during the integration.

Simplifying Integrals

After parameterizing the curve and computing the derivatives, the next task is to simplify the integral expression and solve it.
We substitute the parameterized coordinates and their derivatives into the integral:
\[ \int_{0}^{1} \left( t + 2t \right) 1 dt + \left( 2t + 4t \right) 2 dt + \left( t+ 4t \right) 4 dt \].
Simplify these expressions to: \ \(3t\)\ \(12t\)\ \(20t\).
Finally, we integrate each term from 0 to 1: \ \(\int_{0}^{1} 3t dt = \frac{3}{2}\)\ \(\int_{0}^{1} 12t dt = 6\)\ \(\int_{0}^{1} 20t dt = 10\).
Adding these results provides the total value of the line integral: <\( 16.5 \).
The step-by-step simplification ensures that we handle each part correctly, making it easier to manage more complex integrals.