Domains & Level Sets - Multivariable Calculus

For the function $f(x,y) = e^{-\sqrt{4-x^2-y^2}}$ , the level set $f(x,y)=c$ is a circle for certain values of $c$ . For which of the following values of $c$ is the level set a circle?
Answer: $c = \frac{1}{e}$ First, determine the range of $f$ . The domain is $4-x^2-y^2 \ge 0$ , or $x^2+y^2 \le 4$ . The value of $u = \sqrt{4-x^2-y^2}$ ranges from $0$ (when $x^2+y^2=4$ ) to $2$ (when $x^2+y^2=0$ ). Thus, the range of $f(x,y) = e^{-u}$ is $[e^{-2}, e^0] = [1/e^2, 1]$ . A level set can only exist for $c$ in this range. Now, set $f(x,y)=c$ : $e^{-\sqrt{4-x^2-y^2}} = c$ . This gives $-\sqrt{4-x^2-y^2} = \ln c$ , so $\sqrt{4-x^2-y^2} = -\ln c = \ln(1/c)$ . Squaring both sides yields $4-x^2-y^2 = (\ln(1/c))^2$ , or $x^2+y^2 = 4 - (\ln(1/c))^2$ . For this to be a circle, the radius squared must be positive: $4 - (\ln(1/c))^2 > 0$ . This means $(\ln(1/c))^2 < 4$ , or $|\ln(1/c)| < 2$ . This implies $-2 < \ln(1/c) < 2$ , so $e^{-2} < 1/c < e^2$ , which means $1/e^2 < c < e^2$ . We need a value of $c$ that is in both the range $[1/e^2, 1]$ and the interval $(1/e^2, e^2)$ . The intersection is $(1/e^2, 1]$ . Of the choices, only $c = 1/e$ is in this interval. For $c=1/e^2$ , the radius is 0, which is a point, not a circle.
The function $p(x,y) = \tan^{-1}\left(\frac{y}{x}\right) + \tan^{-1}\left(\frac{x}{y}\right)$ has a domain that excludes:
Answer: Only the coordinate axes where $x = 0$ or $y = 0$ The function p(x,y) = arctan(y/x) + arctan(x/y) is undefined when either denominator is zero. This occurs when x = 0 or y = 0, which are the coordinate axes. The function is defined in all four quadrants where both x ≠ 0 and y ≠ 0. Note that arctan(y/x) is undefined when x = 0, and arctan(x/y) is undefined when y = 0. Choice A incorrectly restricts to positive products, choice C only excludes the origin, and choice D incorrectly excludes entire quadrants.
Consider the function $r(x,y,z) = \frac{1}{\sqrt{(x-1)^2 + (y-2)^2 + (z-3)^2 - 9}}$ . The boundary of the domain of this function is:
Answer: A sphere of radius $3$ centered at $(1,2,3)$ When analyzing the domain of a function with a square root in the denominator, you need to determine where the expression is both positive (for the square root) and non-zero (to avoid division by zero). The function is undefined where the denominator equals zero or becomes imaginary. For $r(x,y,z)$ to be defined, the expression under the square root must be strictly positive: $(x-1)^2 + (y-2)^2 + (z-3)^2 - 9 > 0$ . This can be rewritten as $(x-1)^2 + (y-2)^2 + (z-3)^2 > 9$ . The boundary of the domain occurs where the inequality becomes an equality: $(x-1)^2 + (y-2)^2 + (z-3)^2 = 9$ . This is the standard form of a sphere with center $(1,2,3)$ and radius $\sqrt{9} = 3$ . The function is defined outside this sphere but undefined on and inside it. Looking at the wrong answers: A incorrectly places the center at the origin instead of $(1,2,3)$ . B uses radius $9$ instead of $\sqrt{9} = 3$ - this confuses the value under the square root with the actual radius. D has the wrong center $(-1,-2,-3)$ , likely from a sign error, though it correctly calculates the radius as $\sqrt{9}$ . The correct answer is C: a sphere of radius $3$ centered at $(1,2,3)$ . Study tip: For domain problems involving square roots in denominators, always set the expression equal to zero to find the boundary, and remember that the radius of a sphere equation $(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2$ is $r$ , not $r^2$ .
For the function $g(x,y,z) = \frac{xyz}{x^2+y^2+z^2-1}$ , which statement about its domain is correct?
Answer: The domain includes all points except those on the unit sphere $x^2+y^2+z^2 = 1$ When finding the domain of a multivariable function, you need to identify where the function is undefined. For rational functions like this one, the critical issue is where the denominator equals zero. The function $g(x,y,z) = \frac{xyz}{x^2+y^2+z^2-1}$ is undefined when its denominator equals zero. Setting $x^2+y^2+z^2-1 = 0$ gives us $x^2+y^2+z^2 = 1$ , which is the equation of a unit sphere centered at the origin. The function is defined everywhere except on this sphere, making answer D correct. Let's see why the other options are wrong. Option A suggests the function is undefined only at the origin $(0,0,0)$ . However, substituting the origin into the denominator gives $0^2+0^2+0^2-1 = -1 \neq 0$ , so the function is actually defined at the origin. Option B claims the domain is only the interior of the unit ball ( $x^2+y^2+z^2 < 1$ ). This excludes points outside the unit sphere where the function is perfectly well-defined. Option C makes the opposite error, claiming the domain is only the exterior ( $x^2+y^2+z^2 > 1$ ), which incorrectly excludes the interior points where the function is also defined. Remember: for rational functions, always set the denominator equal to zero to find excluded points. The domain includes all other points in the appropriate space (here, all of $\mathbb{R}^3$ except the unit sphere).
The domain of the function $f(x, y) = \frac{1}{\cos(\pi x) - 1}$ is the entire $xy$ -plane with a set of points removed. What is the geometric description of the removed set?
Answer: A set of vertical lines of the form $x=2k$ for any integer $k$ . The function is defined everywhere except where the denominator is zero. The denominator is $\cos(\pi x) - 1 = 0$ , which means $\cos(\pi x) = 1$ . The cosine function equals 1 when its argument is an even multiple of $\pi$ . So, we must have $\pi x = 2k\pi$ for any integer $k$ . Dividing by $\pi$ gives $x = 2k$ . This condition depends only on $x$ ; the value of $y$ can be anything. Therefore, the set of points removed from the domain is the set of all points $(x,y)$ where $x$ is an even integer. This corresponds to an infinite set of vertical lines of the form $x=..., -4, -2, 0, 2, 4, ...$ .
Consider the function $f(x,y) = \frac{\ln(x+y-1)}{\sqrt{4-x^2-y^2}}$ . Which of the following best describes the domain of $f$ ?
Answer: The interior of the disk $x^2 + y^2 < 4$ where $x + y > 1$ When finding the domain of a multivariable function, you need to identify where each component is mathematically defined and then find their intersection. For $f(x,y) = \frac{\ln(x+y-1)}{\sqrt{4-x^2-y^2}}$ , examine each part separately. The natural logarithm $\ln(x+y-1)$ requires its argument to be strictly positive, so you need $x+y-1 > 0$ , which means $x+y > 1$ . The denominator $\sqrt{4-x^2-y^2}$ requires the expression under the square root to be strictly positive (since it's in the denominator, it cannot equal zero). This gives you $4-x^2-y^2 > 0$ , or equivalently $x^2+y^2 < 4$ . The domain is where both conditions hold simultaneously: the interior of the disk $x^2+y^2 < 4$ intersected with the half-plane $x+y > 1$ . Choice A incorrectly includes the boundary circle $x^2+y^2 = 4$ , but this would make the denominator zero (undefined). Choice B has two errors: it includes the boundary circle and uses $x+y \geq 1$ , but the logarithm requires $x+y > 1$ strictly. Choice C correctly excludes the boundary circle but redundantly states this exclusion - the main issue is that it doesn't clearly specify that we want the interior of the disk. Study tip: For domain problems, systematically check each function component's requirements, paying special attention to whether inequalities are strict or non-strict, especially when expressions appear in denominators or inside logarithms.
What is the level set of the function $f(x,y) = \frac{(x-y)^2}{x^2+y^2}$ corresponding to the level $c=0$ ?
Answer: The line $y=x$ , with the origin excluded. The level set for $c=0$ is the set of all points $(x,y)$ such that $f(x,y) = 0$ . This requires the numerator to be zero and the denominator to be non-zero. The numerator is $(x-y)^2 = 0$ , which implies $x-y=0$ , or $y=x$ . The denominator is $x^2+y^2$ . For the function to be defined, $x^2+y^2 \ne 0$ , which means $(x,y)$ cannot be the origin $(0,0)$ . Therefore, the level set is the line $y=x$ with the point $(0,0)$ removed.
Consider the function $f(x, y) = \ln(x^2 + y^2)$ . Which statement best describes the spacing of its level curves $f(x,y)=c$ for equally spaced values of $c$ ?
Answer: The level curves are circles that become more sparsely spaced as $c$ increases. The level curves are given by $\ln(x^2 + y^2) = c$ . Exponentiating both sides gives $x^2 + y^2 = e^c$ . These are concentric circles centered at the origin with radius $r(c) = \sqrt{e^c} = e^{c/2}$ . To analyze the spacing, consider equally spaced values of $c$ , such as $c, c+1, c+2, \dots$ . The radii of the corresponding circles are $e^{c/2}, e^{(c+1)/2}, e^{(c+2)/2}, \dots$ . The distance between two consecutive curves is the difference in their radii: $\Delta r = r(c+1) - r(c) = e^{(c+1)/2} - e^{c/2} = e^{c/2}(e^{1/2} - 1)$ . Since $e^{1/2} - 1$ is a positive constant and $e^{c/2}$ is an increasing function of $c$ , the distance $\Delta r$ increases as $c$ increases. Therefore, the level curves become more sparsely spaced.
Which statement accurately describes the level surfaces $f(x, y, z) = c$ of the function $f(x, y, z) = x^2 + y^2 - z^2$ ?
Answer: They are hyperboloids of one sheet for $c > 0$ , a cone for $c = 0$ , and hyperboloids of two sheets for $c < 0$ . The level surfaces are given by the equation $x^2 + y^2 - z^2 = c$ . This is the standard form for a family of quadric surfaces. When $c > 0$ , we have a hyperboloid of one sheet. For example, if $c=1$ , the surface is connected and opens along the $z$ -axis. When $c = 0$ , the equation is $x^2 + y^2 = z^2$ , which is the equation of a double cone with its vertex at the origin. When $c < 0$ , let $c = -k^2$ for some $k>0$ . The equation becomes $x^2 + y^2 - z^2 = -k^2$ , or $z^2 - x^2 - y^2 = k^2$ , which is the equation of a hyperboloid of two sheets.
Let $D$ be the domain of the function $f(x, y) = \frac{\ln(y - x^2)}{\sqrt{9 - x^2 - y^2}}$ . Which of the following provides the most accurate geometric description of $D$ ?
Answer: The set of points inside the circle $x^2 + y^2 = 9$ and above the parabola $y = x^2$ . The domain of the function is constrained by two conditions. First, the argument of the natural logarithm must be positive, so $y - x^2 > 0$ , which means $y > x^2$ . This describes the region above the parabola $y = x^2$ . Second, the argument of the square root in the denominator must be positive, so $9 - x^2 - y^2 > 0$ , which means $x^2 + y^2 < 9$ . This describes the region inside the circle of radius 3 centered at the origin. The domain $D$ is the intersection of these two regions: the set of points that are both inside the circle and above the parabola.
Consider the function $f(x, y) = e^{y - x^2}$ . Which of the following statements correctly describes the level curves of this function?
Answer: A family of parabolas opening upward, with vertices moving up the $y$ -axis as the level value increases. A level curve is defined by the equation $f(x, y) = c$ for some constant $c$ . Since the range of the exponential function is positive numbers, we must have $c > 0$ . The equation is $e^{y - x^2} = c$ . Taking the natural logarithm of both sides gives $y - x^2 = \ln(c)$ , which can be rewritten as $y = x^2 + \ln(c)$ . This is the equation of a family of parabolas opening upward. The vertex of each parabola is at $(0, \ln(c))$ . As the level value $c$ increases, $\ln(c)$ also increases, so the vertex moves up the $y$ -axis.
The temperature on a rectangular plate defined by $|x| \le 2$ and $|y| \le 3$ is given by the function $T(x,y) = 90 - 5x^2 - 2y^2$ . An isotherm is a curve of constant temperature. Which of the following statements is true?
Answer: The isotherm corresponding to the lowest temperature on the plate passes through the points $(2,3), (-2,3), (2,-3),$ and $(-2,-3)$ . The isotherms are the level curves of $T(x,y)$ , given by $90 - 5x^2 - 2y^2 = c$ , or $5x^2 + 2y^2 = 90 - c$ . For $c < 90$ , these curves are ellipses centered at the origin. The function $T(x,y)$ has its maximum value of 90 at $(0,0)$ and decreases as $(x,y)$ moves away from the origin. The lowest temperature on the plate will therefore occur at the point(s) within the domain $|x| \le 2, |y| \le 3$ that are 'farthest' from the origin in the sense of the elliptical level curves. This occurs at the four corners of the rectangular domain: $(\pm 2, \pm 3)$ . The temperature at these corners is $T(2,3) = 90 - 5(2^2) - 2(3^2) = 90 - 20 - 18 = 52$ . This is the minimum temperature. The isotherm for $T=52$ is $5x^2+2y^2 = 90 - 52 = 38$ , and this ellipse passes through all four corner points.
Let $f(x,y) = \arcsin(x^2 + y^2 - 1)$ . For which value of $c$ does the level set $f(x,y) = c$ consist of a single point?
Answer: $c = -\frac{\pi}{2}$ The level set is given by the equation $\arcsin(x^2 + y^2 - 1) = c$ . Applying the sine function to both sides gives $x^2 + y^2 - 1 = \sin(c)$ , or $x^2 + y^2 = 1 + \sin(c)$ . This equation describes a circle centered at the origin with radius $r = \sqrt{1 + \sin(c)}$ . For the level set to be a single point (the origin), the radius must be zero. This requires $1 + \sin(c) = 0$ , which means $\sin(c) = -1$ . The range of the $\arcsin$ function is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . The only value of $c$ in this interval for which $\sin(c) = -1$ is $c = -\frac{\pi}{2}$ .
For the function $k(x,y) = \sqrt{\ln(xy)}$ , what additional constraint must be satisfied for points in the first quadrant where $xy > 1$ ?
Answer: No additional constraint is needed beyond $xy > 1$ in the first quadrant For k(x,y) = √(ln(xy)) to be defined, we need: (1) xy > 0 (for the logarithm to be defined), and (2) ln(xy) ≥ 0 (for the square root), which means xy ≥ 1. In the first quadrant, xy > 0 is automatically satisfied. The condition xy > 1 ensures ln(xy) > 0 > 0, so the square root is defined. No additional constraints are needed. Choices B, C, and D introduce unnecessary conditions that aren't required for the function to be defined.
Let $D$ be the domain of the function $f(x,y) = \sqrt{1 - x^2} + \sqrt{1 - y^2}$ , and let $C$ be the level set $f(x,y) = 2$ . Which of the following statements is true?
Answer: $C$ is the center point of the domain $D$ . The domain $D$ requires both $1 - x^2 \ge 0$ and $1 - y^2 \ge 0$ . These inequalities simplify to $|x| \le 1$ and $|y| \le 1$ , which define a closed square in the $xy$ -plane with vertices at $(\pm 1, \pm 1)$ . The center of this square is $(0,0)$ . The level set $C$ is defined by $\sqrt{1 - x^2} + \sqrt{1 - y^2} = 2$ . Within the domain, the maximum value of $\sqrt{1 - x^2}$ is 1 (attained only at $x=0$ ), and the maximum value of $\sqrt{1 - y^2}$ is 1 (attained only at $y=0$ ). The sum can equal 2 only if both terms achieve their maximum value simultaneously. This occurs only when $x=0$ and $y=0$ . Thus, the level set $C$ consists of the single point $(0,0)$ , which is the center of the domain $D$ .
The level sets of $F(x,y,z) = x^2 + y^2 - z^2$ are:
Answer: Hyperboloids of one sheet for $c > 0$ , hyperboloids of two sheets for $c < 0$ , and a cone for $c = 0$ When analyzing level sets of functions in three variables, you're looking at the surfaces formed by setting $F(x,y,z) = c$ for different constant values. The function $F(x,y,z) = x^2 + y^2 - z^2$ creates a family of quadric surfaces that change character based on the value of $c$ . For $c > 0$ : Setting $x^2 + y^2 - z^2 = c$ gives us $x^2 + y^2 = z^2 + c$ . Since the right side is always positive, this describes a hyperboloid of one sheet - a surface that extends infinitely and has a "saddle-like" shape that's connected throughout. For $c < 0$ : The equation becomes $x^2 + y^2 = z^2 + c$ . Since $c < 0$ , we need $z^2 \geq |c|$ for real solutions. This creates two separate surfaces (one for $z \geq \sqrt{|c|}$ and one for $z \leq -\sqrt{|c|}$ ), forming a hyperboloid of two sheets. For $c = 0$ : We get $x^2 + y^2 = z^2$ , which is a circular cone with vertex at the origin. This confirms answer C is correct. Answer A incorrectly identifies these as paraboloids, which would involve linear terms. Answer B describes surfaces like $x^2 + y^2 + z^2 = c$ , where all variables are squared with the same sign. Answer D suggests the surfaces don't change character with $c$ , missing the fundamental distinction between the three cases. Strategy tip: For quadric surface problems, focus on the signs of the squared terms and whether the constant is positive, negative, or zero - this determines the surface type.

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