Calculus 1 Quiz: Sketching Slope Fields

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Sketching Slope FieldsQuestion 1 of 20

The differential equation $dy/dx = (y-1)^2$ has an equilibrium solution at $y=1$ . Based on an analysis of its slope field, how is this equilibrium solution best classified?

Stable

Unstable

Semi-stable

Periodic

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Calculus 1 Quiz

Calculus 1 Quiz: Sketching Slope Fields

Practice Sketching Slope Fields in Calculus 1 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 Sketching Slope Fields, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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.

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Question 1

The differential equation $dy/dx = (y-1)^2$ has an equilibrium solution at $y=1$ . Based on an analysis of its slope field, how is this equilibrium solution best classified?

Stable
Unstable
Semi-stable (correct answer)
Periodic

Explanation: An equilibrium solution occurs where

dy/dx = 0

, which is at

y=1

. To classify it, we examine the sign of

dy/dx

on either side of

y=1

. If

y > 1

(y-1)^2

is positive, so

dy/dx > 0

and solutions move away from

y=1

. If

y < 1

(y-1)^2

is also positive, so

dy/dx > 0

and solutions move towards

y=1

. Since solutions approach the equilibrium from one side (below) and move away from it on the other side (above), the equilibrium at

y=1

is classified as semi-stable.

Question 2

In a certain slope field, the slopes of the line segments depend only on the value of the sum $x+y$ . Which of the following differential equations matches this description?

$dy/dx = x^2+y^2$
$dy/dx = (x+y)^2$ (correct answer)
$dy/dx = xy$
$dy/dx = x-y$

Explanation: The condition 'slopes depend only on the value of the sum

x+y

' means that the differential equation can be written in the form

dy/dx = g(x+y)

for some function

g

. We examine the choices: A)

x^2+y^2

is not a function of

(x+y)

alone. B)

(x+y)^2

is a function of

(x+y)

, where

g(u)=u^2

. C)

xy

is not a function of

(x+y)

alone. D)

x-y

is not a function of

(x+y)

alone. Therefore, only

dy/dx = (x+y)^2

matches the description.

Question 3

The slope field for a certain differential equation is constructed. It is observed that for any horizontal line, all the line segments lying on that line are parallel to each other. Which of the following differential equations could have this slope field?

$dy/dx = x^2$
$dy/dx = y^2$ (correct answer)
$dy/dx = x+y$
$dy/dx = x/y$

Explanation: The property that all line segments on any horizontal line (a line of constant

y

) are parallel means that the slope

dy/dx

depends only on the y-coordinate. If the slope were dependent on

x

, the slopes would change as one moves horizontally. Among the choices, only

dy/dx = y^2

has a slope that depends solely on

y

. Such equations are called autonomous. For

dy/dx = x^2

, slopes are constant on vertical lines. For

dy/dx = x+y

and

dy/dx = x/y

, the slope depends on both

x

and

y

Question 4

For the slope field of the differential equation $dy/dx = y \ln y$ , consider a solution curve passing through the point $(0, 2)$ . Which of the following must be true for this solution curve for all $x$ ?

The solution is a strictly increasing function. (correct answer)
The solution is a strictly decreasing function.
The solution has a local maximum.
The solution is a constant function.

Explanation: The slope of the solution curve at any point

(x,y)

is given by

y \ln y

. The initial condition is

y(0)=2

. At this point, the slope is

2 \ln 2

. Since

\ln 2 > 0

, the slope is positive, and the function is initially increasing. Since

y

starts above 1,

y

will increase. For any

y>1

y>0

and

\ln y > 0

, so the product

y \ln y

is always positive. A function whose derivative is always positive is strictly increasing. Therefore, the solution curve starting at

(0,2)

will be strictly increasing for all

x

Question 5

Consider the slope field for the logistic differential equation $dy/dx = y(1 - y/10)$ . If a solution curve passes through the point $(0, 3)$ , what is the limiting value of $y(x)$ as $x \to \infty$ ?

0
3
10 (correct answer)
$\infty$

Explanation: The logistic equation

dy/dx = y(1 - y/10)

has equilibrium solutions when

dy/dx = 0

, which occur at

y=0

and

y=10

. By analyzing the sign of

dy/dx

, we see that for

0 < y < 10

dy/dx > 0

, meaning solutions increase. For

y > 10

dy/dx < 0

, meaning solutions decrease. This indicates that

y=10

is a stable equilibrium (the carrying capacity). A solution starting at

y(0)=3

, which is between 0 and 10, will increase and approach the stable equilibrium. Therefore, the limit of

y(x)

x \to \infty

is 10.

Question 6

The slope field for a differential equation has the following properties: slopes are positive in quadrants I and III, and negative in quadrants II and IV. Slopes are zero on both the x-axis and y-axis (excluding the origin). Which differential equation could produce this slope field?

$dy/dx = x+y$
$dy/dx = x/y$
$dy/dx = xy$ (correct answer)
$dy/dx = x^2+y^2$

Explanation: We need to analyze the sign of

dy/dx

in each quadrant. The product

xy

is positive when

x

and

y

have the same sign (quadrants I and III) and negative when they have opposite signs (quadrants II and IV). Also,

xy=0

x=0

y=0

. This perfectly matches the description. For

dy/dx = x+y

, the sign depends on the line

y=-x

, not the axes. For

dy/dx = x/y

, the sign pattern is positive in I and III, negative in II and IV, but slopes are zero on

x=0

and undefined on

y=0

, which does not match. For

dy/dx = x^2+y^2

, slopes are always non-negative.

Question 7

A student observes a slope field and notes that all line segments along any line of the form $y=mx$ are parallel to each other. Which of the following differential equations could have produced this slope field?

$dy/dx = x+y$
$dy/dx = x^2$
$dy/dx = xy$
$dy/dx = y/x$ (correct answer)

Explanation: When you encounter slope field problems, you need to understand how the differential equation determines the slope at each point. The key insight here is recognizing what it means for line segments to be parallel along lines of the form

y = mx

. If all line segments are parallel along any line

y = mx

, this means the slope must be constant along each such line. Let's test this condition with answer choice D:

\frac{dy}{dx} = \frac{y}{x}

. Along the line

y = mx

, we substitute to get

\frac{dy}{dx} = \frac{mx}{x} = m

. Since

m

is constant for any given line

y = mx

, the slope is indeed constant along each line, making all segments parallel along that line. Answer choice A (

\frac{dy}{dx} = x + y

) fails because along

y = mx

, the slope becomes

x + mx = x(1 + m)

, which varies with

x

. Answer choice B (

\frac{dy}{dx} = x^2

) gives slopes that depend only on

x

-coordinate, not maintaining constant slope along lines

y = mx

. Answer choice C (

\frac{dy}{dx} = xy

) produces slopes of

x \cdot mx = mx^2

along

y = mx

, which again varies with

x

. When analyzing slope fields, always substitute the given condition into each differential equation to see which one satisfies the described pattern. Look for equations where the slope depends on ratios of variables (like

y/x

) when dealing with radial or linear patterns through the origin.

Question 8

Consider the slope field for $dy/dx = f(x,y)$ . If Euler's method with a small step size $h$ is used to approximate a solution starting at $(x_0, y_0)$ , the first step results in the point $(x_1, y_1)$ . How does the segment connecting $(x_0, y_0)$ to $(x_1, y_1)$ relate to the slope field?

The segment is parallel to the slope field line segment at $(x_1, y_1)$ .
The segment is perpendicular to the slope field line segment at $(x_0, y_0)$ .
The segment is parallel to the slope field line segment at the midpoint $(\frac{x_0+x_1}{2}, \frac{y_0+y_1}{2})$ .
The segment is parallel to the slope field line segment at $(x_0, y_0)$ . (correct answer)

Explanation: When you encounter Euler's method problems, focus on understanding how the method constructs its approximation step by step using the slope field information. Euler's method approximates solutions to differential equations by following the slope field in small, straight-line segments. At each point, the method uses the slope given by the differential equation to determine the direction of the next step. Specifically, from starting point

(x_0, y_0)

, the method calculates the slope

m = f(x_0, y_0)

and then moves to

(x_1, y_1) = (x_0 + h, y_0 + h \cdot f(x_0, y_0))

. The segment connecting

(x_0, y_0)

(x_1, y_1)

has slope

\frac{y_1 - y_0}{x_1 - x_0} = \frac{h \cdot f(x_0, y_0)}{h} = f(x_0, y_0)

. This is exactly the same as the slope of the slope field line segment at the starting point

(x_0, y_0)

. Therefore, answer D is correct. Answer A is wrong because the segment's slope equals

f(x_0, y_0)

, not

f(x_1, y_1)

. Answer B is incorrect since perpendicular lines would have slopes that are negative reciprocals, but our segment has slope

f(x_0, y_0)

, the same as the slope field at

(x_0, y_0)

. Answer C is wrong because Euler's method doesn't use the slope at the midpoint—it uses the slope at the starting point throughout each step. Remember: Euler's method always uses the slope information from where you currently are, not where you're going. Each step follows the current slope field direction exactly.

Question 9

For the differential equation $dy/dx = x^2 - y$ , the set of points in the plane where all solution curves have the same slope $c$ is called an isocline. What is the geometric shape of the isocline corresponding to a slope of $c=2$ ?

A parabola opening upward. (correct answer)
A parabola opening downward.
A line with a positive slope.
A circle centered at the origin.

Explanation: An isocline is a curve where the slope

dy/dx

is constant. To find the isocline for a slope of

c=2

, we set the differential equation equal to 2:

x^2 - y = 2

. To identify the geometric shape, we can solve for

y

y = x^2 - 2

. This is the equation of a parabola that opens upward, with its vertex at

(0, -2)

Question 10

Consider the slope fields for the two differential equations: (I) $dy/dx = -x/y$ and (II) $dy/dx = y/x$ . Which statement best describes the relationship between the two slope fields at any point $(x, y)$ where both are defined?

The line segments at the same point are parallel.
The line segments at the same point are perpendicular. (correct answer)
The slope for (I) is the negative of the slope for (II).
The slope fields are identical.

Explanation: Let

m_1

be the slope from equation (I) and

m_2

be the slope from equation (II). So,

m_1 = -x/y

and

m_2 = y/x

. Two lines are perpendicular if the product of their slopes is -1. Let's check:

m_1 \cdot m_2 = (-x/y) \cdot (y/x) = -1

. Since the product of the slopes is -1 at every point

(x,y)

(where defined), the line segments of the two slope fields are perpendicular to each other. This means the families of solution curves are orthogonal trajectories.

Question 11

Consider the differential equation $dy/dx = e^{x-y}$ . What is a notable feature of its slope field along the line $y=x$ ?

All line segments are horizontal.
All line segments are vertical.
All line segments have a slope of 1. (correct answer)
The slopes of the line segments increase as $x$ increases.

Explanation: To find the slope of the line segments along the line

y=x

, we substitute

y=x

into the differential equation:

dy/dx = e^{x-x} = e^0 = 1

. This means that at every point on the line

y=x

, the slope of the tangent line to the solution curve is 1. Consequently, all line segments in the slope field along the line

y=x

have a slope of 1.

Question 12

The line segments in the slope field for $dy/dx = \cos(x+y)$ are horizontal along a set of isoclines. What is the slope of these isocline curves?

0
1
-1 (correct answer)
$\pi/2$

Explanation: The line segments are horizontal where

dy/dx = 0

. Setting the differential equation to zero gives

\cos(x+y) = 0

. This condition is met when the argument of the cosine function is an odd multiple of

\pi/2

, so

x+y = \frac{\pi}{2} + n\pi

for any integer

n

. These are the equations of the isoclines for slope zero. To find the slope of these isoclines, we can write them in the form

y = -x + (\frac{\pi}{2} + n\pi)

. These are equations of lines, and their slope is the coefficient of

x

, which is -1.

Question 13

If the slope field for the differential equation $dy/dx = f(x)$ is symmetric with respect to the y-axis, what property must the function $f(x)$ have?

$f(x)$ must be an even function. (correct answer)
$f(x)$ must be an odd function.
$f(x)$ must be a constant function.
$f(x)$ must be strictly positive.

Explanation: Symmetry with respect to the y-axis means that the slope at point

(-x, y)

is the same as the slope at point

(x, y)

. In this case, the slope

dy/dx

depends only on

x

, so the condition is that the slope at

-x

must equal the slope at

x

. This means

f(-x) = f(x)

for all

x

in the domain. This is the definition of an even function.

Question 14

Consider a differential equation $dy/dx = f(x,y)$ . The line $y=2$ is a solution to the equation. What must be true about the slope field on the line $y=2$ ?

All line segments are vertical.
All line segments have a slope of 2.
All line segments are horizontal. (correct answer)
All line segments point towards the origin.

Explanation: If

y=2

is a solution curve, it means that the function

y(x)=2

satisfies the differential equation for all

x

. The derivative of this function is

dy/dx = 0

. Therefore, for the differential equation to be satisfied, the slope function

f(x,y)

must be equal to 0 for all points on the line

y=2

. In the slope field, a slope of 0 is represented by a horizontal line segment. Thus, all line segments on the line

y=2

must be horizontal.

Question 15

Consider the differential equation $dy/dx = \frac{x+y}{x-y}$ . Which statement accurately describes a property of its slope field?

Along the line $y=x$ (for $x \neq 0$ ), all line segments are vertical. (correct answer)
Along the line $y=-x$ (for $x \neq 0$ ), all line segments are vertical.
In the first quadrant, all line segments have a positive slope.
Along the x-axis (for $x \neq 0$ ), the slope of the line segments is equal to -1.

Explanation: The slope

dy/dx

represents the slope of the line segments in the slope field. A vertical line segment corresponds to an undefined slope. For the equation

dy/dx = \frac{x+y}{x-y}

, the slope is undefined when the denominator is zero, which occurs when

x-y=0

, or

y=x

. Therefore, along the line

y=x

(excluding the origin where it's indeterminate), the line segments are vertical. Choice B is incorrect because along

y=-x

, the numerator is zero, so slopes are zero (horizontal). Choice C is incorrect because in the first quadrant, if

y>x

, the denominator

x-y

is negative, making the slope negative. Choice D is incorrect because along the x-axis (

y=0

), the slope is

x/x = 1

Question 16

The slope field for $dy/dx = 1 - x^2 - y^2$ has horizontal line segments along a specific curve. What is this curve?

The x-axis and the y-axis.
The unit circle $x^2 + y^2 = 1$ . (correct answer)
The parabola $y = 1 - x^2$ .
The lines $y=x$ and $y=-x$ .

Explanation: Horizontal line segments in a slope field occur where the slope,

dy/dx

, is equal to zero. For the given differential equation, we set

dy/dx = 0

1 - x^2 - y^2 = 0

. Rearranging this equation gives

x^2 + y^2 = 1

, which is the equation of the unit circle centered at the origin with a radius of 1. Therefore, on this circle, all line segments of the slope field are horizontal.

Question 17

Consider the slope field for the differential equation $dy/dx = \sqrt{x^2+y^2}$ . Which statement is true about the isoclines of this slope field?

The isoclines are lines passing through the origin.
The isoclines are circles centered at the origin. (correct answer)
The isoclines are parabolas with vertex at the origin.
The isoclines are hyperbolas centered at the origin.

Explanation: Isoclines are curves where the slope

dy/dx

is constant. Let the constant slope be

c

. Then we have

c = \sqrt{x^2+y^2}

. Since

\sqrt{x^2+y^2}

represents the distance from the origin,

c

must be non-negative. Squaring both sides gives

c^2 = x^2+y^2

. This is the equation of a circle centered at the origin with radius

R=c

. Therefore, the isoclines (curves of constant slope) are circles centered at the origin.

Question 18

For which of the following differential equations are the solution curves circles centered at the origin?

$dy/dx = y/x$
$dy/dx = -y/x$
$dy/dx = x/y$
$dy/dx = -x/y$ (correct answer)

Explanation: A circle centered at the origin is given by the equation

x^2 + y^2 = R^2

. We can find the slope of the tangent line to this curve by implicit differentiation:

2x + 2y \frac{dy}{dx} = 0

, which gives

\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}

. Therefore, the differential equation whose solution curves are circles centered at the origin is

dy/dx = -x/y

. The slope field for this equation would consist of line segments that are tangent to these circles.

Question 19

If the slope field for $dy/dx = f(y)$ is symmetric with respect to the x-axis, what property must the function $f(y)$ have?

$f(y)$ must be an even function, i.e., $f(-y) = f(y)$ .
$f(y)$ must be a periodic function.
$f(y)$ must be equal to zero for all $y$ .
$f(y)$ must be an odd function, i.e., $f(-y) = -f(y)$ . (correct answer)

Explanation: When analyzing slope fields and their symmetries, you need to understand how the differential equation

dy/dx = f(y)

creates directional patterns in the coordinate plane. For a slope field to be symmetric with respect to the x-axis, the slopes at points

(x, y)

and

(x, -y)

must be mirror images of each other. At point

(x, y)

, the slope is

f(y)

. At the corresponding point

(x, -y)

, the slope is

f(-y)

. For x-axis symmetry, these slopes must be negatives of each other:

f(-y) = -f(y)

. This is precisely the definition of an odd function. Think about it visually: if you have a slope pointing up and to the right above the x-axis, its mirror image below the x-axis should point down and to the right with the same steepness but opposite vertical direction. Option A is incorrect because if

f(-y) = f(y)

(even function), then points

(x, y)

and

(x, -y)

would have identical slopes, creating symmetry about the y-axis, not the x-axis. Option B is wrong because periodicity has no direct relationship to x-axis symmetry. A periodic function could create various symmetries or none at all. Option C is incorrect because if

f(y) = 0

everywhere, all slopes would be horizontal, creating a trivial case that's technically symmetric but not the general condition required. Study tip: Remember that x-axis symmetry in slope fields always corresponds to odd functions, while y-axis symmetry corresponds to even functions. The axis of symmetry tells you which variable gets negated in the function property.

Question 20

A slope field has the property that along the y-axis (where $x=0, y \neq 0$ ), all line segments are vertical. Which of the following differential equations could correspond to this slope field?

$dy/dx = x/y$
$dy/dx = y/x$ (correct answer)
$dy/dx = x-y$
$dy/dx = xy^2$

Explanation: Vertical line segments correspond to an undefined slope. We need to check which differential equation becomes undefined when

x=0

and

y \neq 0

. For

dy/dx = x/y

, when

x=0

, the slope is 0 (horizontal). For

dy/dx = y/x

, when

x=0

, the denominator is zero, so the slope is undefined (vertical). For

dy/dx = x-y

, when

x=0

, the slope is

-y

, which is not generally undefined. For

dy/dx = xy^2

, when

x=0

, the slope is 0 (horizontal). Therefore, only

dy/dx = y/x

matches the description.