Problem 18 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 20 , evaluate the line integral over the given curve. \(\int_{c} 2 x y d x+\left(6 y^{2}-x z\right) d y+10 z d z ; C:\) the twisted cubic \(R(t)=t i+t^{2} j+t^{3} k, 0 \leq t \leq 1\)

Step-by-Step Solution

Answer

\( \frac{19}{3} \)

Step 1 - Parameterize the Curve

The given twisted cubic is parameterized as:\[ R(t) = t i + t^2 j + t^3 k \]So, in component form, it is:\[ x = t, \, y = t^2, \, z = t^3 \]with the parameter \( t \) ranging from 0 to 1.

Step 2 - Compute Derivatives of the Parameterization

Compute the derivatives of x, y, and z with respect to t:\[ \frac{dx}{dt} = 1 \]\[ \frac{dy}{dt} = 2t \]\[ \frac{dz}{dt} = 3t^2 \]

Step 3 - Substitute the Parameterization into the Integral

Substitute \(x, y, z, dx, dy, dz\) into the integral:\[ \int_{0}^{1} 2(t)(t^2) \, dt + (6(t^2)^2 - t(t^3))(2t) \, dt + 10(t^3)(3t^2) \, dt \]Simplify each term step by step.

Step 4 - Evaluate Each Term Separately

Evaluate the first term:\[ \int_{0}^{1} 2t^3 \, dt = 2 \int_{0}^{1} t^3 \, dt = 2 \left[ \frac{t^4}{4} \right]_{0}^{1} = 2 \cdot \frac{1}{4} = \frac{1}{2} \]The second term:\[ \int_{0}^{1} (6t^4 - t^4)(2t) \, dt = \int_{0}^{1} 5t^5 \, dt = 5 \left[ \frac{t^6}{6} \right]_{0}^{1} = 5 \cdot \frac{1}{6} = \frac{5}{6} \]And the third term:\[ \int_{0}^{1} 30t^5 \, dt = 30 \left[ \frac{t^6}{6} \right]_{0}^{1} = 30 \cdot \frac{1}{6} = 5 \]

Step 5 - Sum All Terms

Sum the results of all integrals:\[ \frac{1}{2} + \frac{5}{6} + 5 \]Combine terms over a common denominator (12):\[ \frac{6}{12} + \frac{10}{12} + \frac{60}{12} = \frac{76}{12} = \frac{19}{3} \]

Key Concepts

Parameterization of Curves

When dealing with line integrals, it often becomes necessary to parameterize the given curve. This means representing the curve with a parameter, usually denoted by t. For the given twisted cubic curve in the problem, we use the parameterization: \[ R(t) = t i + t^2 j + t^3 k \]This translates to the component form:\[ x = t, \, y = t^2, \, z = t^3 \]where the parameter t ranges from 0 to 1. Essentially, we are expressing the coordinates x, y, and z as functions of t.
Parameterization simplifies the process of evaluating the line integral by transforming the multidimensional curve into a one-dimensional problem. By substituting these parameterized equations into the integral, we turn the challenging task of evaluating in three dimensions into a more manageable computation in terms of t.

Integral Calculus

Integral calculus is a cornerstone of calculus concerned with finding the integral or antiderivative. In this context, we evaluate the line integral over a parameterized curve. Given the integral \[ \int_{c} 2 x y d x + \left(6 y^{2}-x z\right) d y + 10 z d z \]we substitute the parameterized expressions: \[ x = t, \, y = t^2, \, z = t^3 \]as well as their derivatives with respect to t: \[ \frac{dx}{dt} = 1, \, \frac{dy}{dt} = 2t, \, \frac{dz}{dt} = 3t^2 \]Then, we rewrite the integral in terms of t: \[ \int_{0}^{1} \left[2(t)(t^2) \cdot 1 + (6(t^2)^2 - t(t^3))(2t) + 10(t^3)(3t^2) \right] dt \]By simplifying, we get: \[ \int_{0}^{1} 2t^3 \, dt + \int_{0}^{1}(6t^4 - t^4)(2t) \, dt + \int_{0}^{1} 30t^5 \, dt \]Evaluating these individual integrals step by step and summing them gives the final result.

Vector Calculus

Vector calculus extends concepts from calculus to vector fields. It is crucial for understanding physical phenomena where quantities depend on both spatial position and direction. A line integral in vector calculus involves integrating a vector field along a curve. For this problem, we dealt with a vector field composed of three components: \[ F = 2xy \, i + (6y^2 - xz) \, j + 10z \, k \]which we integrate over the parameterized curve \( R(t) = t i + t^2 j + t^3 k \).
The line integral is given by:\[ \int_{c} F \cdot dR \]Here, \( dR \) or \( R'(t) dt \) represents the differential element along the curve. The dot product \( F \cdot dR \) simplifies to evaluating the vector components with respect to t. Performing the dot product and simplifying yields a new integral in terms of t, ready for evaluation with the procedures of integral calculus. This combination of parameterization and integration of vectors is central to solving real-world problems in physics and engineering.