Surfaces & Contour Maps - Multivariable Calculus

1. A particle moves in the $xy$-plane along the path given by the vector function $\vec{r}(t) = \langle 3\cos(t), 5\sin(t) \rangle$ for $t \ge 0$. The height of the surface above the point $(x,y)$ is given by $z = f(x, y) = x^2 + y^2$. What is the maximum height reached by the particle on the surface?
   Answer: 25 The particle's path is given by $x(t) = 3\cos(t)$ and $y(t) = 5\sin(t)$. The height at time $t$ is $z(t) = (3\cos(t))^2 + (5\sin(t))^2 = 9\cos^2(t) + 25\sin^2(t)$. Using the identity $\cos^2(t) = 1 - \sin^2(t)$: $z(t) = 9(1 - \sin^2(t)) + 25\sin^2(t) = 9 + 16\sin^2(t)$. The maximum occurs when $\sin^2(t) = 1$, giving $z_{\text{max}} = 9 + 16(1) = 25$.
2. In multivariable calculus, a 'cylinder' or 'cylindrical surface' is a surface that consists of all lines (rulings) that are parallel to a given line and pass through a given plane curve. Which of the following equations describes a cylinder that is NOT a circular cylinder?
   Answer: $z = y^3$ A cylinder is formed by a generating curve in one plane and rulings parallel to the axis of the missing variable. Choice A, $x^2 + z^2 = 4$, describes a circular cylinder, as the generating curve is a circle in the $xz$-plane and the rulings are parallel to the $y$-axis. Choice B, $z = y^3$, describes a parabolic cylinder. The generating curve is the cubic curve $z=y^3$ in the $yz$-plane, and the rulings are parallel to the $x$-axis. This is not a circular cylinder. Choice C, $z = x^2 + y^2$, is an elliptic paraboloid, not a cylinder. All three variables are present in a non-linear way that doesn't allow for rulings. Choice D, $x^2 + y^2 + z^2 = 9$, is a sphere, not a cylinder.
3. The graph of a function $z = f(x,y)$ has the following properties: (1) its cross-section with any plane $x=c$ (where $c$ is a constant) is a parabola opening upward; (2) its cross-section with any plane $y=c$ is also a parabola opening upward; (3) its level curve for $z=0$ consists of a single point, the origin. Which of the following is a possible formula for $f(x,y)$?
   Answer: $f(x,y) = x^2 + y^2$ Let's test each option against the given properties. (A) $f(x,y)=xy$. The cross-section $z=cy$ for $x=c$ is a line, not a parabola. This fails property (1). (B) $f(x,y)=x^2-y^2$. The cross-section $z=c^2-y^2$ for $x=c$ is a parabola opening downward. This fails property (1). (C) $f(x,y)=x^2+y^2$. For $x=c$, $z=c^2+y^2$ is a parabola in $y$ opening upward. For $y=c$, $z=x^2+c^2$ is a parabola in $x$ opening upward. The level curve for $z=0$ is $x^2+y^2=0$, which is true only for the point $(0,0)$. All three properties are satisfied. This surface is an elliptic (in fact, circular) paraboloid. (D) $f(x,y)=\sqrt{x^2+y^2}$. The cross-section $z=\sqrt{c^2+y^2}$ for $x=c$ is the upper half of a hyperbola, not a parabola. This fails property (1). This surface is a cone.
4. A surface has the property that all of its contour lines are rectangular hyperbolas with the same asymptotes. If the asymptotes are the lines $y = x$ and $y = -x$, and one contour line passes through the point $(2,1)$, what type of surface is this?
   Answer: The surface $z = xy$, which has contour lines $xy = k$ forming rectangular hyperbolas When analyzing surfaces through their contour lines, you need to understand how level curves $f(x,y) = k$ behave for different functions. The key insight here is recognizing what type of equation produces rectangular hyperbolas with specific asymptotes. For rectangular hyperbolas with asymptotes $y = x$ and $y = -x$, the standard form is $xy = k$. These curves have the property that as you move along them, the product of coordinates remains constant. When $k > 0$, the hyperbola sits in the first and third quadrants; when $k < 0$, it sits in the second and fourth quadrants. Let's verify: if the surface is $z = xy$, then contour lines are $xy = k$. The point $(2,1)$ gives us $k = 2 \cdot 1 = 2$, so one contour line is $xy = 2$. To check asymptotes, note that as $x \to \infty$ along the curve, $y \to 0$, and the curve approaches the coordinate axes rotated 45°, which are indeed $y = x$ and $y = -x$. Choice A ($z = x^2 - y^2$) gives contour lines $x^2 - y^2 = k$, which are hyperbolas with asymptotes $y = \pm x$, but these aren't rectangular hyperbolas. Choice B ($z = x^2 + y^2 - 2xy = (x-y)^2$) produces contour lines $(x-y)^2 = k$, which are pairs of parallel lines, not hyperbolas. Choice C is similar to A with a scaling factor but still doesn't produce rectangular hyperbolas. Remember: $xy = k$ always gives rectangular hyperbolas with asymptotes along the coordinate axes rotated 45°. This pattern appears frequently in multivariable calculus problems.
5. The temperature on a square metal plate defined by $0 \le x \le 4$ and $0 \le y \le 4$ is given by $T(x, y) = (x-2)^2 + (y-2)^2$. An ant starts at the point $(4, 4)$ and moves along a path that keeps it on the circle of radius 2 centered at $(2, 2)$. Which of the following is true about the temperature experienced by the ant?
   Answer: The temperature remains constant. The temperature function is $T(x, y) = (x-2)^2 + (y-2)^2$. The level curves of this function are given by setting $T(x,y)$ to a constant $k$. The equation $k = (x-2)^2 + (y-2)^2$ represents a circle of radius $\sqrt{k}$ centered at $(2, 2)$. The ant's path is described as a circle of radius 2 centered at $(2, 2)$, which is the curve defined by the equation $(x-2)^2 + (y-2)^2 = 2^2 = 4$. Since the ant's entire path lies on this curve, it is moving along the level curve where the temperature is constantly $T=4$. Therefore, the temperature experienced by the ant remains constant.
6. A weather map displays isotherms (temperature contour lines) with values 10°C, 15°C, 20°C, and 25°C. At location R, you observe that moving 1 km north increases temperature by approximately 3°C, while moving 1 km east decreases temperature by approximately 4°C. Which isotherm does location R most likely lie on?
   Answer: The 15°C isotherm, calculated using the gradient magnitude and direction relative to nearby isotherms The gradient vector at R is approximately $\nabla T \approx (\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}) = (-4, 3)$ in units of °C/km. The gradient magnitude is $|\nabla T| = \sqrt{16 + 9} = 5$ °C/km. This gradient points in direction $\arctan(\frac{3}{-4}) \approx 143°$ from east (toward northwest). For evenly spaced 5°C isotherms, a gradient of 5°C/km suggests R lies roughly midway between isotherms, making 15°C most likely (midway between 10°C and 20°C). The gradient magnitude alone doesn't determine whether R is in a warm or cool region - choices A and D make this error. Choice C doesn't provide sufficient calculation detail.
7. A surface in $\mathbb{R}^3$ is defined by the following properties of its traces: 1. Its trace in any plane $z=k$ with $k>0$ is an ellipse. 2. Its trace in the $xz$-plane ($y=0$) is a parabola opening upward. 3. Its trace in the $yz$-plane ($x=0$) is a parabola opening upward. Which of the following equations represents this surface?
   Answer: $z = x^2 + y^2$ When analyzing surfaces through their traces, you're essentially taking cross-sections and examining the resulting curves. This approach helps you visualize complex 3D surfaces by understanding their behavior in familiar 2D planes. Let's examine what each condition tells us. For traces in planes $z = k$ (where $k > 0$) to be ellipses, we need an equation where fixing $z$ gives us a relationship of the form $x^2 + y^2 = \text{constant}$. For the traces in both the $xz$-plane and $yz$-plane to be upward-opening parabolas, we need $z$ to equal a quadratic expression in the remaining variable. Option D, $z = x^2 + y^2$, satisfies all conditions perfectly. When $z = k > 0$, we get $k = x^2 + y^2$, which is a circle (a special case of an ellipse). In the $xz$-plane ($y = 0$), we get $z = x^2$, an upward parabola. In the $yz$-plane ($x = 0$), we get $z = y^2$, also an upward parabola. Option A represents a sphere, whose traces are circles in all planes, not parabolas in coordinate planes. Option B, $z^2 = x^2 + y^2$, is a cone where traces in $z = k$ planes are circles, but traces in coordinate planes are hyperbolas, not parabolas. Option C, $z = y^2 - x^2$, is a saddle surface where traces in coordinate planes include both upward and downward parabolas. Remember: when identifying surfaces from traces, systematically substitute the constraint into the equation and check if the resulting curve matches the described shape.
8. Consider the function $W(x, y, z) = x^2 + y^2 - z^2$. The level surfaces of this function are surfaces in $\mathbb{R}^3$ given by $W(x, y, z) = k$ for some constant $k$. Which of the following values of $k$ results in a level surface that is a cone?
   Answer: $k = 0$ The level surfaces are defined by the equation $x^2 + y^2 - z^2 = k$. We analyze the surface for different values of $k$. For $k = -1$, the equation is $x^2 + y^2 - z^2 = -1$, or $z^2 - x^2 - y^2 = 1$. This is the equation of a hyperboloid of two sheets. For $k = 1$, the equation is $x^2 + y^2 - z^2 = 1$. This is the equation of a hyperboloid of one sheet. For $k = 0$, the equation is $x^2 + y^2 - z^2 = 0$, or $z^2 = x^2 + y^2$. This is the standard equation of a double cone with its vertex at the origin and axis along the z-axis. Therefore, $k=0$ results in a level surface that is a cone.
9. Consider the function $f(x, y, z) = x^2 - y^2 - z^2$. How does the geometric shape of the level surface $f(x, y, z) = k$ depend on the value of the constant $k$?
   Answer: For $k>0$, it is a hyperboloid of two sheets; for $k<0$, it is a hyperboloid of one sheet; for $k=0$, it is a cone. The equation of the level surface is $x^2 - y^2 - z^2 = k$. This is a quadric surface. We can rewrite it as $x^2 = y^2 + z^2 + k$. If $k > 0$, we can write it in the standard form $\frac{x^2}{(\sqrt{k})^2} - \frac{y^2}{(\sqrt{k})^2} - \frac{z^2}{(\sqrt{k})^2} = 1$. This is a hyperboloid of two sheets opening along the x-axis. If $k < 0$, let $k = -c^2$ where $c>0$. The equation is $x^2 - y^2 - z^2 = -c^2$, or $y^2 + z^2 - x^2 = c^2$. This is a hyperboloid of one sheet opening along the x-axis. If $k = 0$, the equation is $x^2 = y^2 + z^2$, which is the equation of a double cone with its axis along the x-axis.
10. The surface $z = \ln(x^2 + y^2)$ is defined for $x^2 + y^2 > 0$. What distinctive feature characterizes all contour lines of this surface?
    Answer: All contour lines are concentric circles centered at the origin with radii that increase exponentially For contour lines $\ln(x^2 + y^2) = k$, we have $x^2 + y^2 = e^k$. This gives circles centered at origin with radius $r = \sqrt{e^k} = e^{k/2}$. As k increases arithmetically (k, k+1, k+2, ...), the radius increases as $(e^{k/2}, e^{(k+1)/2}, e^{(k+2)/2}, ...) = (e^{k/2}, e^{k/2} \cdot e^{1/2}, e^{k/2} \cdot e, ...)$, which is exponential growth. Choice B incorrectly suggests arithmetic growth of radii. Choice C incorrectly identifies ellipses instead of circles. Choice D incorrectly suggests spiral shapes when the contours are actually perfect circles.
11. Consider the surface $z = x^2 - y^2 + 4xy$. At which point does this surface have a contour line that forms a pair of intersecting straight lines?
    Answer: At the origin $(0,0)$, where the contour $z = 0$ degenerates into lines $y = (2+\sqrt{5})x$ and $y = (2-\sqrt{5})x$ Setting $z = 0$: $x^2 - y^2 + 4xy = 0$. This is a quadratic form that can be factored. Treating as quadratic in $y$: $-y^2 + 4xy + x^2 = 0$, or $y^2 - 4xy - x^2 = 0$. Using quadratic formula: $y = \frac{4x \pm \sqrt{16x^2 + 4x^2}}{2} = \frac{4x \pm 2x\sqrt{5}}{2} = x(2 \pm \sqrt{5})$. So the degenerate contour consists of lines $y = (2+\sqrt{5})x$ and $y = (2-\sqrt{5})x$ intersecting at origin. Choice A incorrectly factors as difference of squares, ignoring the $4xy$ term. Choice B uses wrong point and value. Choice D uses wrong point.
12. The level curves of the surface $z = f(x, y)$ are described by the equation $9x^2 + 4y^2 - 18x + 16y = k$ for some constant $k$. Which of the following statements accurately describes this surface?
    Answer: The surface is an elliptic paraboloid with a minimum at $(1, -2)$. To identify the shape of the level curves, we must complete the square for $x$ and $y$. $9(x^2 - 2x) + 4(y^2 + 4y) = k$ $9(x^2 - 2x + 1) + 4(y^2 + 4y + 4) = k + 9(1) + 4(4)$ $9(x-1)^2 + 4(y+2)^2 = k + 25$ For any $k > -25$, this equation describes an ellipse centered at $(1, -2)$. A surface whose level curves are ellipses is an elliptic paraboloid. Since the coefficients of the squared terms (9 and 4) are both positive, the paraboloid opens upwards, and its vertex represents a local minimum. The vertex occurs at the center of the ellipses, which is $(1, -2)$. Thus, the surface is an elliptic paraboloid with a local minimum at $(1, -2)$.
13. A topographic map shows circular contour lines with values 100, 200, 300, and 400 meters, all centered at the same point. The radius of the 200m contour is twice the radius of the 300m contour. What can you conclude about the shape of this hill?
    Answer: The hill is steeper in the region between 200m and 300m elevation than between 100m and 200m When analyzing topographic maps, the spacing between contour lines reveals crucial information about terrain steepness. Closer contour lines indicate steeper slopes, while wider spacing shows gentler gradients. Let's examine what the given information tells us. You're told that the 200m contour has twice the radius of the 300m contour. This means as you move from the 300m level down to the 200m level, the horizontal distance increases dramatically – the contour lines are spreading apart significantly. Conversely, between the 200m and 100m levels, if the pattern continued uniformly, you'd expect the 100m contour to have twice the radius of the 200m contour. But since the radius is already quite large at 200m, this creates much tighter spacing between 200m and 300m compared to 100m and 200m. Choice D correctly identifies that the hill is steeper between 200m and 300m than between 100m and 200m, because the horizontal distance covered while dropping 100m in elevation is much smaller in the upper region. Choice A is wrong because uniform conical shape would require constant spacing ratios, which we don't have. Choice B incorrectly suggests the hill becomes less steep at higher elevations – it's actually the opposite. Choice C misunderstands the relationship; while radius does change systematically with elevation, this doesn't create a parabolic profile, and the slope variation contradicts uniform geometry. Remember: on contour maps, steep slopes pack contour lines together, while gentle slopes spread them apart. Always compare horizontal distances for equal elevation changes.
14. Which of the following statements about the interpretation of contour maps for a function $z=f(x,y)$ is necessarily correct?
    Answer: If one follows a path that is everywhere orthogonal to the contour lines, one is moving along a path of steepest ascent or descent. The gradient vector, $\nabla f$, points in the direction of steepest ascent and is always orthogonal to the level curves. A path in the direction of $-\nabla f$ is the path of steepest descent. Therefore, a path always orthogonal to contour lines is a path of steepest ascent or descent. Choice A is false; if two level curves for different values intersected, the function would have two different $z$-values at the same $(x,y)$ point, violating the definition of a function. Choice B is not necessarily true; it could represent a plain or a plateau of constant height if the region inside the innermost contour is flat. Choice D is false; it simply means that the change in elevation within that region is less than the contour interval. The surface could still be sloped.
15. What is the equation of the line tangent to the level curve of the function $f(x, y) = x^3 - xy^2$ at the point $(1, 2)$?
    Answer: $y - 2 = -\frac{1}{4}(x - 1)$ The gradient vector $\nabla f$ is perpendicular to the level curve at any point. First, find the partial derivatives: $f_x = 3x^2 - y^2$ and $f_y = -2xy$. At $(1, 2)$: $\nabla f(1, 2) = \langle 3(1)^2 - (2)^2, -2(1)(2) \rangle = \langle -1, -4 \rangle$. The slope of the tangent line to the level curve is given by $-\frac{f_x}{f_y} = -\frac{-1}{-4} = -\frac{1}{4}$. Using point-slope form: $y - 2 = -\frac{1}{4}(x - 1)$.
16. The level curves of a function $z = f(x, y)$ have the following properties: for a constant $k=0$, the level curve is the pair of lines $y = x$ and $y = -x$. For any constant $k \neq 0$, the level curve $f(x, y) = k$ is a hyperbola. For $k > 0$, the hyperbolas have vertices on the $y$-axis. For $k < 0$, the hyperbolas have vertices on the $x$-axis. Which of the following surfaces is the graph of $z = f(x, y)$?
    Answer: A hyperbolic paraboloid Let's analyze the descriptions. The level curves are given by $f(x,y)=k$. For $k=0$, we have $y = \pm x$, which can be written as $y^2 - x^2 = 0$. For $k>0$, the level curves are hyperbolas with vertices on the y-axis, which have the form $y^2/a^2 - x^2/b^2 = 1$. For $k<0$, the level curves are hyperbolas with vertices on the x-axis, which have the form $x^2/a^2 - y^2/b^2 = 1$. A simple function that fits all these descriptions is $f(x,y) = c(y^2-x^2)$ for some constant $c>0$. For instance, if $f(x,y) = y^2-x^2$, the level curves are $y^2-x^2=k$. This equation matches all the given properties. The graph of $z = y^2-x^2$ is a hyperbolic paraboloid, also known as a saddle surface.
17. The intersection of the hyperboloid of one sheet $x^2 + y^2 - z^2 = 1$ and a plane is a curve. Which of the following statements gives an incorrect description of such a curve?
    Answer: The intersection with the plane $z = x$ is a parabola. To find the curve of intersection, we substitute the equation of the plane into the equation of the surface. A) Plane $z=c$: $x^2 + y^2 - c^2 = 1 \implies x^2 + y^2 = 1+c^2$. Since $1+c^2 > 0$ for any real $c$, this is always the equation of a circle. This statement is correct. B) Plane $x=1$: $1^2 + y^2 - z^2 = 1 \implies y^2 - z^2 = 0 \implies (y-z)(y+z)=0$. This represents the two lines $y=z$ and $y=-z$. This statement is correct. C) Plane $x=c$ with $|c|>1$: $c^2 + y^2 - z^2 = 1 \implies y^2 - z^2 = 1-c^2$. Since $|c|>1$, $c^2>1$, so $1-c^2 < 0$. Let $1-c^2 = -k$ where $k>0$. Then $y^2 - z^2 = -k$, or $z^2 - y^2 = k$. This is the equation of a hyperbola. This statement is correct. D) Plane $z=x$: $x^2 + y^2 - x^2 = 1 \implies y^2 = 1 \implies y = \pm 1$. This represents a pair of parallel lines, not a parabola. This statement is incorrect.
18. A function $f(x,y)$ has the property that its contour line $f(x,y) = 0$ consists of two intersecting curves: the parabola $y = x^2 - 1$ and the line $y = -x$. If the function has the form $f(x,y) = (y - x^2 + 1)(y + x + a)$ for some constant $a$, what is the value of $a$?
    Answer: $a = 0$, since the line component is $y = -x$ which gives $y + x = 0$ For the contour $f(x,y) = 0$, we need $(y - x^2 + 1)(y + x + a) = 0$. This equation is satisfied when either factor equals zero: $y - x^2 + 1 = 0$ (giving $y = x^2 - 1$, the parabola) or $y + x + a = 0$ (giving $y = -x - a$, the line). Since the specified line is $y = -x$, we need $-x - a = -x$, which requires $a = 0$. Choice B and C give incorrect line equations $y = -x - 1$ and $y = -x + 1$ respectively. Choice D gives $y = -x - 2$, also incorrect.

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