/*! 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 48 Find an equation of the plane ta... [FREE SOLUTION] | 91Ó°ÊÓ

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Find an equation of the plane tangent to the following surfaces at the given point. $$(x+z) /(y-z)=2 ;(4,2,0)$$

Short Answer

Expert verified
** The equation of the tangent plane to the given surface at the point (4, 2, 0) is: $$\frac{1}{2}x + y + z = 4$$

Step by step solution

01

Expand the equation

First, we re-write the equation of the given surface in a more convenient form. We can multiply both sides of the equation by \((y-z)\) to obtain: $$x+y-z = 2y-2z$$ Now, we can simplify the equation further by moving terms to the other side: $$x-2z+y+2z = 3y$$ $$x = 3y-y$$ $$x = 2y$$
02

Rewrite the equation in terms of z

We can solve for z to obtain the equation of the surface in terms of x and y: $$z = \frac{1}{2}x - y$$
03

Find partial derivatives

Now, we will find the partial derivative of the z equation with respect to \(x\) and \(y\). The gradient is the normal vector of the tangent plane, so we need the partial derivatives to find the normal vector. $$\frac{\partial z}{\partial x} = \frac{1}{2}$$ $$\frac{\partial z}{\partial y} = -1$$
04

Substitute the given point to find the normal vector

The normal vector of the tangent plane is given by the gradient at the given point, which has components \((\frac{\partial z}{\partial x}(4,2,0), \frac{\partial z}{\partial y}(4,2,0),1)\). To find the normal vector \(n = (A, B, C)\) at the point \((4,2,0)\), we can assign the constant values: $$A = \frac{1}{2}$$ $$B = -1$$ $$C = 1$$ So the normal vector is \(n = (\frac{1}{2}, -1, 1)\).
05

Find the equation of the tangent plane

Using the normal vector from step 4, we can now write the equation of the tangent plane as: $$\frac{1}{2}(x - 4) -1(y - 2) + 1(z - 0) = 0$$ Finally, we expand and simplify the equation of the tangent plane: $$\frac{1}{2}x - 2 + y -2 +1z = 0$$ $$\frac{1}{2}x + y + z = 4$$ The equation of the tangent plane to the given surface at the point \((4, 2, 0)\) is: $$\frac{1}{2}x + y + z = 4$$

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Key Concepts

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

Partial Derivatives
Partial derivatives are like regular derivatives, but they're applied to functions of multiple variables.
Instead of measuring how a single variable changes, partial derivatives tell us how a function changes when one particular variable in a multi-variable function varies, while all other variables are held constant.
To compute the partial derivatives, we focus on one variable at a time. For example, with a function like \( z = \frac{1}{2}x - y \), it's understood as a function of both \(x\) and \(y\).
  • For \( \frac{\partial z}{\partial x} \), we treat \(y\) as a constant and differentiate with respect to \(x\), giving us the expression \( \frac{1}{2} \).
  • For \( \frac{\partial z}{\partial y} \), \(x\) is the constant, and differentiation with respect to \(y\) results in \(-1\).
These derivatives provide us with important detail: they allow us to determine the slope of the tangent line to the surface in the directions of the \(x\) and \(y\) axes.
Normal Vector
A normal vector is a vector that is perpendicular to a surface.
For a given surface, the normal vector is a crucial part in defining the orientation of that surface in space.
In the context of tangent planes, the normal vector is derived from the gradient of the function that defines the surface. The gradient is a vector that points in the direction of the greatest rate of increase of the function.
Given a function \(z = \frac{1}{2}x - y\), the normal vector \(n = (A, B, C)\) is found using:
  • \(A = \frac{\partial z}{\partial x} = \frac{1}{2}\)
  • \(B = \frac{\partial z}{\partial y} = -1\)
  • \(C = 1\)
This gives us the normal vector \((\frac{1}{2}, -1, 1)\).
This vector tells us not just how the surface is oriented at a specific point but, when used in conjunction with a point on the surface, it helps to construct the equation of the tangent plane.
Equation of Surface
To understand the equation of the surface, visualize how a geometric figure can be described using mathematical expressions.
Usually, a surface can be represented by an equation involving \(x\), \(y\), and \(z\) coordinates.
Rewriting the original question equation \((x+z)/(y-z)=2\) as \(z = \frac{1}{2}x - y\) helps us frame this surface in terms of a more direct relationship between these variables.
This simplified form is crucial in the process of finding the tangent plane equations which we use to get a linear approximation of the surface near a point.The equation of the tangent plane at a specific point, here \((4, 2, 0)\), uses the point itself and the normal vector to assert a planarity over a local region of the surface.
By incorporating the normal vector components into the plane equation \((\frac{1}{2}(x - 4) - (y - 2) + z = 0)\), we can simplify to get \(\frac{1}{2}x + y + z = 4\).
This simplified equation represents the flat plane that is just tangent to the surface precisely at our chosen point.

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Most popular questions from this chapter

Generalize Exercise 75 by considering a wave described by the function \(z=A+\sin (a x-b y)\) where \(a, b,\) and \(A\) are real numbers. a. Find the direction in which the crests and troughs of the wave are aligned. Express your answer as a unit vector in terms of \(a\) and \(b\) b. Find the surfer's direction - that is, the direction of steepest descent from a crest to a trough. Express your answer as a unit vector in terms of \(a\) and \(b\)

Describe the set of all points (if any) at which all three planes \(x+3 z=3, y+4 z=6,\) and \(x+y+6 z=9\) intersect.

Identify and briefly describe the surfaces defined by the following equations. $$-y^{2}-9 z^{2}+x^{2} / 4=1$$

Identify and briefly describe the surfaces defined by the following equations. $$x^{2} / 4+y^{2}-2 x-10 y-z^{2}+41=0$$

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Let the equation of the best-fit line be \(y=m x+b,\) where the slope \(m\) and the \(y\) -intercept \(b\) must be determined using the least squares condition. First assume that there are three data points \((1,2),(3,5),\) and \((4,6) .\) Show that the function of \(m\) and \(b\) that gives the sum of the squares of the vertical distances between the line and the three data points is $$ \begin{aligned} E(m, b)=&((m+b)-2)^{2}+((3 m+b)-5)^{2} \\ &+((4 m+b)-6)^{2} \end{aligned}. $$ Find the critical points of \(E\) and find the values of \(m\) and \(b\) that minimize \(E\). Graph the three data points and the best-fit line.

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