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Find an equation of the tangent plane to the surface at the given point.\(x^{2}+2 z^{2}=y^{2}, \quad(1,3,-2)\)

Short Answer

Expert verified
The equation of the tangent plane to the surface \(x^{2}+2 z^{2}=y^{2}\) at the point (1,3,-2) is \(2x - 6y - 8z = 2\).

Step by step solution

01

Find the gradient of the function

The given surface is \(x^{2}+2 z^{2}=y^{2}\). We can write it in the form of \(f(x, y, z) = 0\), where \(f(x, y, z) = x^{2}+2z^{2}-y^{2}\). Then find the gradient \(\nabla f = (f_{x}, f_{y}, f_{z})\), which gives us the normal vector to the tangent plane. After calculation, we have \(f_{x} = 2x\) , \(f_{y} = -2y\), and \(f_{z} = 4z\).
02

Evaluate the gradient at the given point

Evaluate \(f_{x}, f_{y}, f_{z}\) at the given point \((1,3,-2)\). After evaluation, we get \(f_{x}(1,3,-2) = 2\), \(f_{y}(1,3,-2) = -6\), and \(f_{z}(1,3,-2) = -8\), so the normal vector to the tangent plane at \((1,3,-2)\) is \(\nabla f(1,3,-2) = (2, -6, -8)\).
03

Write down the equation of the tangent plane

The equation of the tangent plane is given by \(A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0}) = 0\), where \((A, B, C) = \nabla f(1,3,-2) = (2, -6, -8)\) and \((x_{0}, y_{0}, z_{0}) = (1, 3, -2)\). Therefore, we have \(2(x - 1) - 6(y - 3) - 8(z + 2) = 0\). Simplify this equation, we get the final answer \(2x - 6y - 8z = 2\).

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Key Concepts

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

Gradient of a Function
The gradient of a function is a fundamental concept in multivariable calculus that is particularly useful when analyzing surfaces and their tangent planes. For a function with several variables, the gradient is a vector that points in the direction of the steepest ascent of the function at a particular point. It is composed of partial derivatives with respect to each variable.

For instance, the function from the exercise, written as \(f(x, y, z) = x^{2} + 2z^{2} - y^{2}\), has a gradient \(abla f = (f_x, f_y, f_z)\), where \(f_x, f_y, f_z\) are the partial derivatives of \(f\) with respect to \(x, y, z\), respectively. The gradient at a specific point gives the normal vector to the tangent plane at that point, which is essential in writing the equation of the tangent plane.
Normal Vector
A normal vector is a vector that is perpendicular to a surface at a particular point. In the context of finding the equation for a tangent plane, the normal vector is vital as it dictates the orientation of the plane. It is derived from the gradient of the function, evaluated at the point of tangency, as shown in the exercise.

After evaluating the gradient at the point \( (1, 3, -2) \), the resultant normal vector is \( abla f(1, 3, -2) = (2, -6, -8) \). This normal vector is orthogonal to the tangent plane and helps define the plane's equation by giving the coefficients for the variables \(x, y, z\) in the plane equation.
Partial Derivatives
Partial derivatives represent the rate at which a function changes as one of its variables changes, while all other variables remain constant. In multivariable functions, this concept is crucial for understanding how the function behaves in relation to each of its variables independently.

The partial derivatives in the exercise are denoted by \(f_x, f_y, f_z\) and calculated as \(2x, -2y, 4z\) respectively. When evaluating these at a particular point, they contribute to the components of the gradient vector, which then is used to determine the normal vector for the tangent plane equation.
Multivariable Calculus
Multivariable calculus extends calculus to functions of several variables. It involves concepts like partial derivatives, gradient vectors, and tangent planes, which are all intricately linked. For example, the equation of a tangent plane in three-dimensional space relies on understanding how a function changes in multiple directions, which is where multivariable calculus comes in.

The step-by-step solution provided for the exercise is a practical application of multivariable calculus, involving the use of partial derivatives to find the gradient, which leads to the normal vector that is essential for writing the tangent plane's equation. Such applications are important in fields ranging from engineering to economics, where surfaces and their tangent planes can represent many physical phenomena or optimization problems.

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Most popular questions from this chapter

Use Lagrange multipliers to find the minimum distance from the curve or surface to the indicated point. [Hint: In Exercise 23, minimize \(f(x, y)=x^{2}+y^{2}\) subject to the constraint \(2 x+3 y=-1 .]\) $$ \begin{array}{ll} \underline{\text { Curve }} & \underline{\text {Point}} \\ \text { Line: } 2 x+3 y=-1 \quad (0,0) \end{array} $$

Find the angle of inclination \(\theta\) of the tangent plane to the surface at the given point.\(3 x^{2}+2 y^{2}-z=15, \quad(2,2,5)\)

The endpoints of the interval over which distinct vision is possible are called the near point and far point of the eye. With increasing age, these points normally change. The table shows the approximate near points \(y\) in inches for various ages \(x\) (in years). $$ \begin{array}{|l|c|c|c|c|c|} \hline \text { Age, } x & 16 & 32 & 44 & 50 & 60 \\ \hline \text { Near Point, } y & 3.0 & 4.7 & 9.8 & 19.7 & 39.4 \\ \hline \end{array} $$ (a) Find a rational model for the data by taking the reciprocal of the near points to generate the points \((x, 1 / y)\). Use the regression capabilities of a graphing utility to find a least squares regression line for the revised data. The resulting line has the form \(\frac{1}{y}=a x+b\) Solve for \(y\). (b) Use a graphing utility to plot the data and graph the model. (c) Do you think the model can be used to predict the near point for a person who is 70 years old? Explain.

Prove that \(\lim _{y y \rightarrow(a, b)}[f(x, y)+g(x, y)]=L_{1}+L_{2}\) where \(f(x, y)\) approaches \(L_{1}\) and \(g(x, y)\) approaches \(L_{2}\) as \((x, y) \rightarrow(a, b) .\)

Find the absolute extrema of the function over the region \(R .\) (In each case, \(R\) contains the boundaries.) Use a computer algebra system to confirm your results. \(f(x, y)=\frac{4 x y}{\left(x^{2}+1\right)\left(y^{2}+1\right)}\) \(R=\left\\{(x, y): x \geq 0, y \geq 0, x^{2}+y^{2} \leq 1\right\\}\)

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