Problem 3 - Advanced Engineering Mathematics

Question

\(\int_{C} 2 z d z\), where \(C\) is \(z(t)=2 t^{3}+i\left(t^{4}-4 t^{3}+2\right),-1 \leq t \leq 1\)

Step-by-Step Solution

Answer

The integral evaluates to zero due to the symmetry of the polynomial terms.

Step 1: Parameterize the curve

We are given the curve parameterization by the function \( z(t) = 2t^3 + i(t^4 - 4t^3 + 2) \). For \( -1 \leq t \leq 1 \), this function describes a path \( C \) in the complex plane.

Step 2: Find the derivative of the parameterization

Calculate the derivative \( \frac{dz}{dt} \) of the parameterized function \( z(t) = 2t^3 + i(t^4 - 4t^3 + 2) \). Thus, \( \frac{dz}{dt} = \frac{d}{dt}(2t^3) + i \frac{d}{dt}(t^4 - 4t^3 + 2) = 6t^2 + i(4t^3 - 12t^2) \).

Step 3: Setup the integral with respect to the parameter

Substitute \( z = z(t) \) and \( dz = \frac{dz}{dt} dt \) into the integral: \( \int_{-1}^{1} 2z(t) \frac{dz}{dt} dt \). Substituting, this becomes \( \int_{-1}^{1} 2(2t^3 + i(t^4 - 4t^3 + 2))(6t^2 + i(4t^3 - 12t^2)) dt \).

Step 4: Simplify the integrand

Distribute and simplify the expression inside the integral: \( 2(2t^3 + i(t^4 - 4t^3 + 2))(6t^2 + i(4t^3 - 12t^2)) \). This will involve multiplying and combining like terms, simplifying real and imaginary components separately.

Step 5: Evaluate the integral

After simplifying, integrate the expression from \(-1\) to \(1\) with respect to \(t\). The integration involves straightforward polynomial integration. Evaluate the definite integral over the symmetric interval \([-1, 1]\).

Step 6: Analyze the symmetry to simplify integration

Since the limits are symmetric and the integrand might be odd in nature (specifically power terms with respect to \( t \)), check for odd functions, because they integrate to zero over symmetric limits.

Key Concepts

Parameterization of Curves

When we parameterize a curve, we represent it using a function of a single variable, often time denoted by \( t \). Parameterization allows us to describe complex geometric paths with simple equations. In our exercise, the curve is given by the parameterized function \( z(t) = 2t^3 + i(t^4 - 4t^3 + 2) \). This complex function describes a path \( C \) in the complex plane as \( t \) varies from \(-1\) to \(1\). By breaking down the equation, the real part \( 2t^3 \) and the imaginary part \( i(t^4 - 4t^3 + 2) \) each control how the curve behaves in their respective dimensions.

• The real part, \( 2t^3 \), affects the horizontal movement of the path.
• The imaginary part, \( i(t^4 - 4t^3 + 2) \), controls the vertical deviation.

Understanding parameterization aids in analyzing complex paths by offering a practical way to work with curves algebraically rather than geometrically.

Complex Differentiation

Complex differentiation determines how functions change as inputs change. Differentiation in the complex plane follows principles similar to real calculus, but deals with complex variables. For the curve \( z(t) = 2t^3 + i(t^4 - 4t^3 + 2) \), we find the derivative \( \frac{dz}{dt} \). Differentiation is performed separately for the real and imaginary components of the complex function, generating a derivative:

• For \( 2t^3 \), the derivative is \( 6t^2 \).
• For \( i(t^4 - 4t^3 + 2) \), the derivative is \( i(4t^3 - 12t^2) \).

When combined, the full derivative is: \( \frac{dz}{dt} = 6t^2 + i(4t^3 - 12t^2) \). This expression helps describe how fast and in what way the path \( C \) is changing as \( t \) varies. Differentiation is crucial for understanding how parameters impact movement across the path.

Polynomial Integration

Polynomial integration involves finding integrals of polynomial expressions, which are sums of terms consisting of variables raised to whole number powers, like \( ax^n \). The exercise includes integrating a polynomial formed from the parameterized path and its derivative. In particular, we look at:\[ \int_{-1}^{1} 2z(t) \cdot \frac{dz}{dt} dt \]To integrate, first simplify the polynomial expression by expanding terms and combining like parts. Polynomial integrals typically involve:

• Raising the power of each term by one.
• Dividing by the new power.
• Evaluating at boundaries to get the integral's definite value.

Polynomial integration is straightforward with practice. Once simplified, it provides a way to find areas or net changes over a range, especially when dealing with definite integrals over symmetric limits like our example from \(-1\) to \(1\).

Definite Integrals

Definite integrals compute the accumulated quantity, like area or total change, over an interval. In our case, we integrate between symmetric boundaries \([-1, 1]\). A definite integral with symmetric limits allows for checking the symmetry of the function within that interval.

• If a function is even, it is symmetric about the y-axis and contributes to the net integral from \(-1\) to \(1\).
• If a function is odd, it is symmetric about the origin, meaning the integral over symmetric bounds is zero.

Understanding these properties can vastly simplify the computation. By recognizing patterns or properties in the polynomials involved, we can efficiently determine the final result.