/*! 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 4 Write an equation for the plane ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 4

Write an equation for the plane tangent to the surface \(z=f(x, y)\) at the point \((a, b, f(a, b))\).

Short Answer

Expert verified
Question: Write the equation of the tangent plane to the surface z = f(x, y) at the point (a, b, f(a, b)), given the partial derivatives of f with respect to x and y. Answer: The equation of the tangent plane to the surface z = f(x, y) at the point (a, b, f(a, b)) is as follows: $$ z - f(a, b) = \frac{\partial f}{\partial x}(a, b) (x - a) + \frac{\partial f}{\partial y}(a, b) (y - b) $$

Step by step solution

01

Find the partial derivatives of f with respect to x and y

To find the partial derivatives of the function f(x, y) with respect to x and y, we will use the following notation: $$ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial x}(x, y) $$ and $$ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial y}(x, y) $$ It's important to note that these derivatives will give us the slopes of the tangent plane in the x and Y direction respectively.
02

Evaluate the partial derivatives at the given point (a, b, f(a, b))

Now we need to find the values of the two derivatives at the given point. To do this, we will replace x with a and y with b in both partial derivatives: $$ \frac{\partial f}{\partial x}(a, b) $$ and $$ \frac{\partial f}{\partial y}(a, b) $$
03

Formulate the equation of the tangent plane

Now we have everything we need to write the equation of the tangent plane. The equation of the tangent plane to the surface z=f(x, y) at the point (a, b, f(a, b)) is given by: $$ z - f(a, b) = \frac{\partial f}{\partial x}(a, b) (x - a) + \frac{\partial f}{\partial y}(a, b) (y - b) $$ In summary, we first found the partial derivatives of f with respect to x and y, then we evaluated these derivatives at the given point (a, b, f(a, b)), and finally, we formulated the equation of the tangent plane using the values of the partial derivatives at the given point.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Partial Derivatives
Understanding partial derivatives is essential when venturing into the world of multivariable calculus. These derivatives are a measure of how a function changes as one of its variables changes while all other variables are held constant. Imagine you're on a hillside, the slope of the hill will change differently if you walk directly north as opposed to walking east. In mathematical terms, for a function like f(x, y), the partial derivative with respect to x is denoted as \(\frac{\partial f}{\partial x}\), and similarly for y as \(\frac{\partial f}{\partial y}\).

When computing \(\frac{\partial f}{\partial x}\), treat y as a constant, and vice versa. These derivatives are fundamental when trying to find the slope in the direction of either axis - which eventually helps us determine the slope of the tangent plane at a particular point on a surface. While this may sound complicated, it's similar to finding the single derivative of a function in one variable, but doing it for each variable independently in a multivariable context.
Multivariable Calculus
Multivariable calculus extends single variable calculus to higher dimensions. In single variable calculus, you work with functions of one variable and analyse things like the derivative and the integral of these functions. However, once we step into the domain of functions with two or more variables, like f(x, y) for a two-dimensional plane, we need to consider how the function changes in multiple directions simultaneously.

One of the main applications is finding and understanding gradients, directional derivatives, and, as it relates to our context, the equations of tangent planes. These are tools that help us touch the surface of a hill, not with just a point, but with an entire plane that is perfectly aligned at just one point on our multivariable terrain. The complex beauty of multivariable calculus lies in examining how changing different variables affects the outcome, much like adjusting different settings on a camera to get the perfect picture.
Tangent Plane
The concept of the tangent plane is a natural extension of the tangent line from single variable calculus. A tangent line touches a curve at a single point, and the tangent plane does the same but for a surface. More specifically, for a surface defined by z = f(x, y), the tangent plane is the plane that touches the surface at a single point, without cutting through it.

Equipped with partial derivatives that we calculated earlier, we can establish the equation for this plane. Take the point (a, b, f(a, b)) on the surface - our touchpoint. At this point, the slopes in the direction of x and y are given by the partial derivatives we evaluated at a and b. These slopes guide us to the correct orientation of the plane so that it just grazes the surface at the point. The final equation \(z - f(a, b) = \frac{\partial f}{\partial x}(a, b) (x - a) + \frac{\partial f}{\partial y}(a, b) (y - b)\) is a compact way to carry all this information. It tells us that if we know a point on the surface and how the surface is tilting at that point, we can describe the entire plane that hugs the surface at that exact location.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose \(n\) houses are located at the distinct points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right) .\) A power substation must be located at a point such that the sum of the squares of the distances between the houses and the substation is minimized. a. Find the optimal location of the substation in the case that \(n=3\) and the houses are located at \((0,0),(2,0),\) and (1,1) b. Find the optimal location of the substation in the case that \(n=3\) and the houses are located at distinct points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right),\) and \(\left(x_{3}, y_{3}\right)\) c. Find the optimal location of the substation in the general case of \(n\) houses located at distinct points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots\) \(\left(x_{n}, y_{n}\right)\) d. You might argue that the locations found in parts (a), (b), and (c) are not optimal because they result from minimizing the sum of the squares of the distances, not the sum of the distances themselves. Use the locations in part (a) and write the function that gives the sum of the distances. Note that minimizing this function is much more difficult than in part (a).

Potential functions arise frequently in physics and engineering. A potential function has the property that \(a\) field of interest (for example, an electric field, a gravitational field, or a velocity field is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter 14 .) The gravitational potential associated with two objects of mass \(M\) and \(m\) is \(\varphi=-G M m / r,\) where \(G\) is the gravitational constant. If one of the objects is at the origin and the other object is at \(P(x, y, z),\) then \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between the objects. The gravitational field at \(P\) is given by \(\mathbf{F}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. Show that the force has a magnitude \(|\mathbf{F}|=G M m / r^{2}\) Explain why this relationship is called an inverse square law.

Given a differentiable function \(w=f(x, y, z),\) the goal is to find its maximum and minimum values subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0\) and \(h(x, y, z)=0\).

When two electrical resistors with resistance \(R_{1}>0\) and \(R_{2}>0\) are wired in parallel in a circuit (see figure), the combined resistance \(R,\) measured in ohms \((\Omega),\) is given by \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}.\) a. Estimate the change in \(R\) if \(R_{1}\) increases from \(2 \Omega\) to \(2.05 \Omega\) and \(R_{2}\) decreases from \(3 \Omega\) to \(2.95 \Omega\) b. Is it true that if \(R_{1}=R_{2}\) and \(R_{1}\) increases by the same small amount as \(R_{2}\) decreases, then \(R\) is approximately unchanged? Explain. c. Is it true that if \(R_{1}\) and \(R_{2}\) increase, then \(R\) increases? Explain. d. Suppose \(R_{1}>R_{2}\) and \(R_{1}\) increases by the same small amount as \(R_{2}\) decreases. Does \(R\) increase or decrease?

Determine whether the following statements are true and give an explanation or counterexample. a. Suppose you are standing at the center of a sphere looking at a point \(P\) on the surface of the sphere. Your line of sight to \(P\) is orthogonal to the plane tangent to the sphere at \(P\). b. At a point that maximizes \(f\) on the curve \(g(x, y)=0,\) the dot product \(\nabla f \cdot \nabla g\) is zero.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.