/*! 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 73 Consider the following surfaces ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 73

Consider the following surfaces \(f(x, y, z)=0,\) which may be regarded as a level surface of the function \(w=f(x, y, z) . A\) point \(P(a, b, c)\) on the surface is also given. a. Find the (three-dimensional) gradient of \(f\) and evaluate it at \(P\). b. The set of all vectors orthogonal to the gradient with their tails at \(P\) form a plane. Find an equation of that plane (soon to be called the tangent plane). $$f(x, y, z)=e^{x+y-z}-1=0 ; P(1,1,2)$$

Short Answer

Expert verified
Answer: The equation of the tangent plane at point \(P\) is \(x + y - z = 0\).

Step by step solution

01

Find the gradient of f(x, y, z)

To find the gradient of the function \(f(x, y, z)=e^{x+y-z}-1\), we need to find the partial derivatives with respect to each variable. The gradient vector \(\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\). Calculate the partial derivatives: $$\frac{\partial f}{\partial x}=e^{x+y-z}$$ $$\frac{\partial f}{\partial y}=e^{x+y-z}$$ $$\frac{\partial f}{\partial z}=-e^{x+y-z}$$ So, the gradient of \(f\) is: $$\nabla f = (e^{x+y-z}, e^{x+y-z}, -e^{x+y-z})$$
02

Evaluate the gradient at point P

Now, we will evaluate the gradient at point \(P(1, 1, 2)\). $$\nabla f(1, 1, 2) = (e^{1+1-2}, e^{1+1-2}, -e^{1+1-2}) = (1, 1, -1)$$
03

Find the equation of the tangent plane at P

The gradient at the point \(P\), \(\nabla f(1, 1, 2) = (1, 1, -1)\), is normal to the tangent plane. Thus, we can use the point-normal form for the equation of the tangent plane: $$(x - a, y - b, z - c) \cdot (1, 1, -1) = 0$$ Where \((x-a, y-b, z-c)\) is a position vector on the plane and \((1, 1, -1)\) is the normal vector. The point P \((1, 1, 2)\) is on the plane, so we can substitute its coordinates and obtain the equation: $$(x - 1, y - 1, z - 2) \cdot (1, 1, -1) = 0$$ Multiply the vectors and simplify: $$(x - 1)(1) + (y - 1)(1) + (z - 2)(-1) = 1(x-1) + 1(y-1) - 1(z-2) = 0$$ Thus, the equation of the tangent plane at point \(P\) is: $$x + y - z = 0$$

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Key Concepts

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

Gradient Vector
The gradient vector is a crucial concept in multivariable calculus. It acts much like a compass, pointing in the direction of the greatest rate of increase of a function. For a function of three variables, like our example function \( f(x, y, z) = e^{x+y-z} - 1 \), the gradient is a vector consisting of the function's partial derivatives:
  • \( \frac{\partial f}{\partial x} = e^{x+y-z} \)
  • \( \frac{\partial f}{\partial y} = e^{x+y-z} \)
  • \( \frac{\partial f}{\partial z} = -e^{x+y-z} \)
The gradient vector, denoted \( abla f \), is thus \( (e^{x+y-z}, e^{x+y-z}, -e^{x+y-z}) \).
At any point, such as \( P(1, 1, 2) \), substituting the coordinates into the gradient gives the vector \( (1, 1, -1) \).
This vector is very informative: beyond pointing to the steepest ascent, it is perpendicular to level surfaces like our given surface \( f(x, y, z) = 0 \).
Tangent Plane
A tangent plane to a surface at a given point is an important geometric concept. It can be understood as the best linear approximation of the surface at that point.
To find the equation of a tangent plane, we use the property that it is perpendicular to the gradient vector.
Given a point \( P(a,b,c) \) on the surface and its gradient \( abla f(a,b,c) \), the equation of the tangent plane can be expressed as:
  • \( (x-a, y-b, z-c) \cdot abla f(a,b,c) = 0 \)
The dot product here represents how aligned the position vector on the plane is with the gradient vector.
In our example, substituting the point \( P(1,1,2) \) and \( abla f(1,1,2) = (1,1,-1) \), we compute:
\[ (x-1)(1) + (y-1)(1) - (z-2)(-1) = 0 \]Simplifying this, the equation \( x + y - z = 0 \), describes the tangent plane at \( P \).
Partial Derivatives
Partial derivatives give us the rate of change of a function with respect to one variable while keeping others constant.
They are fundamental in defining the gradient vector and understanding changes within multivariable functions.
When finding partial derivatives for a function \( f(x, y, z) \), you treat all variables other than the one you're differentiating with respect to as constants.
In our function \( f(x, y, z) = e^{x+y-z} - 1 \), this gives:
  • For \( x \), \( \frac{\partial f}{\partial x} = e^{x+y-z} \)
  • For \( y \), \( \frac{\partial f}{\partial y} = e^{x+y-z} \)
  • For \( z \), \( \frac{\partial f}{\partial z} = -e^{x+y-z} \), given the negative sign due to \( f(x, y, z) \)'s composition.
These derivatives highlight how the function behaves along each axis.
Combining them into the gradient gives us a comprehensive peek into the function's overall variability. Thus, recognizing and interpreting partial derivatives is key in handling complex multivariable calculus problems.

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Most popular questions from this chapter

Given the production function \(P=f(K, L)=K^{a} L^{1-a}\) and the budget constraint \(p K+q L=B,\) where \(a, p, q,\) and \(B\) are given, show that \(P\) is maximized when \(K=a B / p\) and \(L=(1-a) B / q\).

Given positive numbers \(x_{1}, \ldots, x_{n},\) prove that the geometric mean \(\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n}\) is no greater than the arithmetic mean \(\left(x_{1}+\cdots+x_{n}\right) / n\) in the following cases. a. Find the maximum value of \(x y z,\) subject to \(x+y+z=k\) where \(k\) is a real number and \(x>0, y>0\), and \(z>0 .\) Use the result to prove that $$(x y z)^{1 / 3} \leq \frac{x+y+z}{3}.$$ b. Generalize part (a) and show that $$\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n} \leq \frac{x_{1}+\cdots+x_{n}}{n}.$$

Find the dimensions of the rectangular box with maximum volume in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid \(36 x^{2}+4 y^{2}+9 z^{2}=36\).

Let \(w=f(x, y, z)=2 x+3 y+4 z\) which is defined for all \((x, y, z)\) in \(\mathbb{R}^{3}\). Suppose that we are interested in the partial derivative \(w_{x}\) on a subset of \(\mathbb{R}^{3}\), such as the plane \(P\) given by \(z=4 x-2 y .\) The point to be made is that the result is not unique unless we specify which variables are considered independent. a. We could proceed as follows. On the plane \(P\), consider \(x\) and \(y\) as the independent variables, which means \(z\) depends on \(x\) and \(y,\) so we write \(w=f(x, y, z(x, y)) .\) Differentiate with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial w}{\partial x}\right)_{y}=18,\) where the subscript \(y\) indicates that \(y\) is held fixed. b. Alternatively, on the plane \(P,\) we could consider \(x\) and \(z\) as the independent variables, which means \(y\) depends on \(x\) and \(z,\) so we write \(w=f(x, y(x, z), z)\) and differentiate with respect to \(x\) holding \(z\) fixed. Show that \(\left(\frac{\partial w}{\partial x}\right)_{z}=8,\) where the subscript \(z\) indicates that \(z\) is held fixed. c. Make a sketch of the plane \(z=4 x-2 y\) and interpret the results of parts (a) and (b) geometrically. d. Repeat the arguments of parts (a) and (b) to find \(\left(\frac{\partial w}{\partial y}\right)_{x}\) \(\left(\frac{\partial w}{\partial y}\right)_{z},\left(\frac{\partial w}{\partial z}\right)_{x},\) and \(\left(\frac{\partial w}{\partial z}\right)_{y}\)

Given a differentiable function \(w=f(x, y, z),\) the goal is to find its maximum and minimum values subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0\) and \(h(x, y, z)=0\).
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