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Chapter 11: Problem 12

Find and sketch the domain of the function. $$f(x, y, z)=\ln \left(16-4 x^{2}-4 y^{2}-z^{2}\right)$$

Short Answer

Expert verified
The domain is the interior of an ellipsoid: \(\frac{x^2}{4} + \frac{y^2}{4} + \frac{z^2}{16} < 1\).

Step by step solution

01

Understanding the Domain of a Logarithmic Function

The function given is a logarithmic function:\[f(x, y, z) = \ln(16 - 4x^2 - 4y^2 - z^2)\]The domain of a logarithmic function includes all input values for which the argument (inside of the log) is positive. This means we need to determine where the expression inside the logarithm is greater than zero.
02

Setting the Argument Greater Than Zero

To satisfy the condition for the domain, we set:\[16 - 4x^2 - 4y^2 - z^2 > 0\]We will simplify this inequality to find the values of \((x, y, z)\) that satisfy it.
03

Rearranging the Inequality

Rearrange the inequality as follows:\[4x^2 + 4y^2 + z^2 < 16\]Divide each term by 16 to simplify further:\[\frac{x^2}{4} + \frac{y^2}{4} + \frac{z^2}{16} < 1\]This is the inequality that describes an ellipsoid.
04

Identifying the Domain as an Ellipsoid

The inequality \(\frac{x^2}{4} + \frac{y^2}{4} + \frac{z^2}{16} < 1\) represents the interior of an ellipsoid centered at the origin \((0, 0, 0)\). This ellipsoid is elongated along the z-axis because the coefficient of \(z^2\) is smallest.
05

Sketching the Domain Ellipsoid

The ellipsoid's domain can be sketched in a 3D coordinate system:- Along the x-axis and y-axis, the limits are \(-2 \leq x \leq 2\) and \(-2 \leq y \leq 2\) respectively.- Along the z-axis, the limits extend from \(-4 \leq z \leq 4\). Sketch this elongated sphere in the 3D coordinate system.

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Key Concepts

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

Domain of a Function
In multivariable calculus, understanding the domain of a function is crucial as it tells us which input values are allowed. The domain is essentially the set of all possible inputs
  • For which the function is defined.
  • It ensures that calculations involving the function do not run into undefined situations.
When dealing with functions that are more complex, such as those involving variables in multiple dimensions, it's important to identify these input values carefully. In the context of a function involving real numbers, the domain will often be a particular region or shape in three-dimensional space. To solve for the domain of a function, identify the conditions that the variables must satisfy for the function to operate without errors.
Logarithmic Functions
Logarithmic functions like \[ f(x, y, z) = \ln(16 - 4x^2 - 4y^2 - z^2) \] require specific conditions to be met.
  • The argument, which is the expression inside the logarithm, must be positive.
Here's why: \( \ln(x) \) is only defined for \( x > 0 \). For our function, this means solving the inequality \[ 16 - 4x^2 - 4y^2 - z^2 > 0 \] to ensure the argument inside the logarithm remains positive for all values of \( x, y, \) and \( z \). Break this down to understand:
  • We derive an inequality which directs our understanding to the shape and size of regions where the function is defined.
  • In simpler terms, the graph of the function will only exist where these conditions are met.
Ellipsoid
The solution to the inequality \[ \frac{x^2}{4} + \frac{y^2}{4} + \frac{z^2}{16} < 1 \] defines a shape known as an ellipsoid. Think of it as a stretched sphere.
  • Just like an egg, it has different dimensions along different axes.
  • The ellipsoid is centered at the origin (0, 0, 0) and stretches further along one axis compared to others.
In this case:
  • The x-axis and y-axis have the same length, extending from -2 to 2.
  • The z-axis is longer, ranging from -4 to 4, making the ellipsoid elongated along the z-axis.
This ellipsoid represents all the points where the original function is defined, providing us the complete domain in three-dimensional space.

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Most popular questions from this chapter

Show that the ellipsoid \(3 x^{2}+2 y^{2}+z^{2}=9\) and the sphere \(x^{2}+y^{2}+z^{2}-8 x-6 y-8 z+24=0\) are tangent to each other at the point \((1,1,2) .\) (This means that they have a common tangent plane at the point.)

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. $$ f(x, y)=y^{2}-x^{2} ; \quad \frac{1}{4} x^{2}+y^{2}=1 $$

Suppose that a scientist has reason to believe that two quantities \(x\) and \(y\) are related linearly, that is, \(y=m x+b,\) at least approximately, for some values of \(m\) and \(b\) . The scientist performs an experiment and collects data in the form of points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) and then plots these points. The points don't lie exactly on a straight line, so the scientist wants to find constants \(m\) and \(b\) so that the line \(y=m x+b\) "fits" the points as well as possible (see the figure). Let \(d_{i}=y_{i}-\left(m x_{i}+b\right)\) be the vertical deviation of the point \(\left(x_{i}, y_{i}\right)\) from the line. The method of least squares determines \(m\) and \(b\) so as to minimize \(\Sigma_{i-1}^{n} d_{i}^{2},\) the sum of the squares of these deviations. Show that, according to this method, the line of best fit is obtained when $$\begin{array}{c}{m \sum_{i=1}^{n} x_{i}+b n=\sum_{i=1}^{n} y_{i}} \\ {m \sum_{i=1}^{n} x_{i}^{2}+b \sum_{l=1}^{n} x_{i}=\sum_{i=1}^{n} x_{i} y_{i}}\end{array}$$ Thus the line is found by solving these two equations in the two unknowns \(m\) and \(b .\)

Find the extreme values of \(f\) on the region described by the inequality. $$ f(x, y)=e^{-x y}, \quad x^{2}+4 y^{2} \leq 1 $$

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. $$ f(x, y, z)=x^{2}+y^{2}+z^{2} ; \quad x+y+z=12 $$
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