Problem 12
Question
In Problems 9-14, compute the directional derivative of \(f(x, y)\) at the given point in the indicated direction. $$ f(x, y)=x^{3} y^{2} \text { at }(2,3) \text { in the direction }\left[\begin{array}{r} -2 \\ 1 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The directional derivative is \(-36\).
1Step 1: Calculate Gradient of f(x, y)
First, we need to find the gradient \( abla f(x, y) \) of the function \( f(x, y) = x^3 y^2 \). The gradient \( abla f(x, y) \) is a vector of partial derivatives, \( abla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right).\) Calculate each partial derivative:- \( \frac{\partial f}{\partial x} = 3x^2 y^2 \)- \( \frac{\partial f}{\partial y} = 2x^3 y \)Thus, \( abla f(x, y) = \left(3x^2 y^2, 2x^3 y \right).\)
Key Concepts
GradientPartial DerivativesCalculus
Gradient
In calculus, the gradient of a function provides a multi-dimensional direction that indicates where a function increases the most. When discussing functions like \( f(x, y) = x^3 y^2 \), the gradient is crucial in finding directional derivatives, which tell us the rate of change of the function in a specific direction.
The gradient, represented as \( abla f(x, y) \), is a vector consisting of partial derivatives of the function with respect to each variable. For our function \( f(x, y) \), we calculate these partial derivatives:
So, the gradient vector becomes \( abla f(x, y) = (3x^2 y^2, 2x^3 y) \).
This vector is immensely helpful because it provides not just a direction, but a magnitude that reflects how steeply the function \( f(x, y) \) changes in that direction. The gradient always points towards the greatest rate of increase of the function, and it's perpendicular to level curves where the function value remains constant.
The gradient, represented as \( abla f(x, y) \), is a vector consisting of partial derivatives of the function with respect to each variable. For our function \( f(x, y) \), we calculate these partial derivatives:
- \( \frac{\partial f}{\partial x} = 3x^2 y^2 \)
- \( \frac{\partial f}{\partial y} = 2x^3 y \)
So, the gradient vector becomes \( abla f(x, y) = (3x^2 y^2, 2x^3 y) \).
This vector is immensely helpful because it provides not just a direction, but a magnitude that reflects how steeply the function \( f(x, y) \) changes in that direction. The gradient always points towards the greatest rate of increase of the function, and it's perpendicular to level curves where the function value remains constant.
Partial Derivatives
Partial derivatives are an essential element in multivariable calculus, enabling us to explore how functions change with respect to one variable while keeping the others constant. This concept is especially important when dealing with functions of more than one variable, like our function \( f(x, y) = x^3 y^2 \).
To compute the partial derivative of \( f \) with respect to \( x \) or \( y \), we differentiate the function while treating the other variable as a constant:
These derivatives help construct the gradient. In practical terms, they tell us how the function \( f(x, y) \) changes as we alter one variable, while the other remains unchanged, enabling more precise understanding and control of the behavior of \( f \). Partial derivatives are the building blocks for more advanced concepts like directional derivatives, which depend directly on the gradient.
To compute the partial derivative of \( f \) with respect to \( x \) or \( y \), we differentiate the function while treating the other variable as a constant:
- The partial derivative with respect to \( x \) is given by: \( \frac{\partial f}{\partial x} = 3x^2 y^2 \)
- The partial derivative with respect to \( y \) is given by: \( \frac{\partial f}{\partial y} = 2x^3 y \)
These derivatives help construct the gradient. In practical terms, they tell us how the function \( f(x, y) \) changes as we alter one variable, while the other remains unchanged, enabling more precise understanding and control of the behavior of \( f \). Partial derivatives are the building blocks for more advanced concepts like directional derivatives, which depend directly on the gradient.
Calculus
Calculus is the mathematical study of change, crucial for understanding how functions behave and change over different variables. It provides tools like derivatives, integrals, and limits, which are fundamental for analyzing both single-variable and multi-variable functions.
In the context of this task, we use calculus to navigate and compute the rate of change of functions. A focal point here is the computation of derivatives—particularly partial derivatives—of multi-variable functions like \( f(x, y) = x^3 y^2 \). Through calculus, we aim to understand both the individual contributions of different variables to the function's rate of change and their combined effects.
The core of our current discussion is the gradient, a concept derived from calculus, that leverages partial derivatives to express the direction and magnitude of maximal function increase. Calculus, thus, provides not just the theoretical foundation but practical tools to evaluate and resolve complex problems involving rates and directions of change, pivotal in diverse fields like physics, engineering, and economics.
In the context of this task, we use calculus to navigate and compute the rate of change of functions. A focal point here is the computation of derivatives—particularly partial derivatives—of multi-variable functions like \( f(x, y) = x^3 y^2 \). Through calculus, we aim to understand both the individual contributions of different variables to the function's rate of change and their combined effects.
The core of our current discussion is the gradient, a concept derived from calculus, that leverages partial derivatives to express the direction and magnitude of maximal function increase. Calculus, thus, provides not just the theoretical foundation but practical tools to evaluate and resolve complex problems involving rates and directions of change, pivotal in diverse fields like physics, engineering, and economics.
Other exercises in this chapter
Problem 12
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