Problem 6 - Advanced Engineering Mathematics

Question

Find the gradient of the given function at the indicated point. $$ f(x, y)=\sqrt{x^{3} y-y^{4}} ;(3,2) $$

Step-by-Step Solution

Answer

The gradient at (3,2) is \( \left( \frac{27}{\sqrt{38}}, \frac{-5}{2\sqrt{38}} \right) \).

Step 1: Find the partial derivative with respect to x

To find the gradient, we first need to determine the partial derivative of the function with respect to x. Consider the function: \[ f(x, y) = \sqrt{x^3 y - y^4} \]The partial derivative with respect to x is:\[ \frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x^3 y - y^4}} \cdot \left(3x^2 y\right) \]

Step 2: Find the partial derivative with respect to y

Next, we find the partial derivative of the function with respect to y. Using the same function:\[ f(x, y) = \sqrt{x^3 y - y^4} \]The partial derivative with respect to y is:\[ \frac{\partial f}{\partial y} = \frac{1}{2\sqrt{x^3 y - y^4}} \cdot \left(x^3 - 4y^3\right) \]

Step 3: Evaluate the partial derivatives at the given point

Now that we have both partial derivatives, we evaluate them at the point \((3, 2)\). For the partial derivative with respect to x:\[ \frac{\partial f}{\partial x}(3,2) = \frac{1}{2\sqrt{3^3 (2) - 2^4}} \cdot (3 \cdot 3^2 \cdot 2) = \frac{1}{2\sqrt{54 - 16}} \cdot 54 \]This simplifies to \[ \frac{1}{2\sqrt{38}} \cdot 54 \]For the partial derivative with respect to y:\[ \frac{\partial f}{\partial y}(3,2) = \frac{1}{2\sqrt{3^3 (2) - 2^4}} \cdot (3^3 - 4\cdot 2^3) = \frac{1}{2\sqrt{38}} \cdot (27 - 32) \]This simplifies to \[ \frac{1}{2\sqrt{38}} \cdot (-5) \]

Step 4: Comprehend the gradient vector

The gradient vector \(abla f\) is comprised of these partial derivatives:\[ abla f = \left( \frac{\partial f}{\partial x}(3,2), \frac{\partial f}{\partial y}(3,2) \right) \]Putting both values together:\[ abla f = \left( \frac{54}{2\sqrt{38}}, \frac{-5}{2\sqrt{38}} \right) \]This can also be expressed as:\[ abla f = \frac{1}{2\sqrt{38}} (54, -5) \]

Step 5: Simplify the gradient

To express the gradient in a more concise manner, we rewrite it using a common coefficient:\[ abla f = \left( \frac{27}{\sqrt{38}}, \frac{-5}{2\sqrt{38}} \right) \]

Key Concepts

Partial Derivatives

Partial derivatives are a key concept in multivariable calculus. They provide us with the rate of change of a function concerning one variable while keeping other variables constant. In the context of the original exercise, we have a function \( f(x, y) = \sqrt{x^3 y - y^4} \). To find the gradient at a specific point, we need to calculate its partial derivatives.

• The partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), gives us the rate at which \( f \) changes as \( x \) changes, keeping \( y \) constant.
• Similarly, the partial derivative with respect to \( y \), denoted as \( \frac{\partial f}{\partial y} \), indicates how \( f \) changes as \( y \) varies while \( x \) remains constant.

The process generally involves applying rules of differentiation. For the given function, these derivatives help us understand how each variable contributes to changes in \( f \). After computing these derivatives, we evaluate them at the designated point \((3, 2)\), letting us observe the directional changes at that specific location.

Multivariable Calculus

Multivariable calculus expands the concepts of calculus into functions of several variables, like the one we encountered in this exercise. Unlike single-variable calculus dealing with curves, multivariable calculus helps analyze surfaces and higher dimensional analogs.In our case, the function \( f(x, y) = \sqrt{x^3 y - y^4} \) is a representation of a surface in three-dimensional space. Multivariable calculus allows us to explore many real-world phenomena, such as:

• Temperature variations across a geographical location.
• Pressure changes in different regions of the atmosphere.
• Market trends depending on various economic parameters.

Navigating these complex shifts requires tools like partial derivatives to capture how each dimension impacts another. In multivariable calculus, understanding these connections points us towards a greater insight into the function's overall behavior.

Gradient Vector

The gradient vector, often symbolized as \( abla f \), is an essential tool when working with multivariable functions. It combines all the partial derivatives into a single entity, thereby giving a direction in which the function \( f \) increases most rapidly.In the original problem, after calculating the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \), the gradient vector at point \((3, 2)\) is constructed as:\[abla f = \left( \frac{54}{2\sqrt{38}}, \frac{-5}{2\sqrt{38}} \right)\]The gradient vector tells us:

• The direction in which increases in the input yield the steepest increase in \( f \).
• The magnitude of \( abla f \) signifies the steepness of this increase.

This vector is foundational in optimization problems, helping identify paths of steepest ascent or descent. Thus, understanding it is not just about finding derivatives but interpreting them in the context of real-world scenarios and applications.