Problem 5 - The Calculus with Analytic Geometry

Question

In the following exercises, determine if the vector is a gradient. If it is, find a function having the given gradient \(\left(6 x^{2} y^{2}-14 x y+3\right) \mathrm{i}+\left(4 x^{3} y-7 x^{2}-8\right) \mathbf{j}\)

Step-by-Step Solution

Answer

Yes, it is a gradient. The function is \(f(x, y) = 2 x^{3} y^{2} - 7 x^{2} y + 3 x - 8 y + C\).

Step 1: Recall the Definition of a Gradient Vector

A vector field \(\textbf{F}(x, y) = P(x, y)\textbf{i} + Q(x, y)\textbf{j}\) is a gradient field if there exists a scalar function \(f(x, y)\) such that \(\textbf{F}(x, y) = abla f(x, y) = \frac{\text{d} f}{\text{d} x} \textbf{i} + \frac{\text{d} f}{\text{d} y} \textbf{j}\). In other words, \(P = \frac{\text{d} f}{\text{d} x}\) and \(Q = \frac{\text{d} f}{\text{d} y}\).

Step 2: Compute the Partial Derivatives

Given \(P = 6 x^{2} y^{2} - 14 x y + 3\) and \(Q = 4 x^{3} y - 7 x^{2} - 8\), find \(\frac{\text{d} P}{\text{d} y}\) and \(\frac{\text{d} Q}{\text{d} x}\). \(\frac{\text{d} P}{\text{d} y} = 12 x^{2} y - 14 x\) and \(\frac{\text{d} Q}{\text{d} x} = 12 x^{2} y - 14 x\).

Step 3: Verify the Equality of Mixed Partial Derivatives

Check if \( \frac{\text{d} P}{\text{d} y} = \frac{\text{d} Q}{\text{d} x} \). Since \( \frac{\text{d} P}{\text{d} y} = 12 x^{2} y - 14 x \) and \( \frac{\text{d} Q}{\text{d} x} = 12 x^{2} y - 14 x \), the mixed partial derivatives are equal. Thus, the vector is a gradient.

Step 4: Integrate to Find the Potential Function

First, integrate \(P = 6 x^{2} y^{2} - 14 x y + 3\) with respect to \(x\) to get \(f(x, y) = 2 x^{3} y^{2} - 7 x^{2} y + 3 x + g(y)\), where \(g(y)\) is a function of \(y\).

Step 5: Determine the Function \(g(y)\)

Differentiate \(f(x, y)\) with respect to \(y\) to compare with \(Q\). This gives \( \frac{\text{d} f}{\text{d} y} = 4 x^{3} y - 7 x^{2} + g'(y)\). Set this equal to \(Q = 4 x^{3} y - 7 x^{2} - 8\). Thus, \(g'(y) = -8\), integrating gives \(g(y) = -8y + C\).

Step 6: Write the Final Potential Function

Substitute \(g(y)\) back into \(f(x, y)\) to get \(f(x, y) = 2 x^{3} y^{2} - 7 x^{2} y + 3 x - 8 y + C\).

Key Concepts

Vector Field

A vector field is a mathematical construction where each point in a space is assigned a vector. In two-dimensional space, a vector field can be written as \(\textbf{F}(x, y) = P(x, y)\textbf{i} + Q(x, y)\textbf{j}\). Here, \(\textbf{i}\) and \(\textbf{j}\) are unit vectors in the direction of the \(x\) and \(y\) axes, respectively. \(\textbf{F}(x, y)\) points towards the direction and has a magnitude dependent on the values of \(P\) and \(Q\) at \((x, y)\). Vector fields are used in physics and engineering to model fields like the gravitational field, electric field, and fluid flow.
Understanding vector fields is crucial for solving problems involving gradients, potential functions, and various types of differential equations in multivariable calculus.

Partial Derivatives

Partial derivatives measure how a function changes as each variable is varied while keeping the other variables constant. For example, if you have a function \(f(x, y)\), the partial derivative of \(f\) with respect to \(x\) is denoted as \(\frac{\text{d} f}{\text{d} x}\), and it tells you how \(f\)'s value changes as \(x\) changes, with \(y\) held fixed. Similarly, \(\frac{\text{d} f}{\text{d} y}\) measures the rate of change of \(f\) in the \(y\) direction.
Calculating partial derivatives is essential when working with vector fields because you often need to check if a vector field is conservative by verifying the equality of mixed partial derivatives. In this specific exercise, you computed \(\frac{\text{d} P}{\text{d} y}\) and \(\frac{\text{d} Q}{\text{d} x}\) to determine if the given vector field is the gradient of a potential function.

Potential Function

A potential function, \(f(x, y)\), plays a significant role in determining whether a vector field \(\textbf{F} = P(x, y)\textbf{i} + Q(x, y)\textbf{j}\) is conservative. If \(\textbf{F}\) is a gradient vector field, then there exists a scalar function \(f \) such that \(\textbf{F} = abla f\). This means that both \(P\) and \(Q\) can be viewed as partial derivatives of \(f\): \(P = \frac{\text{d} f}{\text{d} x}\) and \(Q = \frac{\text{d} f}{\text{d} y}\).
In the given exercise, you integrated \(P\) with respect to \(x\) and later matched \(Q\) by differentiating \(f\) with respect to \(y\) to find an unknown function \(g(y)\). Once you identified \(g(y)= -8y + C\), you could write the full potential function \(f(x, y) = 2 x^{3} y^{2} - 7 x^{2} y + 3 x - 8 y + C\).

Mixed Partial Derivatives

Mixed partial derivatives involve taking derivatives of a function with respect to more than one variable, in different orders. For functions of two variables, \(( x, y)\), the mixed partial derivatives would be \(\frac{\text{d}}{\text{d} y} \frac{\text{d} f}{\text{d} x}\) and \(\frac{\text{d}}{\text{d} x} \frac{\text{d} f}{\text{d} y}\). Clairaut's Theorem states that if all these partial derivatives are continuous, then these mixed partial derivatives are equal. This is important in verifying if a vector field is conservative.
In the vector field exercise, after computing \(\frac{\text{d} P}{\text{d} y} = 12 x^{2} y - 14 x\) and \(\frac{\text{d} Q}{\text{d} x} = 12 x^{2} y - 14 x\), the equality of these mixed partial derivatives validates that the given vector field can indeed be expressed as the gradient of some potential function.