Multivariable Functions & Level Sets - Calculus 3

1. Consider the function $w(x,y,z) = \sqrt{25 - x^2 - y^2 - z^2} + \sqrt{x^2 + y^2 + z^2 - 9}$. The domain of this function is characterized by which condition?
   Answer: $9 \leq x^2 + y^2 + z^2 \leq 25$, representing the region between two concentric spheres including both boundary surfaces When finding the domain of a multivariable function, you need to identify where each component is mathematically defined. For functions involving square roots, the expression under each radical must be non-negative. For $w(x,y,z) = \sqrt{25 - x^2 - y^2 - z^2} + \sqrt{x^2 + y^2 + z^2 - 9}$ to be defined, both conditions must hold simultaneously: First term: $25 - x^2 - y^2 - z^2 \geq 0$, which gives us $x^2 + y^2 + z^2 \leq 25$ Second term: $x^2 + y^2 + z^2 - 9 \geq 0$, which gives us $x^2 + y^2 + z^2 \geq 9$ The domain is the intersection of these conditions: $9 \leq x^2 + y^2 + z^2 \leq 25$. This represents the region between two concentric spheres centered at the origin—inside the sphere of radius 5 but outside (or on) the sphere of radius 3. The inequalities include equality, so both boundary surfaces are part of the domain. Option A describes two separate regions that don't overlap—this would make the function undefined everywhere. Option B excludes the boundary surfaces where the expressions under the radicals equal zero, but square roots of zero are perfectly valid (they equal zero). Option D only includes the boundary surfaces themselves, ignoring the solid regions between them where the function is also well-defined. Remember: for domain problems with multiple constraints, find where ALL conditions are satisfied simultaneously, and don't exclude boundary points unless the function is actually undefined there.
2. A manufacturer models profit using $P(x,y) = 50\sqrt{xy} - x^2 - 2y^2 + 100$ where $x,y \geq 0$ represent production levels. The level curve $P(x,y) = 100$ represents combinations yielding exactly $100 profit. This level curve is equivalent to which equation after simplification?
   Answer: $50\sqrt{xy} = x^2 + 2y^2$, giving $2500xy = x^4 + 4x^2y^2 + 4y^4$ Setting $P(x,y) = 100$: $50\sqrt{xy} - x^2 - 2y^2 + 100 = 100$, which simplifies to $50\sqrt{xy} = x^2 + 2y^2$. Squaring both sides: $(50\sqrt{xy})^2 = (x^2 + 2y^2)^2$, giving $2500xy = x^4 + 4x^2y^2 + 4y^4$. Choice A incorrectly rearranges the initial equation. Choice C incorrectly manipulates the constraint. Choice D incorrectly claims no real solutions exist, but solutions do exist (for example, when $x = y = 50/3$).
3. The level curves of $G(x,y) = \frac{x^2}{4} - \frac{y^2}{9}$ are studied for different values of $c$. Which statement correctly describes the behavior of these level curves as $c$ varies?
   Answer: For $c > 0$: hyperbolas opening left-right; for $c < 0$: hyperbolas opening up-down; for $c = 0$: intersecting lines $y = \pm \frac{3x}{2}$ Level curves are sets of points where a function equals a constant value, so you need to analyze the equation $\frac{x^2}{4} - \frac{y^2}{9} = c$ for different values of $c$. For $c > 0$, rearrange to get $\frac{x^2}{4} - \frac{y^2}{9} = c > 0$, which can be written as $\frac{x^2}{4c} - \frac{y^2}{9c} = 1$. This is the standard form of a hyperbola that opens horizontally (left-right), since the $x^2$ term is positive and the $y^2$ term is negative. For $c < 0$, you get $\frac{x^2}{4} - \frac{y^2}{9} = c < 0$, which rearranges to $\frac{y^2}{9|c|} - \frac{x^2}{4|c|} = 1$. This is a hyperbola opening vertically (up-down), since now the $y^2$ term is positive. For $c = 0$, the equation becomes $\frac{x^2}{4} = \frac{y^2}{9}$, which gives $\frac{x}{2} = \pm\frac{y}{3}$, or $y = \pm\frac{3x}{2}$. These are two intersecting lines through the origin. Option A incorrectly claims $c = 0$ gives a single point. Option B wrongly identifies positive $c$ values as ellipses and gets the wrong slope for the intersecting lines. Option C has the hyperbola orientations backwards and the wrong slope. Only option D correctly identifies the orientations and gives $y = \pm\frac{3x}{2}$ for $c = 0$. Study tip: Remember that for hyperbolas in the form $\frac{A^2}{a} - \frac{B^2}{b} = 1$, whichever variable has the positive coefficient determines the opening direction.
4. A function is defined as $f(x,y) = \arcsin\left(\frac{x+y-2}{3}\right) + \ln(x-y)$. The domain of $f$ requires both terms to be defined simultaneously. Which inequality system correctly describes this domain?
   Answer: $-3 \leq x + y - 2 \leq 3$ and $x > y$, giving the region $-1 \leq x + y \leq 5$ and $x > y$ For $\arcsin\left(\frac{x+y-2}{3}\right)$ to be defined, we need $-1 \leq \frac{x+y-2}{3} \leq 1$. Multiplying by 3: $-3 \leq x+y-2 \leq 3$. Adding 2: $-1 \leq x+y \leq 5$. For $\ln(x-y)$ to be defined, we need $x-y > 0$, so $x > y$. Therefore, the domain is described by $-1 \leq x+y \leq 5$ and $x > y$. Choice A incorrectly states the first condition as $-1 \leq x+y-2 \leq 3$ instead of $-3 \leq x+y-2 \leq 3$.
5. Consider the function $f(x,y) = \frac{x^2 + y^2 - 4}{x^2 + y^2 - 9}$. Which statement about the domain and level sets of this function is correct?
   Answer: The domain excludes the circle $x^2 + y^2 = 9$, and the level set $f(x,y) = 0$ is the single circle $x^2 + y^2 = 4$ The domain excludes points where the denominator equals zero, so $x^2 + y^2 - 9 = 0$, which means $x^2 + y^2 = 9$ is excluded. For the level set $f(x,y) = 0$, we need the numerator to equal zero while the denominator is nonzero: $x^2 + y^2 - 4 = 0$ gives $x^2 + y^2 = 4$. Since $4 \neq 9$, this circle is in the domain, so the level set $f(x,y) = 0$ is indeed the circle $x^2 + y^2 = 4$.
6. Consider the function $h(x,y,z) = x^2 + 4y^2 + 9z^2 - 2x + 8y - 18z$. The level surface $h(x,y,z) = k$ represents an ellipsoid when $k$ satisfies which condition?
   Answer: $k > -13$, and the center of the ellipsoid is at $(1, -1, 1)$ for all such values Complete the square: $h(x,y,z) = (x^2 - 2x + 1) + 4(y^2 + 2y + 1) + 9(z^2 - 2z + 1) - 1 - 4 - 9 + 1 = (x-1)^2 + 4(y+1)^2 + 9(z-1)^2 - 13$. So $h(x,y,z) = k$ becomes $(x-1)^2 + 4(y+1)^2 + 9(z-1)^2 = k + 13$. This represents an ellipsoid when $k + 13 > 0$, i.e., $k > -13$. The center is at $(1, -1, 1)$ for all such values.
7. Let $F(x,y) = \ln(xy - x - y)$. The domain of $F$ consists of all points $(x,y)$ such that $xy - x - y > 0$. Which region correctly describes this domain?
   Answer: The region where $(x-1)(y-1) > 1$, excluding the lines $x = 1$ and $y = 1$ We need $xy - x - y > 0$. Factor by adding and subtracting 1: $xy - x - y + 1 - 1 = (x-1)(y-1) - 1 > 0$, so $(x-1)(y-1) > 1$. This gives us a hyperbola-like region. The lines $x = 1$ and $y = 1$ are automatically excluded since they make $(x-1)(y-1) = 0 < 1$. Choice B is wrong because it uses $> -1$ instead of $> 1$.
8. Consider the level surfaces of $\Phi(x,y,z) = x^2 + y^2 - z^2$. The level surface $\Phi(x,y,z) = -4$ intersects the plane $z = 3$. What is the geometric shape of this intersection?
   Answer: A circle centered at $(0,0,3)$ with radius $\sqrt{5}$, lying entirely within the plane $z = 3$ When you encounter questions about intersections of level surfaces with planes, you're looking at how a 3D surface cuts through a specific slice of space. The key is to substitute the plane's equation into the level surface equation and see what 2D curve emerges. To find where the level surface $\Phi(x,y,z) = x^2 + y^2 - z^2 = -4$ intersects the plane $z = 3$, substitute $z = 3$ into the level surface equation: $x^2 + y^2 - 3^2 = -4$, which simplifies to $x^2 + y^2 - 9 = -4$, or $x^2 + y^2 = 5$. This is the equation of a circle in the $xy$-plane with radius $\sqrt{5}$, but since we're in the plane $z = 3$, it's a circle centered at $(0,0,3)$ with radius $\sqrt{5}$. Choice A is wrong because $x^2 + y^2 = 5$ describes a circle, not a hyperbola. Hyperbolas have the form $x^2 - y^2 = k$ or similar. Choice B incorrectly calculates the radius as $\sqrt{13}$, which would come from mistakenly adding rather than subtracting: $x^2 + y^2 = 9 + 4 = 13$. Choice C is incorrect because the level surface $x^2 + y^2 - z^2 = -4$ is a hyperboloid of two sheets, which definitely exists in 3D space. Always substitute the plane's constraint directly into the level surface equation, then recognize the resulting 2D curve type. Watch for arithmetic errors when simplifying—they're the most common trap in these intersection problems.

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