/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Let \(f(x, y)=\sqrt{3 x+2 y}\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(f(x, y)=\sqrt{3 x+2 y}\) (a) Find the slope of the surface \(z=f(x, y)\) in the \(x-\) direction at the point \((4,2) .\) (b) Find the slope of the surface \(z=f(x, y)\) in the \(y-\) direction at the point \((4,2) .\)

Short Answer

Expert verified
(a) \( \frac{3}{8} \); (b) \( \frac{1}{4} \).

Step by step solution

01

Understand the Function

We are given the function \( f(x, y) = \sqrt{3x + 2y} \). Our goal is to find the slope of this surface in the directions of \( x \) and \( y \) at the point \( (4, 2) \). This involves calculating the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
02

Compute Partial Derivative with Respect to x

To find the slope in the \( x \)-direction, we compute the partial derivative of \( f \) with respect to \( x \): \[ \frac{\partial f}{\partial x} = \frac{d}{dx}[\sqrt{3x + 2y}] = \frac{1}{2}\frac{1}{\sqrt{3x + 2y}} \cdot 3 = \frac{3}{2\sqrt{3x + 2y}}. \] Evaluate this derivative at the point \( (4, 2) \).
03

Evaluate \( \frac{\partial f}{\partial x} \) at (4,2)

Substitute \( x = 4 \) and \( y = 2 \) into the expression for \( \frac{\partial f}{\partial x} \):\[ \frac{\partial f}{\partial x}(4, 2) = \frac{3}{2\sqrt{3(4) + 2(2)}} = \frac{3}{2\sqrt{12 + 4}} = \frac{3}{2\sqrt{16}} = \frac{3}{2 \cdot 4} = \frac{3}{8}. \] Thus, the slope in the \( x \)-direction at \((4, 2)\) is \( \frac{3}{8} \).
04

Compute Partial Derivative with Respect to y

Next, find the partial derivative of \( f \) with respect to \( y \): \[ \frac{\partial f}{\partial y} = \frac{d}{dy}[\sqrt{3x + 2y}] = \frac{1}{2}\frac{1}{\sqrt{3x + 2y}} \cdot 2 = \frac{2}{2\sqrt{3x + 2y}} = \frac{1}{\sqrt{3x + 2y}}. \] Evaluate this derivative at the point \( (4, 2) \).
05

Evaluate \( \frac{\partial f}{\partial y} \) at (4,2)

Substitute \( x = 4 \) and \( y = 2 \) into the expression for \( \frac{\partial f}{\partial y} \):\[ \frac{\partial f}{\partial y}(4, 2) = \frac{1}{\sqrt{3(4) + 2(2)}} = \frac{1}{\sqrt{12 + 4}} = \frac{1}{\sqrt{16}} = \frac{1}{4}. \] Thus, the slope in the \( y \)-direction at \((4, 2)\) is \( \frac{1}{4} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multivariable Calculus
Multivariable Calculus is an extension of calculus that deals with functions of multiple variables. Unlike single-variable calculus, where functions depend on just one variable, multivariable calculus considers how functions vary with respect to two or more variables.
This type of calculus is crucial in fields such as physics, engineering, and economics, where systems are often dependent on several factors.
In the context of this problem, the function \( f(x, y) \) is a two-variable function, which will be explained more in the next section. Understanding how the surface behaves in different directions involves computing partial derivatives, which are the core tools in multivariable calculus.
Multivariable calculus gives us the power to:
  • Analyze surfaces and curves in higher dimensions.
  • Determine the change of a function in any given direction.
  • Predict the behavior of complex systems by studying gradients and other derivatives.
Function of Two Variables
In our problem, we have a function \( f(x, y) = \sqrt{3x + 2y} \), which depends on two variables, \(x\) and \(y\).
This function can be visualized as a surface in a three-dimensional space, where each pair \( (x, y) \) corresponds to a height determined by the equation. The role of each variable is significant as they influence the value of \(f\) independently.
To understand how this function behaves, we examine its partial derivatives. These derivatives tell us how \(f(x, y)\) changes as each variable \(x\) or \(y\) is varied, while the other is held constant.
  • Partial Derivative w.r.t. \(x\) (\( \frac{\partial f}{\partial x} \)) interprets how the surface slopes as \(x\) varies.
  • Partial Derivative w.r.t. \(y\) (\( \frac{\partial f}{\partial y} \)) shows the slope in the \(y\) direction.
These concepts allow us to explore and understand the shape and gradient of the surface formed by the function, which is readily applicable in real-world scenarios when modeling terrains, optimizing processes, and predicting outcomes.
Gradient in Calculus
The gradient is a vector that shows how a multivariable function changes at any point. If you imagine a landscape or a hill, the gradient points in the direction of the steepest ascent.
For a function of two variables, like our \( f(x, y) = \sqrt{3x + 2y} \), the gradient vector \( abla f \) is composed of its partial derivatives:
\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]
  • This vector captures both the direction and the rate of the steepest increase of the function.
  • For our exercise, we calculated \ \( \frac{\partial f}{\partial x} = \frac{3}{8} \) and \( \frac{\partial f}{\partial y} = \frac{1}{4} \), so at point \((4, 2)\), the gradient vector is \((\frac{3}{8}, \frac{1}{4})\).
  • The gradient is very useful in optimization problems, helping to find maximum or minimum points by guiding adjustments in variables.
Understanding the gradient allows you to analyze and predict how a function behaves, particularly in the context of adjusting inputs for desired outcomes, much like determining the best path up or down a hill.

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Most popular questions from this chapter

Writing Explain how one might use the graph of the equation \(z=f(x, y)\) to determine the signs of \(f_{x}\left(x_{0}, y_{0}\right)\) and \(f_{y}\left(x_{0}, y_{0}\right)\) by inspection.

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