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Find an equation of the tangent plane and find symmetric equations of the normal line to the surface at the given point.\(x^{2}+y^{2}+z=9, \quad(1,2,4)\)

Short Answer

Expert verified
The equation of the tangent plane at point \((1,2,4)\) on the surface \(x^{2}+y^{2}+z=9\) is \(2(x - 1) + 4(y - 2) + 1(z - 4) = 0\). The symmetric equations of the normal line at the same point are \( \frac{x - 1}{2} = \frac{y - 2}{4} = z - 4 \)

Step by step solution

01

Find the gradient of the surface

1. The general form of the surface equation is \(F(x, y, z) = 0\). Here, \(F(x, y, z) = x^{2}+y^{2}+z-9\). So,2. To find the gradient, compute the partial derivatives with respect to each variable. The gradient \(\nabla F\) is given by: \(\nabla F = \frac{dF}{dx}i + \frac{dF}{dy}j + \frac{dF}{dz}k\) 3. Here, \(\frac{dF}{dx}=2x\), \(\frac{dF}{dy}=2y\), \(\frac{dF}{dz}=1\). Note: \(i\), \(j\), \(k\) are unit vectors along x, y and z axis respectively.
02

Evaluate the gradient at the given point

Now, substitute the given point \((1,2,4)\) into the gradient vector i.e. \(\nabla F = 2xi + 2yj + k\), we find, \(\nabla F|_{(1, 2, 4)} = 2(1)i + 2(2)j + k = 2i + 4j + k\). The gradient evaluated at \((1,2,4)\) gives a normal vector to the surface at that point.
03

Write down the equation of the tangent plane

The equation of the tangent plane to a surface at a given point can be given as \(A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0}) = 0\)where\((x_{0}, y_{0}, z_{0})\) is the given point, and \(A\), \(B\), and \(C\) are the components of the normal vector. Plug in \(A = 2\), \(B = 4\), \(C = 1\), and the point \(x_{0} = 1\), \(y_{0} = 2\), \(z_{0} = 4\), we find the equation for the tangent plane is: \(2(x - 1) + 4(y - 2) + 1(z - 4) = 0\)
04

Find symmetric equations for the normal line

The symmetric equations of a line in 3D space can be given as:\( \frac{x - x_{0}}{A} = \frac{y - y_{0}}{B} = \frac{z - z_{0}}{C}\)where\((x_{0}, y_{0}, z_{0})\) is a point on the line, and \(A\), \(B\), and \(C\) are the direction numbers which here are the components of the gradient (normal to the surface). Plug in \(A = 2\), \(B = 4\), \(C = 1\), and the point \(x_{0} = 1\), \(y_{0} = 2\), \(z_{0} = 4\). This gives us the symmetric equations of the normal line: \( \frac{x - 1}{2} = \frac{y - 2}{4} = z - 4 \)

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Key Concepts

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

Gradient Vector
In calculus, the gradient vector is a crucial concept used to find the direction of steepest ascent of a function. Think of it like a compass that points you towards higher ground on a mountain. The gradient vector for a function of three variables, say \( F(x, y, z) \), is denoted by \( abla F \) (read as "del F"). It is composed of the partial derivatives of \( F \) with respect to each variable, forming a vector. If \( F(x, y, z) = x^2 + y^2 + z - 9 \), then the gradient vector \( abla F \) is calculated as:
  • \( \frac{\partial F}{\partial x} = 2x \)
  • \( \frac{\partial F}{\partial y} = 2y \)
  • \( \frac{\partial F}{\partial z} = 1 \)
Thus, we have \( abla F = 2xi + 2yj + k \). Evaluated at a specific point like \((1, 2, 4)\), these values provide the normal vector to the surface, which is crucial when setting equations for tangent planes or normal lines.
Normal Line
The normal line to a surface at a point is the line that is perpendicular to the tangent plane at that point. Imagine it as a stick that pokes directly out of the surface, sticking straight up. To find the normal line, we utilize the components of the gradient vector, as they represent the direction numbers of the line in space.

Finding the Normal Line

To write a symmetric equation for the normal line:
  • Use the formula \( \frac{x - x_0}{A} = \frac{y - y_0}{B} = \frac{z - z_0}{C} \)
  • This is based on a point \((x_0, y_0, z_0)\) on the line and the direction from the gradient vector \( (A, B, C) \)
  • For example, at \((1, 2, 4) \) with \( abla F = 2i + 4j + k \), the symmetric equations are \( \frac{x - 1}{2} = \frac{y - 2}{4} = z - 4 \)
The resulting equations describe the line extending from the surface point in the exact direction of the surface's highest increase, which is essential for various applications in geometry and physics.
Partial Derivatives
Partial derivatives help us understand how a function changes as we slightly tweak just one of its variables while keeping the others constant. It’s like focusing on how changing one ingredient affects the taste of soup while keeping others the same.

Deriving Partial Derivatives

Consider a function \( F(x, y, z) \). To find a partial derivative of \( F \) with respect to a specific variable:
  • Treat all other variables as constants.
  • Differentiate with respect to the chosen variable.
For our example, \( F(x, y, z) = x^2 + y^2 + z - 9 \), the partial derivatives might include:
  • \( \frac{\partial F}{\partial x} = 2x \)
  • \( \frac{\partial F}{\partial y} = 2y \)
  • \( \frac{\partial F}{\partial z} = 1 \)
These derivatives are foundational for building the gradient vector, essential for further analysis, like determining tangent planes and the behavior of surfaces in multidimensional calculus.

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Most popular questions from this chapter

The table shows the world populations \(y\) (in billions) for five different years. (Source: U.S. Bureau of the Census, International Data Base) $$ \begin{array}{|l|c|c|c|c|c|} \hline \text { Year } & 1994 & 1996 & 1998 & 2000 & 2002 \\ \hline \text { Population, } \boldsymbol{y} & 5.6 & 5.8 & 5.9 & 6.1 & 6.2 \\ \hline \end{array} $$ Let \(x=4\) represent the year 1994 . (a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data. (b) Use the regression capabilities of a graphing utility to find the least squares regression quadratic for the data. (c) Use a graphing utility to plot the data and graph the models. (d) Use both models to forecast the world population for the year \(2010 .\) How do the two models differ as you extrapolate into the future?

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Investigation Consider the function \(f(x, y)=\frac{\sin y}{x}\) on the intervals \(-3 \leq x \leq 3\) and \(0 \leq y \leq 2 \pi\). (a) Find a set of parametric equations of the normal line and an equation of the tangent plane to the surface at the point \(\left(2, \frac{\pi}{2}, \frac{1}{2}\right)\) (b) Repeat part (a) for the point \(\left(-\frac{2}{3}, \frac{3 \pi}{2}, \frac{3}{2}\right)\). (c) Use a computer algebra system to graph the surface, the normal lines, and the tangent planes found in parts (a) and (b). (d) Use analytic and graphical analysis to write a brief description of the surface at the two indicated points.

Consider the function \(F(x, y, z)=0\), which is differentiable at \(P\left(x_{0}, y_{0}, z_{0}\right) .\) Give the definition of the tangent plane at \(P\) and the normal line at \(P\).

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