/*! 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 Find equations for the (a) tan... [FREE SOLUTION] | 91Ó°ÊÓ

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Find equations for the (a) tangent plane and (b) normal line at the point \(P_{0}\) on the given surface. $$ 2 z-x^{2}=0, \quad P_{0}(2,0,2) $$

Short Answer

Expert verified
The tangent plane equation is \(z = 2x - 2\), and the normal line equations are \(x = 2 - 4t\), \(y = 0\), \(z = 2 + 2t\).

Step by step solution

01

Identify the Surface Equation

The given surface equation is \(2z - x^2 = 0\). We need to find the tangent plane and the normal line at a specific point on this surface.
02

Find the Gradient of the Surface

To find the tangent plane, we first find the gradient of the surface function \(f(x, y, z) = 2z - x^2\). The gradient vector is \(abla f = \left( -2x, 0, 2\right)\).
03

Evaluate the Gradient at the Given Point

The point given is \(P_0(2, 0, 2)\). Substitute \(x = 2\), \(y = 0\), and \(z = 2\) into the gradient vector: \(abla f = (-4, 0, 2)\).
04

Write the Tangent Plane Equation

The equation of the tangent plane at \(P_0\) is given by: \[ -4(x - 2) + 0(y - 0) + 2(z - 2) = 0 \]Simplifying, the tangent plane equation becomes: \[ -4(x - 2) + 2(z - 2) = 0 \]\[ -4x + 8 + 2z - 4 = 0 \]\[ -4x + 2z = -4 \]\[ 2z = 4x - 4 \]\[ z = 2x - 2 \]
05

Write the Normal Line Equation

The direction vector for the normal line is the same as the gradient vector \((-4, 0, 2)\). Use the point \(P_0(2, 0, 2)\) to write the parametric equations for the normal line:\[ x = 2 - 4t \]\[ y = 0 \]\[ z = 2 + 2t \]

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Key Concepts

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

Normal Line
A normal line to a surface is a line that is perpendicular to the tangent plane at a given point. Understanding the normal line is crucial because it helps define the orientation of the tangent plane in three-dimensional space.

In this exercise, the normal line is derived at the point on the surface determined by the equation \(2z - x^2 = 0\). Given the point \( P_0(2, 0, 2) \), we found the gradient vector at this point, which serves as the direction vector for the normal line. The gradient vector calculated is \((-4, 0, 2)\).

To formulate the normal line, we use the parametric equations which introduce a variable \(t\) that allows for movement along the line:
  • \( x = 2 - 4t \)
  • \( y = 0 \)
  • \( z = 2 + 2t \)
This set of equations fully describes the normal line extending from the point \(P_0\) along the surface.
Gradient Vector
The gradient vector is an essential tool in multivariable calculus, representing the direction of the steepest ascent of a function. When dealing with surfaces, it acts as a normal vector to the surface at a given point.

For the surface equation \(2z - x^2 = 0\), the function to consider is \( f(x, y, z) = 2z - x^2 \). The gradient vector \( abla f \) for this function is calculated by finding the partial derivatives with respect to each variable:
  • \( \frac{\partial f}{\partial x} = -2x \)
  • \( \frac{\partial f}{\partial y} = 0 \)
  • \( \frac{\partial f}{\partial z} = 2 \)
Substituting the coordinates of the point \(P_0(2, 0, 2)\) into the gradient vector formula gives \((-4, 0, 2)\). This vector is used in determining both the tangent plane and the normal line equations.
Surface Equation
A surface equation describes a surface in three-dimensional space. It encapsulates the relationship between the coordinates \((x, y, z)\) that make up points on the surface.

In this particular exercise, the surface equation is given by \(2z - x^2 = 0\). This equation defines a curved surface known as a quadratic surface. The structure indicates that for every value of \(z\), the values \(x\) that satisfy the equation describe the shape of the surface.

With the point \( P_0(2, 0, 2) \) identified, we explored how this surface behaves near this particular point, leading us to the tangent plane and normal line. The surface equation is the foundational component that guides these calculations and helps visualize the geometric properties.
Parametric Equations
Parametric equations are ways of expressing points or paths along a line or curve using a parameter, often denoted as \(t\). Instead of expressing \(y\) directly in terms of \(x\), parametric equations use a third variable to define \(x\), \(y\), and \(z\) separately.

For the normal line in this exercise, we used parametric equations to describe the coordinates of any point on the line extending from \(P_0(2, 0, 2)\) in the direction given by the gradient vector \((-4, 0, 2)\). The parametric form allowed us to define:
  • \(x = 2 - 4t\) - the x-component
  • \(y = 0\) - the y-component, which remains constant
  • \(z = 2 + 2t\) - the z-component
These equations result in a more flexible, continuous representation of the normal line, making it easier to map out or visualize in applications that require detailed analysis of geometric shapes. They are crucial for both theoretical insights and practical computations in mathematics.

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Most popular questions from this chapter

Extrema on a circle of intersection Find the extreme values of the function \(f(x, y, z)=x y+z^{2}\) on the circle in which the plane \(y-x=0\) intersects the sphere \(x^{2}+y^{2}+z^{2}=4\)

In Exercises \(63-66,\) use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$f(x, y)=\sqrt{2 x+3 y-1}, \quad \frac{\partial f}{\partial x} \quad \text { and } \quad \frac{\partial f}{\partial y} \quad \text { at }(-2,3)$$

Let \(f(x, y)=2 x+3 y-4 .\) Find the slope of the line tangent tothis surface at the point \((2,-1)\) and lying in the a. plane \(x=2\) b. plane \(y=-1 .\)

Least squares and regression lines When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$ \begin{array}{l}{m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}}} \\\ {b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right)}\end{array} $$ with all sums running from \(k=1\) to \(k=n .\) Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\) 3\. handle data analytically.

a. Around the point \((1,0),\) is \(f(x, y)=x^{2}(y+1)\) more sensitive to changes in \(x\) or to changes in \(y ?\) Give reasons for your answer. b. What ratio of \(d x\) to \(d y\) will make \(d f\) equal zero at \((1,0) ?\)

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