Calculus 1 Quiz: Reasoning With Slope Fields

20 questions · exam conditions

0:00

Reasoning With Slope FieldsQuestion 1 of 20

A solution $y(x)$ to the differential equation $dy/dx = \sin(x) - y$ passes through the origin $(0,0)$ . What feature does the solution curve have at this point?

A local maximum.

A local minimum.

A point of inflection.

A vertical tangent.

← Back to quizzes

Calculus 1 Quiz

Calculus 1 Quiz: Reasoning With Slope Fields

Practice Reasoning With 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 Reasoning With 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.

All questions

Question 1

A solution $y(x)$ to the differential equation $dy/dx = \sin(x) - y$ passes through the origin $(0,0)$ . What feature does the solution curve have at this point?

A local maximum.
A local minimum. (correct answer)
A point of inflection.
A vertical tangent.

Explanation: To analyze the behavior at

(0,0)

, we check the first and second derivatives. First derivative (slope):

dy/dx|_{(0,0)} = \sin(0) - 0 = 0

. A zero slope indicates a horizontal tangent, meaning there is a local extremum or a horizontal inflection point. To distinguish these, we use the second derivative test.

d^2y/dx^2 = d/dx(\sin(x) - y) = \cos(x) - dy/dx

. Evaluating at

(0,0)

, and using the fact that

dy/dx=0

there, we get

d^2y/dx^2|_{(0,0)} = \cos(0) - 0 = 1

. Since the second derivative is positive, the curve is concave up at

(0,0)

. A horizontal tangent combined with concave up behavior signifies a local minimum.

Question 2

A slope field is generated by a differential equation $dy/dx = g(x)$ , where $g(x)$ is a continuous function. Which statement describes the fundamental relationship between the solution curves $y(x)$ ?

All solution curves are always increasing.
All solution curves have the same concavity everywhere.
All solution curves are vertical shifts of one another. (correct answer)
All solution curves intersect at exactly one point.

Explanation: The differential equation is

y'(x) = g(x)

. The general solution is

y(x) = \int g(x) dx = G(x) + C

, where

G(x)

is an antiderivative of

g(x)

and

C

is the constant of integration. The family of solutions

y(x) = G(x) + C

represents curves that are all vertical shifts of each other by the constant

C

. This is the fundamental structural property of differential equations of the form

dy/dx = g(x)

Question 3

Consider a solution $y(x)$ to $dy/dx = (y-1)(y-3)$ with initial condition $y(0)=2$ . Another function $z(x)$ is defined as $z(x) = 1/y(x)$ . Which statement describes the initial behavior of $z(x)$ at $x=0$ ?

$z(x)$ is increasing at $x=0$ . (correct answer)
$z(x)$ is decreasing at $x=0$ .
$z(x)$ has a local minimum at $x=0$ .
$z(x)$ has a local maximum at $x=0$ .

Explanation: First, analyze

y(x)

x=0

. The initial value is

y(0)=2

. The slope is

dy/dx|_{(0,2)} = (2-1)(2-3) = (1)(-1) = -1

. So,

y(x)

is decreasing at

x=0

. Now, we analyze

z(x) = 1/y(x)

. To find its initial behavior, we find its derivative,

dz/dx

, using the chain rule:

dz/dx = d/dx(y^{-1}) = -1 \cdot y^{-2} \cdot (dy/dx) = -\frac{1}{y^2}\frac{dy}{dx}

. Now, evaluate

dz/dx

x=0

. We know

y(0)=2

and

dy/dx|_{(x=0)} = -1

. So,

dz/dx|_{(x=0)} = -\frac{1}{(2)^2}(-1) = -\frac{1}{4}(-1) = \frac{1}{4}

. Since

dz/dx > 0

x=0

, the function

z(x)

is increasing at

x=0

Question 4

A slope field is generated for the differential equation $dy/dx = x - y$ . For a particular solution curve that passes through a point in the second quadrant (where $x < 0$ and $y > 0$ ), what must be true about the concavity of this solution curve while it remains in the second quadrant?

The curve is always concave up. (correct answer)
The curve is always concave down.
The curve changes concavity from up to down.
The concavity cannot be determined from the information given.

Explanation: To determine concavity, we find the second derivative,

d^2y/dx^2

. Given

dy/dx = x - y

, we differentiate with respect to

x

d^2y/dx^2 = d/dx(x - y) = 1 - dy/dx

. Substituting the original equation gives

d^2y/dx^2 = 1 - (x - y) = 1 - x + y

. In the second quadrant,

x < 0

and

y > 0

. Therefore,

-x > 0

and

y > 0

. The expression for the second derivative,

1 + (-x) + y

, is the sum of three positive terms and thus must be positive. A positive second derivative implies the solution curve is concave up.

Question 5

The slope field for an autonomous differential equation $dy/dx = f(y)$ has horizontal line segments on the lines $y=1$ and $y=4$ . For values of $y$ between 1 and 4, the slopes are positive. For values of $y$ greater than 4 or less than 1, the slopes are negative. If a solution curve $y(x)$ passes through the point $(0, 2)$ , what is $\lim_{x \to \infty} y(x)$ ?

1
2
4 (correct answer)
$\infty$

Explanation: The horizontal segments at

y=1

and

y=4

represent equilibrium solutions. The initial condition is

y(0)=2

, which is between these equilibria. In this region (

1 < y < 4

), the slopes are positive, so the solution

y(x)

will increase as

x

increases. As an increasing function bounded above by the equilibrium solution

y=4

, the solution curve must approach

y=4

as a horizontal asymptote. Therefore, the limit as

x \to \infty

is 4. The equilibrium at

y=1

is unstable, while

y=4

is a stable equilibrium for solutions starting below it.

Question 6

The slope field for an autonomous differential equation $dy/dx = f(y)$ has positive slopes for all $y$ . Furthermore, for any given $x$ , the slopes of the line segments are observed to decrease as $y$ increases. What must be true about any solution curve $y(x)$ ?

It is always increasing and always concave up.
It is always increasing and always concave down. (correct answer)
It is always decreasing and always concave up.
Its concavity cannot be determined from the information given.

Explanation: The statement 'positive slopes for all

y

' means

dy/dx = f(y) > 0

for all

y

. This implies that any solution

y(x)

is always increasing. The statement 'slopes decrease as

y

increases' means that

f(y)

is a decreasing function of

y

, so its derivative

f'(y)

must be negative. To find the concavity, we calculate the second derivative:

d^2y/dx^2 = d/dx(f(y)) = f'(y) \cdot dy/dx = f'(y) \cdot f(y)

. Since

f'(y) < 0

and

f(y) > 0

, their product

d^2y/dx^2

is always negative. Therefore, any solution curve is always concave down.

Question 7

The slope field for $dy/dx = f(x, y)$ is symmetric with respect to the origin, meaning the slope at $(-x, -y)$ is the same as at $(x, y)$ . Additionally, slopes are zero on the x-axis (for $x \ne 0$ ) and undefined on the y-axis (for $y \ne 0$ ). Which equation could be $f(x, y)$ ?

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

Explanation: The condition of symmetry with respect to the origin means

f(-x, -y) = f(x, y)

. Let's test the options. A)

(-x)(-y) = xy

. B)

(-x)^2 + (-y)^2 = x^2+y^2

. C)

-x-y = -(x+y) \ne x+y

. D)

-y)/(-x) = y/x

. So A, B, and D satisfy the symmetry. Now check the axis conditions. A)

xy

is zero on both axes. B)

x^2+y^2

is zero only at the origin. D)

y/x

is zero when

y=0

(x-axis) and undefined when

x=0

(y-axis). Only

y/x

satisfies all three given properties.

Question 8

The slope field for $dy/dx = y(L-y)$ for some constant $L>0$ has equilibrium solutions at $y=0$ and $y=L$ . Let $y_1(x)$ be the solution with initial condition $y_1(0) = L/4$ and $y_2(x)$ be the solution with initial condition $y_2(0)=y_1(1)$ . What is the long-term behavior of $y_2(x)$ ?

$\lim_{x \to \infty} y_2(x) = 0$
$\lim_{x \to \infty} y_2(x) = L/4$
$\lim_{x \to \infty} y_2(x) = L$ (correct answer)
$\lim_{x \to \infty} y_2(x) = \infty$

Explanation: First, analyze the stability of the equilibria. For

0 < y < L

, both

y

and

L-y

are positive, so

dy/dx > 0

. This means solutions starting between 0 and L will increase. For

y > L

y < 0

dy/dx < 0

. This means

y=L

is a stable equilibrium and

y=0

is an unstable equilibrium. The first solution,

y_1(x)

, starts at

y_1(0) = L/4

. Since

0 < L/4 < L

y_1(x)

will be an increasing function. Therefore, at

x=1

, its value

y_1(1)

will be greater than its starting value, so

L/4 < y_1(1) < L

. The second solution,

y_2(x)

, starts at

y_2(0) = y_1(1)

. Since this starting value is also between 0 and L,

y_2(x)

will also increase and approach the stable equilibrium at

y=L

. Thus,

\lim_{x \to \infty} y_2(x) = L

Question 9

Let $S_1$ be the slope field for the differential equation $dy/dx = x/y$ and $S_2$ be the slope field for $dy/dx = -y/x$ . What is the geometric relationship between the line segments at any corresponding point $(x, y)$ (where $x, y \ne 0$ ) in the two slope fields?

The line segments are parallel.
The line segments are perpendicular. (correct answer)
The line segments are reflections across the line $y=x$ .
One slope is the negative of the other.

Explanation: Let

m_1

be the slope at a point in

S_1

and

m_2

be the slope at the same point in

S_2

. We have

m_1 = x/y

and

m_2 = -y/x

. To check for perpendicularity, we multiply the slopes:

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

. Two lines are perpendicular if and only if the product of their slopes is -1. Therefore, the line segments at any corresponding point in the two slope fields are perpendicular.

Question 10

The first step of Euler's method to approximate a solution for $dy/dx = f(x,y)$ with step size $h$ from an initial point $(x_0, y_0)$ is $y_1 = y_0 + h \cdot f(x_0, y_0)$ . How does this step conceptually relate to the slope field of the differential equation?

It finds the average slope in the neighborhood of $(x_0, y_0)$ .
It calculates the concavity of the solution curve at $(x_0, y_0)$ .
It follows the tangent line segment from the slope field at $(x_0, y_0)$ for a horizontal distance $h$ . (correct answer)
It identifies the location of the nearest equilibrium solution to $(x_0, y_0)$ .

Explanation: The slope field at the point

(x_0, y_0)

consists of a short line segment with slope

m = f(x_0, y_0)

. This slope is the same as the slope of the tangent line to the true solution curve at that point. The equation of this tangent line is

L(x) = y_0 + m(x-x_0)

. Euler's method approximates the solution at

x_1 = x_0+h

by finding the value on this tangent line:

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

. This is precisely a linear step along the direction indicated by the slope field's line segment.

Question 11

Consider the slope field for the differential equation $dy/dx = 1 - e^{-y}$ . Let $y_1(x)$ be the solution with $y_1(0) = 1$ and $y_2(x)$ be the solution with $y_2(0) = -1$ . Which of the following statements is true for $x > 0$ ?

Both $y_1(x)$ and $y_2(x)$ are increasing.
Both $y_1(x)$ and $y_2(x)$ are decreasing.
$y_1(x)$ is increasing and $y_2(x)$ is decreasing. (correct answer)
$y_1(x)$ is decreasing and $y_2(x)$ is increasing.

Explanation: We need to determine the sign of

dy/dx

for the initial conditions. For

y_1(x)

, the initial value is

y=1

. At this value,

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

. Since

e \approx 2.718 > 1

0 < 1/e < 1

, so

dy/dx > 0

. Therefore,

y_1(x)

is increasing. For

y_2(x)

, the initial value is

y=-1

. At this value,

dy/dx = 1 - e^{-(-1)} = 1 - e

. Since

e > 1

1-e < 0

, so

dy/dx < 0

. Therefore,

y_2(x)

is decreasing. The correct statement is that

y_1(x)

is increasing and

y_2(x)

is decreasing.

Question 12

The slope field for $dy/dx = f(x, y)$ has the property that for any point $(x, y)$ with $y > 0$ , the slope is equal to the slope at $(x, -y)$ . What does this imply about the function $f(x, y)$ ?

$f(x, -y) = f(x, y)$ ; $f$ is an even function of $y$ . (correct answer)
$f(x, -y) = -f(x, y)$ ; $f$ is an odd function of $y$ .
$f(-x, y) = f(x, y)$ ; $f$ is an even function of $x$ .
$f(-x, y) = -f(x, y)$ ; $f$ is an odd function of $x$ .

Explanation: The slope at a point

(x, y)

is given by the value of

f(x, y)

. The problem states that the slope at

(x, -y)

is the same as the slope at

(x, y)

. This translates directly to the mathematical statement

f(x, -y) = f(x, y)

. A function that satisfies this condition is said to be an even function with respect to the variable

y

. This property corresponds to symmetry in the slope field across the x-axis.

Question 13

A slope field for $dy/dx = f(x, y)$ shows that all solution curves in the upper half-plane ( $y>0$ ) are parabolas of the form $y = c - x^2$ for different constants $c$ . What must $f(x, y)$ be?

$-2x$ (correct answer)
$y + x^2$
$-2x/y$
$c - 2x$

Explanation: If

y = c - x^2

is the general form of the solution curves, we can find the differential equation by differentiating this expression with respect to

x

. Differentiating

y = c - x^2

gives

dy/dx = -2x

. The resulting differential equation,

dy/dx = -2x

, does not depend on

y

or the constant

c

. This means that for any point

(x, y)

on any of these solution curves, the slope is determined solely by the x-coordinate and is equal to

-2x

. Therefore,

f(x, y) = -2x

. The other options would arise from incorrect differentiation or algebraic manipulation.

Question 14

The slope field for a differential equation has the property that all line segments along any non-vertical line passing through the origin (i.e., any line $y=mx$ ) have the same slope. Which of the following differential equations could correspond to this slope field?

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

Explanation: The condition that slopes are constant along any line

y=mx

means that the value of

dy/dx

depends only on the ratio

m = y/x

, not on the specific values of

x

and

y

. The differential equation must be a function of the form

dy/dx = g(y/x)

. Let's check the options: A)

x+y = x+mx = x(1+m)

depends on

x

. B)

xy = x(mx) = mx^2

depends on

x

. C)

y/x = mx/x = m

depends only on the ratio

y/x

. D)

y-x^2 = mx-x^2 = x(m-x)

depends on

x

. Only

dy/dx = y/x

satisfies the property.

Question 15

A slope field has positive slopes in the first and third quadrants and negative slopes in the second and fourth quadrants. Slopes are zero on both the x-axis and y-axis. A solution curve $y(x)$ passes through the point $(-2, 1)$ . Which statement best describes the behavior of this solution?

The solution has a local minimum on the y-axis. (correct answer)
The solution has a local maximum on the y-axis.
The solution is always decreasing for all $x$ .
The solution approaches a horizontal asymptote as $x \to \infty$ .

Explanation: The point

(-2, 1)

is in the second quadrant, where slopes are negative. Thus, the solution

y(x)

is decreasing at

x=-2

. As

x

increases from -2, the curve moves to the right and down. It will cross the y-axis at some point

(0, c)

where

0 < c < 1

. On the y-axis, the slope is zero, so the function has a horizontal tangent at

x=0

. For

x > 0

, the solution curve enters the first quadrant (since

y

is still positive), where slopes are positive. This means the solution begins to increase. A function that is decreasing before

x=0

and increasing after

x=0

has a local minimum at

x=0

Question 16

The line $y=2x$ is a solution to a certain differential equation. The slope field for this differential equation is observed to have slopes that depend only on the difference $y-2x$ . Which of the following could be the differential equation?

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

Explanation: Let the differential equation be

dy/dx = f(y-2x)

. We are given that

y=2x

is a solution. We can substitute this into the differential equation to check for consistency. The left side is

dy/dx = d/dx(2x) = 2

. The right side is

f(y-2x) = f(2x-2x) = f(0)

. For

y=2x

to be a solution, we must have

2 = f(0)

. Now we check which of the given options satisfies this condition. A)

f(u) = u^2+2

f(0)=0^2+2=2

. This works. B)

f(u) = u

f(0)=0 \ne 2

. This fails. C)

f(u) = \cos(u)

f(0)=\cos(0)=1 \ne 2

. This fails. D) This is

dy/dx=2

. The solution is

y=2x+C

y=2x

is one of these solutions. The slopes are constant, which is a function of

y-2x

, e.g.,

f(u)=2

. But

f(u)=u^0+1

is not

f(u)=(y-2x)^0+1

. The description says slopes DEPEND on

y-2x

dy/dx=2

implies the slopes are constant and do not depend on anything. Option A has slopes that explicitly depend on the value of

y-2x

and satisfies the condition that

y=2x

is a solution.

Question 17

The slope field for the differential equation $dy/dx = \sqrt{x^2+y^2} - 2$ has line segments with a slope of zero along a specific curve. What is the shape of this curve?

A parabola.
A circle. (correct answer)
A hyperbola.
A pair of intersecting lines.

Explanation: Slopes of zero occur where

dy/dx = 0

. Setting the given differential equation to zero gives

\sqrt{x^2+y^2} - 2 = 0

. Adding 2 to both sides gives

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

. Squaring both sides yields

x^2 + y^2 = 4

. This is the equation of a circle centered at the origin with a radius of 2. This curve is the isocline for a slope of zero.

Question 18

A slope field displays vertical line segments at every point on the parabola $y = x^2$ and nowhere else. Which of the following differential equations could generate this slope field?

$dy/dx = y - x^2$
$dy/dx = \frac{y-x^2}{x}$
$dy/dx = \frac{x}{y-x^2}$ (correct answer)
$dy/dx = \frac{y}{x^2}$

Explanation: Vertical line segments in a slope field correspond to points where the slope

dy/dx

is undefined. This typically occurs when the denominator of the expression for

dy/dx

is zero. We are looking for an equation where the denominator is zero precisely when

y = x^2

. The expression

y - x^2

is zero on this parabola. Option C,

dy/dx = x/(y-x^2)

, has a denominator of

y-x^2

, which causes undefined slopes exactly on the parabola

y = x^2

. Option A has zero slopes there. Option B has zero slopes there and vertical slopes on the y-axis. Option D has vertical slopes on the y-axis.

Question 19

A solution $y(x)$ to the differential equation $dy/dx = \cos(\pi y) + x$ passes through the point $(1, 0.5)$ . Which of the following statements is true about the solution curve at this point?

It is increasing and concave up.
It is increasing and concave down. (correct answer)
It is decreasing and concave up.
It is decreasing and concave down.

Explanation: First, evaluate the slope

dy/dx

(1, 0.5)

dy/dx = \cos(\pi \cdot 0.5) + 1 = \cos(\pi/2) + 1 = 0 + 1 = 1

. Since the slope is positive, the solution is increasing. Next, find the second derivative to determine concavity:

d^2y/dx^2 = d/dx(\cos(\pi y) + x) = -\sin(\pi y) \cdot (\pi \cdot dy/dx) + 1

. Now evaluate this at

(1, 0.5)

, using the fact that

dy/dx = 1

at this point:

d^2y/dx^2 = -\sin(\pi \cdot 0.5) \cdot (\pi \cdot 1) + 1 = -\sin(\pi/2) \cdot \pi + 1 = -1 \cdot \pi + 1 = 1 - \pi

. Since

\pi > 1

1 - \pi < 0

, so the second derivative is negative. This means the solution is concave down. Therefore, the solution is increasing and concave down at

(1, 0.5)

Question 20

In the slope field for an autonomous differential equation $dy/dx = f(y)$ , the slopes of the line segments depend only on $y$ . What does this property imply about the family of solution curves?

All solution curves are horizontal shifts of one another. (correct answer)
All solution curves are symmetric with respect to the y-axis.
Every solution curve must be a periodic function.
The x-axis must be an equilibrium solution.

Explanation: The property that

dy/dx

does not depend on

x

is the definition of an autonomous equation. If

y(x)

is a solution, consider the function

z(x) = y(x-c)

for some constant

c

. Using the chain rule,

dz/dx = y'(x-c) \cdot 1

. Since

y

is a solution,

y'(x-c) = f(y(x-c)) = f(z)

. So,

dz/dx = f(z)

, which means

z(x)

is also a solution. The graph of

z(x) = y(x-c)

is a horizontal shift of the graph of

y(x)

. Therefore, the entire family of non-constant solutions consists of horizontal shifts of any single solution curve.