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Chapter 12: Problem 72

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). $$f(x, y, z)=8-x y z=0 ; P(2,2,2)$$

Short Answer

Expert verified
Question: Find the equation of the tangent plane to the function f(x, y, z) = 8 - xyz at the point P(2, 2, 2). Solution: After finding the gradient of the function and evaluating it at the given point, we determined the equation of the tangent plane to be x + y + z = 6.

Step by step solution

01

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

Calculate the partial derivatives of the function f(x, y, z) with respect to x, y, and z. The gradient vector will be the vector of these partial derivatives: $$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)$$ For the given function f(x, y, z) = 8 - xyz, calculate the partial derivatives: $$\frac{\partial f}{\partial x} = -yz, \frac{\partial f}{\partial y} = -xz, \frac{\partial f}{\partial z} = -xy$$ So, the gradient of the function f(x, y, z) is: $$\nabla f = (-yz, -xz, -xy)$$
02

Evaluate the gradient at point P(a, b, c)

Substitute the given point P(2, 2, 2) into the gradient vector: $$\nabla f_P = (-2(2), -2(2), -2(2)) = (-4, -4, -4)$$
03

Find the equation of the plane orthogonal to the gradient vector with its tail at point P

Now, since the normal vector to the plane is given by the gradient vector, and the plane passes through the point P, we use the general formula for the equation of a plane, which is: $$A(x - a) + B(y - b) + C(z - c) = 0$$ Substitute the components of the gradient vector for A, B, and C, and the coordinates of point P for a, b, and c: $$-4(x - 2) + (-4)(y - 2) + (-4)(z - 2) = 0$$ Simplify the above equation to obtain the equation of the tangent plane: $$-4x + 8 - 4y + 8 - 4z + 8 = 0$$ $$-4x -4y - 4z + 24 = 0$$ $$x + y + z = 6$$

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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 central concept in multivariable calculus that points in the direction of the greatest rate of increase of the function. For a function with three variables, say, w=f(x, y, z), the gradient is a three-dimensional vector given by the partial derivatives of the function with respect to each variable. We write this as:

abla f = \(\bigg(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\bigg)\).

The gradient can tell us a lot about a function's behavior. At the point P(a, b, c), evaluating the gradient gives us specific information on how the function changes if we move away from this point. If the calculated gradient vector at P is nonzero, it will always be perpendicular to the level surface of the function at P. This perpendicular property is used to define the tangent plane to the surface at P.

In the given exercise, the gradient of f(x, y, z) = 8 - xyz at point P(2, 2, 2) is determined to be (-4, -4, -4). This vector not only characterizes the steepest ascension path at P, but also remarkably simplifies the search for the equation of the tangent plane.
Tangent Plane
The concept of a tangent plane might be familiar from single-variable calculus, where the tangent line touches a curve at a point without crossing it. Extending this to functions of three variables, the tangent plane does a similar job, touching a surface at a single point. This plane can be understood as the best linear 'approximation' of the surface near that point.

The equation for the tangent plane to a surface defined by w=f(x, y, z) at a given point P(a, b, c) can be found using the gradient vector \(abla f_P\). Since this vector is normal to the plane, the general equation is:\(A(x - a) + B(y - b) + C(z - c) = 0\), where A, B, and C are the components of the gradient vector \(abla f_P\) and (a, b, c) are the coordinates of the point P.

Using the gradient calculated in the exercise, the equation for the tangent plane at P(2, 2, 2) on the surface f(x, y, z) = 8 - xyz turns out to be x + y + z = 6. This equation is crucial, as it defines a flat surface tangential to the curved surface at point P and orthogonal to the gradient vector.
Partial Derivatives
Understanding partial derivatives is essential when dealing with functions of multiple variables. They measure how a function changes as only one variable changes while the others remain fixed. Partial derivatives are the building blocks of the gradient vector and several key concepts in multivariable calculus, such as directional derivatives and tangent planes.

For a function f(x, y, z), the partial derivatives with respect to x, y, and z are denoted by \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\), respectively. Calculating these derivatives involves differentiating f with respect to one variable while treating the other variables as constants.

In our exercise, the partial derivatives of f(x, y, z) = 8 - xyz with respect to x, y, and z are -yz, -xz, and -xy, respectively. These are essential in defining the slope of the function in the direction of each variable and play a critical role in finding the gradient vector necessary for determining the tangent plane at a specific point.

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Most popular questions from this chapter

a. Show that the point in the plane \(a x+b y+c z=d\) nearest the origin is \(P\left(a d / D^{2}, b d / D^{2}, c d / D^{2}\right),\) where \(D^{2}=a^{2}+b^{2}+c^{2} .\) Conclude that the least distance from the plane to the origin is \(|d| / D\). (Hint: The least distance is along a normal to the plane.) b. Show that the least distance from the point \(P_{0}\left(x_{0}, y_{0}, z_{0}\right)\) to the plane \(a x+b y+c z=d\) is \(\left|a x_{0}+b y_{0}+c z_{0}-d\right| / D\) (Hint: Find the point \(P\) on the plane closest to \(P_{0}\).)

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but that the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) b. \(f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}}\) This property has the following interpretation. Suppose that a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum. (Source: Mathematics Magazine, May 1985, and Calculus and Analytical Geometry, 2nd ed., Philip Gillett, 1984)

Recall that Cartesian and polar coordinates are related through the transformation equations $$\left\\{\begin{array}{l} x=r \cos \theta \\ y=r \sin \theta \end{array} \quad \text { or } \quad\left\\{\begin{array}{l} r^{2}=x^{2}+y^{2} \\ \tan \theta=y / x \end{array}\right.\right.$$ a. Evaluate the partial derivatives \(x_{r}, y_{r}, x_{\theta},\) and \(y_{\theta}\) b. Evaluate the partial derivatives \(r_{x}, r_{y}, \theta_{x},\) and \(\theta_{y}\) c. For a function \(z=f(x, y),\) find \(z_{r}\) and \(z_{\theta},\) where \(x\) and \(y\) are expressed in terms of \(r\) and \(\theta\) d. For a function \(z=g(r, \theta),\) find \(z_{x}\) and \(z_{y},\) where \(r\) and \(\theta\) are expressed in terms of \(x\) and \(y\) e. Show that \(\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}=\left(\frac{\partial z}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial z}{\partial \theta}\right)^{2}\)

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 \(\left(\frac{\partial w}{\partial z}\right)_{y}\)

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=e^{a x} \cos a y, \text { for any real number } a$$
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