Multivariable Calculus Quiz: Surfaces And Contour Maps

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Surfaces And Contour MapsQuestion 1 of 18

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?

x^2 + z^2 = 4

z = y^3

z = x^2 + y^2

x^2 + y^2 + z^2 = 9

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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Surfaces And Contour Maps

Practice Surfaces And Contour Maps in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Surfaces And Contour Maps, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$x^2 + z^2 = 4$
$z = y^3$ (correct answer)
$z = x^2 + y^2$
$x^2 + y^2 + z^2 = 9$

Explanation: 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.

Question 2

A surface in $\mathbb{R}^3$ is defined by the following properties of its traces:

Its trace in any plane $z=k$ with $k>0$ is an ellipse.
Its trace in the $xz$ -plane ( $y=0$ ) is a parabola opening upward.
Its trace in the $yz$ -plane ( $x=0$ ) is a parabola opening upward. Which of the following equations represents this surface?

$x^2 + y^2 + z^2 = 1$
$z^2 = x^2 + y^2$
$z = y^2 - x^2$
$z = x^2 + y^2$ (correct answer)

Explanation: 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.

Question 3

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?

9
16
25 (correct answer)
34

Explanation: The particle's path is given by

x(t) = 3\cos(t)

and

y(t) = 5\sin(t)

. The height at time

t

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

Question 4

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?

$k = -1$
$k = 0$ (correct answer)
$k = 1$
No value of $k$ results in a cone.

Explanation: 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.

Question 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?

The temperature continuously increases.
The temperature continuously decreases.
The temperature remains constant. (correct answer)
The temperature increases then decreases.

Explanation: 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.

Question 6

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 $f(x,y)$ ?

$f(x,y) = xy$
$f(x,y) = x^2 - y^2$
$f(x,y) = x^2 + y^2$ (correct answer)
$f(x,y) = \sqrt{x^2+y^2}$

Explanation: 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

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.

Question 7

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 $z = f(x, y)$ ?

A cone
An elliptic paraboloid
A hyperbolic paraboloid (correct answer)
A hyperboloid of one sheet

Explanation: 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.

Question 8

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?

The hill has a uniform conical shape with constant slope angle throughout
The hill becomes less steep at higher elevations, creating a rounded summit profile
The hill has a parabolic profile since the radius varies linearly with elevation
The hill is steeper in the region between 200m and 300m elevation than between 100m and 200m (correct answer)

Explanation: 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.

Question 9

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 $a$ ?

$a = 0$ , since the line component is $y = -x$ which gives $y + x = 0$ (correct answer)
$a = 1$ , found by ensuring the curves intersect at the correct points
$a = -1$ , determined by the requirement that both curves pass through their intersection points
$a = 2$ , calculated from the condition that the parabola vertex aligns with the line

Explanation: 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.

Question 10

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?

The 10°C isotherm, since the large temperature gradients indicate R is in a region of low temperature
The 15°C isotherm, calculated using the gradient magnitude and direction relative to nearby isotherms (correct answer)
The 20°C isotherm, determined by analyzing the directional derivatives given in the problem
The 25°C isotherm, since the strong gradients suggest R is near the warmest region shown

Explanation: 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.

Question 11

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?

The surface $z = x^2 - y^2$ , which has contour lines $x^2 - y^2 = k$ for constant k
The surface $z = x^2 + y^2 - 2xy$ , which simplifies to hyperbolic level curves
The surface $z = \frac{x^2 - y^2}{2}$ , which produces hyperbolic contours with the given asymptotes
The surface $z = xy$ , which has contour lines $xy = k$ forming rectangular hyperbolas (correct answer)

Explanation: 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.

Question 12

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?

All contour lines are concentric circles centered at the origin with radii that increase exponentially (correct answer)
All contour lines are concentric circles centered at the origin with radii that increase arithmetically
All contour lines are ellipses with the same center but different eccentricities depending on the contour value
All contour lines are logarithmic spirals that approach the origin asymptotically

Explanation: 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.

Question 13

Which of the following statements about the interpretation of contour maps for a function $z=f(x,y)$ is necessarily correct?

Two distinct level curves, $f(x,y)=c_1$ and $f(x,y)=c_2$ with $c_1 \neq c_2$ , can intersect at a point.
A set of closed, concentric contour lines always indicates the location of a local maximum or minimum.
If one follows a path that is everywhere orthogonal to the contour lines, one is moving along a path of steepest ascent or descent. (correct answer)
The absence of contour lines in a region implies that the surface is perfectly flat and horizontal in that region.

Explanation: 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.

Question 14

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?

The surface is a hyperbolic paraboloid with a saddle point at $(1, -2)$ .
The surface is an elliptic paraboloid with a minimum at $(1, -2)$ . (correct answer)
The surface is a circular cone with vertex at $(1, -2)$ .
The surface is a tilted plane with parallel level curves.

Explanation: 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)

Question 15

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 $k$ ?

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. (correct answer)
For $k>0$ , it is a hyperboloid of one sheet; for $k<0$ , it is a hyperboloid of two sheets; for $k=0$ , it is a cone.
For all $k \neq 0$ , the level surfaces are hyperboloids of one sheet, and for $k=0$ it is a cone.
For all $k \neq 0$ , the level surfaces are elliptic paraboloids, and for $k=0$ it is a single point.

Explanation: 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.

Question 16

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 $(1, 2)$ ?

$y - 2 = -\frac{1}{4}(x - 1)$ (correct answer)
$y - 2 = \frac{1}{4}(x - 1)$
$y - 2 = -4(x - 1)$
$y - 2 = 4(x - 1)$

Explanation: 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)

Question 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?

The intersection with the plane $z = c$ for any real constant $c$ is a circle.
The intersection with the plane $x = 1$ is a pair of intersecting lines.
The intersection with the plane $x = c$ for any constant $c$ with $|c| > 1$ is a hyperbola.
The intersection with the plane $z = x$ is a parabola. (correct answer)

Explanation: 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.

Question 18

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?

At the origin $(0,0)$ , where the contour $z = 0$ degenerates into lines $y = x$ and $y = -x$
At the point $(1,1)$ , where the contour $z = 4$ degenerates into perpendicular lines
At the origin $(0,0)$ , where the contour $z = 0$ degenerates into lines $y = (2+\sqrt{5})x$ and $y = (2-\sqrt{5})x$ (correct answer)
At the point $(2,-2)$ , where the contour $z = 0$ degenerates into intersecting lines

Explanation: 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.