Problem 35 - Advanced Engineering Mathematics

Question

(a) If \(f(x, y)=x^{3}-3 x^{2} y^{2}+y^{3}\), find the directional derivative of \(f\) at a point \((x, y)\) in the direction of \(\mathbf{u}=(1 / \sqrt{10})(3 \mathbf{i}+\mathbf{j})\). (b) If \(F(x, y)=D_{\mathrm{u}} f(x, y)\) in part (a), find \(D_{\mathrm{u}} F(x, y)\).

Step-by-Step Solution

Answer

(a) \( D_\mathbf{u}f(x, y) = \frac{1}{\sqrt{10}}(9x^2 - 6x^2y - 18xy^2 + 3y^2) \). (b) \( D_\mathbf{u}F(x, y) = \frac{1}{10}(54x - 72xy - 54y^2 - 6x^2 + 6y) \).

Step 1: Calculate the Gradient of f

To begin solving the exercise, find the gradient \( abla f(x, y) \). The function is given as \( f(x, y) = x^3 - 3x^2y^2 + y^3 \). Calculate the partial derivatives:\( \frac{\partial f}{\partial x} = 3x^2 - 6xy^2 \)\( \frac{\partial f}{\partial y} = -6x^2y + 3y^2 \)Thus,\( abla f(x, y) = \left( 3x^2 - 6xy^2, -6x^2y + 3y^2 \right) \).

Step 2: Normalize the Directional Vector

The directional vector is given as \( \mathbf{u} = \frac{1}{\sqrt{10}}(3\mathbf{i} + \mathbf{j}) \). This vector is already normalized because its magnitude is 1. Therefore, we can continue using it directly for our calculations.

Step 3: Calculate the Directional Derivative of f

The directional derivative is calculated using the formula: \( D_\mathbf{u}f(x, y) = abla f(x, y) \cdot \mathbf{u} \).Compute:\[ D_\mathbf{u}f(x, y) = \left( 3x^2 - 6xy^2, -6x^2y + 3y^2 \right) \cdot \frac{1}{\sqrt{10}}(3, 1) \]Substitute and compute the dot product:\[ D_\mathbf{u}f(x, y) = \frac{1}{\sqrt{10}}((3x^2 - 6xy^2) \cdot 3 + (-6x^2y + 3y^2) \cdot 1) \]\[ = \frac{1}{\sqrt{10}}(9x^2 - 18xy^2 - 6x^2y + 3y^2) \]\[ = \frac{1}{\sqrt{10}}(9x^2 - 6x^2y - 18xy^2 + 3y^2) \]

Step 4: Define F(x, y) Using the Directional Derivative

For part (b), define \( F(x, y) = D_\mathrm{u}f(x, y) \), which is the expression we calculated earlier:\[ F(x, y) = \frac{1}{\sqrt{10}}(9x^2 - 6x^2y - 18xy^2 + 3y^2) \].

Step 5: Compute the Gradient of F

To find \( D_\mathrm{u}F(x, y) \), first calculate the partial derivatives of \( F(x, y) \):\( \frac{\partial F}{\partial x} = \frac{1}{\sqrt{10}}(18x - 12xy - 18y^2) \)\( \frac{\partial F}{\partial y} = \frac{1}{\sqrt{10}}(-6x^2 - 36xy + 6y) \)The gradient is:\[ abla F(x, y) = \left( \frac{1}{\sqrt{10}}(18x - 12xy - 18y^2), \frac{1}{\sqrt{10}}(-6x^2 - 36xy + 6y) \right) \].

Step 6: Calculate the Directional Derivative of F

Now, find \( D_\mathrm{u}F(x, y) \) using the previously calculated gradient of \( F(x, y) \) and \( \mathbf{u} \):\[ D_\mathrm{u}F(x, y) = abla F(x, y) \cdot \mathbf{u} \]Calculate the dot product:\[ D_\mathrm{u}F(x, y) = \left( \frac{1}{\sqrt{10}}(18x - 12xy - 18y^2), \frac{1}{\sqrt{10}}(-6x^2 - 36xy + 6y) \right) \cdot \frac{1}{\sqrt{10}}(3, 1) \]\[ = \frac{1}{10}((18x - 12xy - 18y^2) \cdot 3 + (-6x^2 - 36xy + 6y) \cdot 1) \]\[ = \frac{1}{10}(54x - 36xy - 54y^2 - 6x^2 - 36xy + 6y) \]\[ = \frac{1}{10}(54x - 72xy - 54y^2 - 6x^2 + 6y) \].

Key Concepts

Gradient

The gradient is a vector that comprises all the partial derivatives of a multivariable function. It shows the direction in which the function increases most quickly. For a function such as \( f(x, y) \), the gradient is denoted as \( abla f(x, y) \). In this exercise, the gradient is calculated as:

• \( \frac{\partial f}{\partial x} = 3x^2 - 6xy^2 \)
• \( \frac{\partial f}{\partial y} = -6x^2y + 3y^2 \)

The gradient vector thus becomes \( abla f(x, y) = \left( 3x^2 - 6xy^2, -6x^2y + 3y^2 \right) \). This vector not only provides the direction of steepest ascent but its magnitude also indicates how fast the function value increases in that direction. Understanding the gradient helps in moving optimally along the function's surface.

Partial Derivatives

Partial derivatives are an essential concept when dealing with functions of multiple variables. They measure how the function changes as we vary only one of the variables, keeping the others constant. Here, for \( f(x, y) \), we have:

• The partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x}\), reflects how \( f(x, y) \) changes as \( x \) changes while \( y \) remains constant.
• The partial derivative with respect to \( y \), denoted \( \frac{\partial f}{\partial y}\), shows the change in \( f(x, y) \) as \( y \) changes while \( x \) is constant.

To find these derivatives, differentiate the function treating all but one variable as constant for each derivative. This is crucial in constructing the gradient and subsequently computing directional derivatives.

Dot Product

The dot product is a fundamental operation that takes two vectors and returns a scalar. It quantifies the degree to which two vectors point in the same direction. In vector mathematics for directional derivatives, this operation is used to combine the gradient and the directional vector.For this exercise, the dot product of \( abla f(x, y) \) and the normalized vector \( \mathbf{u} \) is:\[ D_\mathbf{u}f(x, y) = abla f(x, y) \cdot \mathbf{u} \]The operation involves multiplying corresponding components of the vectors and summing them up:\[ \left( 3x^2 - 6xy^2, -6x^2y + 3y^2 \right) \cdot \frac{1}{\sqrt{10}}(3, 1) \]This yields a scalar that represents how steep the function \( f(x, y) \) increases in the direction of \( \mathbf{u} \), effectively offering insight into changes of the function in a given direction.

Vector Normalization

Vector normalization is the process of converting a vector into a unit vector, meaning the magnitude of the vector is scaled to one. Normalizing vectors is particularly useful in directional calculations because it ensures that the direction is considered without scaling effects of vector magnitude.For a vector \( \mathbf{a} = (a_1, a_2) \), its normalized form is:\[ \mathbf{a}_{norm} = \frac{1}{\| \mathbf{a} \|}(a_1, a_2) \]In this problem, \( \mathbf{u} = \frac{1}{\sqrt{10}}(3\mathbf{i} + \mathbf{j}) \) is already normalized, with \[ \|\mathbf{u} \| = \sqrt{\left(\frac{3}{\sqrt{10}}\right)^2 + \left(\frac{1}{\sqrt{10}}\right)^2 } = 1 \].This allows us to use \( \mathbf{u} \) directly to calculate the directional derivative, ensuring only the directional component influences the result, not its size.