/*! 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 40 Find an equation of the tangent ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find an equation of the tangent plane to the parametric surface at the stated point. $$ x=u^{2}, y=v^{2}, z=u+v ;(1,4,3) $$

Short Answer

Expert verified
The equation of the tangent plane is \( 2x + 4y + 2z = 25 \).

Step by step solution

01

Understand the Parametric Equations

The given surface is defined parametrically by the equations \( x = u^2 \), \( y = v^2 \), and \( z = u + v \). We need to find the values of \( u \) and \( v \) that correspond to the point \( (1, 4, 3) \) on the surface.
02

Find Parameters \( u \) and \( v \) for the Given Point

From \( x = u^2 = 1 \), we have \( u = 1 \) or \( u = -1 \). Since \( z = u + v = 3 \), we test \( u = 1 \): \( v = 3 - 1 = 2 \). Verify in \( y = v^2 = 4 \), so \( v = 2 \) holds. Thus, \( u = 1 \), \( v = 2 \) are correct.
03

Compute Partial Derivatives

Firstly, compute the partial derivatives of \( x \), \( y \), and \( z \) with respect to \( u \) and \( v \):- \( \frac{\partial x}{\partial u} = 2u \), \( \frac{\partial y}{\partial u} = 0 \), \( \frac{\partial z}{\partial u} = 1 \)- \( \frac{\partial x}{\partial v} = 0 \), \( \frac{\partial y}{\partial v} = 2v \), \( \frac{\partial z}{\partial v} = 1 \).Evaluate these at \( u = 1 \), \( v = 2 \).
04

Evaluate Partial Derivatives at the Point

Substitute \( u = 1 \), \( v = 2 \) into the partial derivatives:- \( \frac{\partial x}{\partial u} = 2(1) = 2 \), \( \frac{\partial y}{\partial u} = 0 \), \( \frac{\partial z}{\partial u} = 1 \)- \( \frac{\partial x}{\partial v} = 0 \), \( \frac{\partial y}{\partial v} = 2(2) = 4 \), \( \frac{\partial z}{\partial v} = 1 \).
05

Write the Tangent Plane Equation

The tangent plane to the surface at \( u = 1, v = 2 \) is given by: \[(x - x_0) \frac{\partial x}{\partial u} + (y - y_0) \frac{\partial y}{\partial u} + (z - z_0) \frac{\partial z}{\partial u} + (x - x_0) \frac{\partial x}{\partial v} + (y - y_0) \frac{\partial y}{\partial v} + (z - z_0) \frac{\partial z}{\partial v} = 0\]Substitute \( (x_0, y_0, z_0) = (1, 4, 3) \), we obtain:\( 2(x - 1) + 0 + 1(z - 3) + 0 + 4(y - 4) + 1(z - 3) = 0 \).
06

Simplify the Tangent Plane Equation

Expand and simplify the equation:\[ 2(x - 1) + 4(y - 4) + 2(z - 3) = 0 \]This simplifies to:\[ 2x + 4y + 2z = 25 \].

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Key Concepts

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

parametric surface
A parametric surface is a type of surface in three-dimensional space where the coordinates
  • x
  • y
  • z
are expressed as functions of two parameters, usually denoted as \(u\) and \(v\). For instance, in the given problem, these equations are specified as \(x = u^2\), \(y = v^2\), and \(z = u + v\).
Parametric equations are beneficial when modeling complex surfaces because they let us use two simple parameters to navigate any point on the surface.Reducing complex shapes into simple, navigable forms can simplify the process of finding related attributes such as tangent planes.
The primary task is to find the values of the parameters \(u\) and \(v\) such that the given parametric equations coincide with known points on the surface. In this exercise, by substituting values and solving these relations, parameters \(u = 1\) and \(v = 2\) were determined.
partial derivatives
Partial derivatives are a fundamental concept in calculus, especially in dealing with functions of multiple variables. They measure the change in a function concerning one variable while keeping other variables constant.
In this exercise, we need the partial derivatives of our parametric equations with respect to the parameters \(u\) and \(v\).
  • The partial derivatives of the coordinates with respect to \(u\) are:
    • \( \frac{\partial x}{\partial u} = 2u \)
    • \( \frac{\partial y}{\partial u} = 0 \)
    • \( \frac{\partial z}{\partial u} = 1 \)
  • Those with respect to \(v\) are:
    • \( \frac{\partial x}{\partial v} = 0 \)
    • \( \frac{\partial y}{\partial v} = 2v \)
    • \( \frac{\partial z}{\partial v} = 1 \)

These calculations provide us with the rate of change of each coordinate of the surface with respect to one parameter at a time.
equation of a surface
The equation of a tangent plane to a surface at a given point provides a flat approximation of the surface near that point. For a parametric surface, this requires the calculated partial derivatives and the known point on the surface.
In this problem, given the partial derivatives and point \((1, 4, 3)\), we can write a formula for the tangent plane using:
\[(x - x_0) \frac{\partial x}{\partial u} + (y - y_0) \frac{\partial y}{\partial u} + (z - z_0) \frac{\partial z}{\partial u} + (x - x_0) \frac{\partial x}{\partial v} + (y - y_0) \frac{\partial y}{\partial v} + (z - z_0) \frac{\partial z}{\partial v} = 0\]
This formula accounts for the rate of change in \(x\), \(y\), and \(z\) concerning changes in \(u\) and \(v\).
Substituting with the given values, simplifying, and rearranging, we derive the equation for the tangent plane:\[2x + 4y + 2z = 25\]
which gives us a precise linear equation representing the plane tangent to our parametric surface at the specified point.

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Most popular questions from this chapter

Express each integral as an equivalent integral in which the \(z\) -integration is performed first, the \(y\) -integration second, and the \(x\) -integration last. (a) \(\int_{0}^{3} \int_{0}^{\sqrt{9-z^{2}}} \int_{0}^{\sqrt{9-y^{2}-z^{2}}} f(x, y, z) d x d y d z\) (b) \(\int_{0}^{4} \int_{0}^{2} \int_{0}^{x / 2} f(x, y, z) d y d z d x\) (c) \(\int_{0}^{4} \int_{0}^{4-y} \int_{0}^{\sqrt{z}} f(x, y, z) d x d z d y\)

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