Problem 291 - Calculus Volume 3

Question

For the following exercises, find the gradient vector at the indicated point. $$\quad f(x, y)=x e^{y}-\ln (x), P(-3,0)$$

Step-by-Step Solution

Answer

The gradient vector at (-3, 0) is \( (\frac{4}{3}, -3) \).

Step 1: Understand the Gradient Vector

The gradient vector of a function \( f(x, y) \) is given by \( abla f(x, y) \), which is composed of its partial derivatives with respect to each variable. Specifically, \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Thus, to find the gradient vector, we need to differentiate \( f(x, y) \) with respect to both \( x \) and \( y \).

Step 2: Compute Partial Derivative with Respect to x

Differentiate \( f(x, y) = x e^y - \ln(x) \) with respect to \( x \). The derivative of \( x e^y \) with respect to \( x \) is \( e^y \) and the derivative of \( -\ln(x) \) with respect to \( x \) is \( -\frac{1}{x} \). Thus, \( \frac{\partial f}{\partial x} = e^y - \frac{1}{x} \).

Step 3: Compute Partial Derivative with Respect to y

Differentiate \( f(x, y) = x e^y - \ln(x) \) with respect to \( y \). The derivative of \( x e^y \) with respect to \( y \) is \( x e^y \), and since \( \ln(x) \) is treated as a constant regarding \( y \), its derivative is 0. Thus, \( \frac{\partial f}{\partial y} = x e^y \).

Step 4: Evaluate the Gradient at Point (-3, 0)

Substitute \( x = -3 \) and \( y = 0 \) into the partial derivatives. For \( \frac{\partial f}{\partial x} = e^y - \frac{1}{x} \), substitute to get \( e^0 - \frac{1}{-3} = 1 + \frac{1}{3} = \frac{4}{3} \). For \( \frac{\partial f}{\partial y} = x e^y \), substitute to get \( -3 \cdot e^0 = -3 \).

Step 5: Combine Partial Derivatives for Gradient Vector

The gradient vector \( abla f(x, y) \) evaluated at \( (-3, 0) \) is \( \left( \frac{4}{3}, -3 \right) \). Thus, the gradient vector at the point \( P(-3, 0) \) is \( \left( \frac{4}{3}, -3 \right) \).

Key Concepts

Partial Derivatives

The concept of partial derivatives arises when you work with functions of multiple variables, say \( f(x, y) \). In essence, partial derivatives help us understand how a function changes as you vary one variable, keeping others constant.
This is fundamental in analyzing multivariable functions, ensuring we pinpoint how precisely each variable influences the overall function. For example, to find \( \frac{\partial f}{\partial x} \), you differentiate while treating \( y \) as a constant. Similarly, for \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant.

• Partial derivatives form the building blocks of the gradient vector.
• They can be thought of as the function's sensitivity to each individual variable.
• In the exercise, differentiating \( f(x, y)=x e^{y}-\ln(x) \) with respect to each variable is key.

Understanding partial derivatives allows you to articulate how functions decorate curves and surfaces in multidimensional spaces.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of several variables. It’s essential for analyzing systems where multiple factors change simultaneously. Functions like \( f(x, y) \) involve more than one variable, requiring strategies beyond simple differentiation.
The primary tools in this realm include studying partial derivatives, as they represent the rate of change in a function along one dimension while holding others steady.
In this domain, the gradient vector offers a powerful way to encapsulate changes in all dimensions at once.

• Helps to understand complex motions, models, and systems that depend on multiple factors.
• Integrates partial derivatives to build comprehensive models and predictions.
• Underpins physics, engineering, economics, and other fields involving dynamic changes.

Grasping multivariable calculus equips you to handle real-world problems where single-variable calculus falls short.

Differentiation

Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. It is the backbone of calculus, allowing us to examine instantaneous rates of change.
With differentiation, you can ascertain a function's slope at any point, vital for sketching curves and solving optimization problems.
For multivariable functions, differentiation is slightly more intricate. You distinguish changes respective to each variable individually, employing partial derivatives:

• Find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) to assess how each input variable influences the output.
• The integration of these derivatives forms the gradient, offering a holistic view of the function's behavior.

Through diligent differentiation, we unearth the nuances of function behaviors and interactions.

Vector Calculus

Vector calculus is a branch of mathematics concerned with differentiation and integration of vector fields, primarily in multidimensional spaces. It's pivotal in fields from electromagnetism to fluid dynamics.
One of vector calculus's essential aspects is the gradient vector, which embodies the direction and rate of fastest increase of a scalar field.
For function \( f(x, y) \), the gradient is a vector \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \) and shows how \( f \) changes:

• This vector points in the direction where \( f \) increases most rapidly.
• The length of this vector represents the steepness of that incline.
• In the provided exercise, \( abla f(x, y) \) was evaluated to give \( \left( \frac{4}{3}, -3 \right) \), guiding us to comprehend how function behaves at point \( P(-3,0) \).

Vector calculus combines these ideas to analyze complex fields and lines energetically directing actions like heat, sound, and light.