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In Exercises \(13-18\) , find parametric equations for the line tangent to the curve of intersection of the surfaces at the given point. $$ \begin{array}{l}{\text { Surfaces: } x^{2}+2 y+2 z=4, \quad y=1} \\ {\text { Point: } \quad(1,1,1 / 2)}\end{array} $$

Short Answer

Expert verified
Parametric equations: \(x = 1 - 2t\), \(y = 1\), \(z = \frac{1}{2} + 2t\).

Step by step solution

01

Understand the Surfaces

We have two surfaces: the first is the equation of a surface given by \( x^2 + 2y + 2z = 4 \), and the second is a plane \( y = 1 \). The point of intersection on these surfaces is \((1, 1, 1/2)\).
02

Compute Gradient Vectors

The gradient vector for the first surface, \( f(x, y, z) = x^2 + 2y + 2z - 4 \), is \( abla f = (2x, 2, 2) \). For the second surface, \( g(x, y, z) = y - 1 \), the gradient vector is \( abla g = (0, 1, 0) \).
03

Evaluate Gradient Vectors at the Point

Evaluate the gradients at the point \((1, 1, 1/2)\). Therefore, \( abla f |_{(1, 1, 1/2)} = (2\cdot1, 2, 2) = (2, 2, 2) \) and \( abla g = (0, 1, 0) \) doesn't depend on \((x, y, z)\).
04

Compute the Cross Product for the Tangent Direction

The direction of the tangent line is given by the cross product \( abla f \times abla g \). Calculate this: \((2, 2, 2) \times (0, 1, 0)\) results in \(( -2, 0, 2 )\).
05

Write Parametric Equations

Using the direction \((-2, 0, 2)\) and point \((1, 1, 1/2)\), the parametric equations are: \(x = 1 - 2t\), \(y = 1\), \(z = \frac{1}{2} + 2t\).

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Key Concepts

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

Gradient Vector
A gradient vector is a fundamental tool in multivariable calculus used to describe how a function changes at any given point in space. For any function of three variables, such as \( f(x, y, z) \), the gradient vector, denoted as \( abla f \), is computed by taking the partial derivatives of \( f \) with respect to \( x \), \( y \), and \( z \):
  • \( \frac{\partial f}{\partial x} \)
  • \( \frac{\partial f}{\partial y} \)
  • \( \frac{\partial f}{\partial z} \)
For the surface described by \( f(x, y, z) = x^2 + 2y + 2z - 4 \), the gradient vector becomes \( abla f = (2x, 2, 2) \). Meanwhile, for a simpler linear surface like \( g(x, y, z) = y - 1 \), the gradient is constant: \( abla g = (0, 1, 0) \).

These vectors provide the direction of steepest ascent on the surface, and by evaluating them at specific points such as \((1, 1, 1/2)\), you get \( abla f = (2, 2, 2) \) and \( abla g = (0, 1, 0) \). Gradient vectors are vital in finding tangent lines and planes to surfaces, as they highlight crucial directional information.
Cross Product
The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors in three-dimensional space, which is especially useful in physics and engineering applications. For vectors \( \mathbf{A} = (a_1, a_2, a_3) \) and \( \mathbf{B} = (b_1, b_2, b_3) \), the cross product, \( \mathbf{A} \times \mathbf{B} \), is defined as:
  • \((a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\)
In the context of finding a tangent line, the cross product of the gradient vectors \( abla f \) and \( abla g \) gives us the direction of the tangent line.

When you take the cross product of \( (2, 2, 2) \) and \( (0, 1, 0) \), you get \( ( -2, 0, 2 ) \). This resulting vector represents the direction in which the tangent line extends. The cross product works efficiently because it combines information from both surfaces to find the common tangent direction.
Parametric Equations
Parametric equations are a way of defining a line or curve by expressing the coordinates as functions of a parameter, often denoted by \( t \). This is particularly useful for describing motion or lines in space, where each equation represents a dimension:
  • For the \( x \)-coordinate: \( x(t) = x_0 + at \)
  • For the \( y \)-coordinate: \( y(t) = y_0 + bt \)
  • For the \( z \)-coordinate: \( z(t) = z_0 + ct \)
In the case of the tangent line problem, given the starting point \((1, 1, 1/2)\) and the direction vector \(( -2, 0, 2 )\), the parametric equations become:

\[x = 1 - 2t,y = 1,z = \frac{1}{2} + 2t\]These equations describe a line in space beginning at the point \((1, 1, 1/2)\) and moving in the direction provided by the cross product. Parametric equations are versatile and can clearly convey how points on a line change with respect to the parameter \( t \).

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Most popular questions from this chapter

In Exercises \(1-10\) , use Taylor's formula for \(f(x, y)\) at the origin to find quadratic and cubic approximations of \(f\) near the origin. $$ f(x, y)=e^{x} \ln (1+y) $$

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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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