/*! 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 47 Find an equation of the plane ta... [FREE SOLUTION] | 91Ó°ÊÓ

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Find an equation of the plane tangent to the following surfaces at the given point. $$z=\tan ^{-1}(x y) ;(1,1, \pi / 4)$$

Short Answer

Expert verified
Answer: The equation of the tangent plane at the given point is $$\frac{1}{2}x + \frac{1}{2}y - z = -\frac{\pi}{4}+\frac{3}{2}$$

Step by step solution

01

Find the partial derivatives

To determine the gradient of the surface at the given point, we need to find the partial derivatives of the function with respect to x and y. The function is given by $$z = \tan^{-1}(xy)$$. Let's find the partial derivatives: $$\frac{\partial z}{\partial x} = \frac{y}{1+(xy)^2}$$ $$\frac{\partial z}{\partial y} = \frac{x}{1+(xy)^2}$$
02

Evaluate the partial derivatives at the given point

Now, let's evaluate these partial derivatives at the given point \((1,1,\pi / 4)\): $$\frac{\partial z}{\partial x} \Big|_{(1,1,\pi/4)} = \frac{1}{1+(1\cdot1)^2} = \frac{1}{2}$$ $$\frac{\partial z}{\partial y} \Big|_{(1,1,\pi/4)} = \frac{1}{1+(1\cdot1)^2} = \frac{1}{2}$$ The gradient at the point \((1,1,\pi / 4)\) is \(\left( \frac{1}{2}, \frac{1}{2} \right)\).
03

Write the equation of the tangent plane

The equation of the tangent plane at the point \((1,1,\pi / 4)\) is given by: $$\frac{\partial z}{\partial x}(x-1) + \frac{\partial z}{\partial y}(y-1) = z - \frac{\pi}{4}$$ Plugging in the values we found in Step 2: $$\frac{1}{2}(x-1) + \frac{1}{2}(y-1) = z - \frac{\pi}{4}$$ Rearrange the equation to make it look like a standard plane equation (in the form of \(Ax + By + Cz = D\)): $$\frac{1}{2}x + \frac{1}{2}y - z = -\frac{\pi}{4}+\frac{3}{2}$$ This is the equation of the plane tangent to the surface $$z = \tan^{-1}(xy)$$ at the point \((1,1,\pi/4)\).

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Key Concepts

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

Partial Derivatives
Partial derivatives are fundamental in understanding how a surface changes in different directions. When dealing with a function of two variables, like \( z = \tan^{-1}(xy) \), you take partial derivatives to see how changes in \(x\) or \(y\) affect \(z\) independently.
To find the partial derivatives, we hold one variable constant while differentiating with respect to the other. Here, the partial derivative with respect to \(x\), denoted \( \frac{\partial z}{\partial x} \), assumes \(y\) is constant. Similarly, \( \frac{\partial z}{\partial y} \) considers \(x\) constant. \(
\)For our surface, \( \frac{\partial z}{\partial x} = \frac{y}{1+(xy)^2} \) and \( \frac{\partial z}{\partial y} = \frac{x}{1+(xy)^2} \). These provide the rates of change in the surface’s height in the \(x\) and \(y\) directions, respectively.
Gradient of a Surface
The gradient of a surface at a point is a vector that combines the partial derivatives. It points in the direction of the steepest ascent on the surface and its magnitude represents how steep this ascent is.
For the function \( z = \tan^{-1}(xy) \), the gradient at a point \((x, y)\) is expressed as the vector \( abla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right) \).
At the point \((1,1)\), both partial derivatives evaluated to \( \frac{1}{2} \). This gives the gradient vector \( \left( \frac{1}{2}, \frac{1}{2} \right) \) at the point \((1,1, \pi/4)\). This vector tells us that the surface rises equally in the \(x\) and \(y\) directions at this point.
Surface Equation
The surface equation describes a surface in three-dimensional space. In this context, the surface is given by \( z = \tan^{-1}(xy) \).
We are interested in finding the equation of the tangent plane to this surface at a specific point \((1,1, \pi/4)\). A tangent plane gives us the linear approximation of the surface near the point. It touches the surface at exactly one point without cutting through it.
To find this plane, we use the partial derivatives at the point, as these are crucial for constructing the equation. The general tangent plane formula is:
  • \( \frac{\partial z}{\partial x} (x - x_0) + \frac{\partial z}{\partial y} (y - y_0) = z - z_0 \)
For our surface, substituting the values gives us:
  • \( \frac{1}{2}(x-1) + \frac{1}{2}(y-1) = z - \frac{\pi}{4} \)
This simplifies to \( \frac{1}{2}x + \frac{1}{2}y - z = -\frac{\pi}{4} + \frac{3}{2} \), which is the precise equation of our tangent plane.

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Most popular questions from this chapter

Identify and briefly describe the surfaces defined by the following equations. $$9 x^{2}+y^{2}-4 z^{2}+2 y=0$$

Suppose \(n\) houses are located at the distinct points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right) .\) A power substation must be located at a point such that the sum of the squares of the distances between the houses and the substation is minimized. a. Find the optimal location of the substation in the case that \(n=3\) and the houses are located at \((0,0),(2,0),\) and (1,1) b. Find the optimal location of the substation in the case that \(n=3\) and the houses are located at distinct points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right),\) and \(\left(x_{3}, y_{3}\right)\) c. Find the optimal location of the substation in the general case of \(n\) houses located at distinct points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots\) \(\left(x_{n}, y_{n}\right)\) d. You might argue that the locations found in parts (a), (b), and (c) are not optimal because they result from minimizing the sum of the squares of the distances, not the sum of the distances themselves. Use the locations in part (a) and write the function that gives the sum of the distances. Note that minimizing this function is much more difficult than in part (a). Then use a graphing utility to determine whether the optimal location is the same in the two cases. (Also see Exercise 75 about Steiner's problem.)

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