Calculus 2 Quiz: Sketching Slope Fields
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Sketching Slope FieldsQuestion 1 of 20

A solution y(x)y(x) to the differential equation dydx=1x−y2\frac{dy}{dx} = \frac{1}{x} - y^2 passes through the point (1,1)(1, 1). What is the value of the second derivative, y′′(1)y''(1)?

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Calculus 2 Quiz

Calculus 2 Quiz: Sketching Slope Fields

Practice Sketching Slope Fields in Calculus 2 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 2.

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

A solution y(x)y(x) to the differential equation dydx=1x−y2\frac{dy}{dx} = \frac{1}{x} - y^2 passes through the point (1,1)(1, 1). What is the value of the second derivative, y′′(1)y''(1)?

  1. -2
  2. -1 (correct answer)
  3. 0
  4. 1
Explanation: First, evaluate the first derivative at the given point: y′(1)=11−(1)2=1−1=0y'(1) = \frac{1}{1} - (1)^2 = 1 - 1 = 0. Next, find the second derivative by differentiating the differential equation with respect to xx: y′′(x)=ddx(1x−y2)=−1x2−2ydydxy''(x) = \frac{d}{dx}(\frac{1}{x} - y^2) = -\frac{1}{x^2} - 2y \frac{dy}{dx}. Now, substitute the known values at x=1x=1: y(1)=1y(1)=1 and y′(1)=0y'(1)=0. y′′(1)=−112−2(1)(0)=−1−0=−1y''(1) = -\frac{1}{1^2} - 2(1)(0) = -1 - 0 = -1.

Question 2

In the slope field for dydx=y(2−y)\frac{dy}{dx} = y(2-y), the slope segments are horizontal when yy equals which values?

  1. y=2y = 2 only
  2. y=1y = 1 only
  3. y=0y = 0, y=1y = 1, and y=2y = 2
  4. y=0y = 0 and y=2y = 2 only (correct answer)
Explanation: When you encounter slope field questions, you're looking at the visual representation of a differential equation. The key insight is that slope segments are horizontal when the slope dydx=0\frac{dy}{dx} = 0. To find when the slope is zero, set the right side of the differential equation equal to zero: y(2−y)=0y(2-y) = 0. Using the zero product property, this equation is satisfied when either factor equals zero. So either y=0y = 0 or 2−y=02-y = 0, which gives us y=0y = 0 and y=2y = 2. At these yy-values, regardless of the xx-coordinate, every slope segment will be perfectly horizontal because the derivative equals zero at every point along these horizontal lines. Looking at the wrong answers: Choice A (y=2y = 2 only) misses the fact that y=0y = 0 also makes the slope zero. Choice B (y=1y = 1 only) represents a common error—when y=1y = 1, we get dydx=1(2−1)=1\frac{dy}{dx} = 1(2-1) = 1, so the slope is actually 1, not 0. Choice C (y=0y = 0, y=1y = 1, and y=2y = 2) incorrectly includes y=1y = 1 and demonstrates the same miscalculation as choice B. The correct answer is D: y=0y = 0 and y=2y = 2 only. Study tip: For slope field problems, always set dydx=0\frac{dy}{dx} = 0 to find horizontal tangent lines, and remember that these create the equilibrium solutions where the function neither increases nor decreases.

Question 3

Consider the slope field for dydx=sin⁡(x)+sin⁡(y)\frac{dy}{dx} = \sin(x) + \sin(y). At which point in the square region [0,2π]×[0,2π][0, 2\pi] \times [0, 2\pi] is the slope maximized?

  1. (π/2,π/2)(\pi/2, \pi/2) (correct answer)
  2. (π,π)(\pi, \pi)
  3. (3π/2,3π/2)(3\pi/2, 3\pi/2)
  4. (0,0)(0, 0)
Explanation: To maximize the slope dydx=sin⁡(x)+sin⁡(y)\frac{dy}{dx} = \sin(x) + \sin(y), we need to maximize both sin⁡(x)\sin(x) and sin⁡(y)\sin(y) independently. The maximum value of the sine function is 1. In the interval [0,2π][0, 2\pi], sin⁡(x)=1\sin(x)=1 occurs at x=π/2x=\pi/2, and sin⁡(y)=1\sin(y)=1 occurs at y=π/2y=\pi/2. Therefore, the maximum slope is 1+1=21+1=2, and it occurs at the point (π/2,π/2)(\pi/2, \pi/2).

Question 4

For the differential equation dydx=x−ln⁡(y)\frac{dy}{dx} = x - \ln(y), let m1m_1 be the slope of the solution curve at the point (2,e)(2, e) and m2m_2 be the slope at the point (3,e2)(3, e^2). Which of the following is true?

  1. m1<m2m_1 < m_2
  2. m1>m2m_1 > m_2
  3. m1=m2m_1 = m_2 (correct answer)
  4. The relationship between m1m_1 and m2m_2 depends on the specific solution curve.
Explanation: The slope at any point is determined directly by the differential equation, not by a specific solution curve. We calculate the slopes at the given points. At (2,e)(2, e), the slope is m1=2−ln⁡(e)=2−1=1m_1 = 2 - \ln(e) = 2 - 1 = 1. At (3,e2)(3, e^2), the slope is m2=3−ln⁡(e2)=3−2=1m_2 = 3 - \ln(e^2) = 3 - 2 = 1. Since both slopes are equal to 1, m1=m2m_1 = m_2. The two points lie on the same isocline.

Question 5

Consider the differential equation dydx=y−x2\frac{dy}{dx} = y - x^2. Solution curves to this equation can have local extrema. Which of the following statements correctly describes where these extrema can occur?

  1. Any local maximum must occur on the parabola y=x2y=x^2 in the region where x>0x > 0. (correct answer)
  2. Any local minimum must occur on the parabola y=x2y=x^2 in the region where x>0x > 0.
  3. Local extrema can occur at any point on the parabola y=x2y=x^2.
  4. Solution curves for this differential equation have no local extrema.
Explanation: Local extrema occur where dydx=0\frac{dy}{dx} = 0, which means y−x2=0y - x^2 = 0, or y=x2y = x^2. To classify these extrema, we use the second derivative test. d2ydx2=ddx(y−x2)=dydx−2x\frac{d^2y}{dx^2} = \frac{d}{dx}(y - x^2) = \frac{dy}{dx} - 2x. At a point on the parabola y=x2y=x^2, we have dydx=0\frac{dy}{dx}=0, so d2ydx2=0−2x=−2x\frac{d^2y}{dx^2} = 0 - 2x = -2x. For a local maximum, we need d2ydx2<0\frac{d^2y}{dx^2} < 0, which means −2x<0-2x < 0, or x>0x > 0. For a local minimum, we need d2ydx2>0\frac{d^2y}{dx^2} > 0, which means −2x>0-2x > 0, or x<0x < 0. Thus, local maxima occur on y=x2y=x^2 for x>0x>0.

Question 6

For the differential equation dydx=x2−y\frac{dy}{dx} = x^2 - y, in which region of the xyxy-plane are the solution curves concave down?

  1. Above the parabola y=x2−2xy = x^2 - 2x
  2. Below the parabola y=x2−2xy = x^2 - 2x (correct answer)
  3. Below the parabola y=x2y = x^2
  4. Wherever x<0x < 0
Explanation: Concavity is determined by the sign of the second derivative, d2ydx2\frac{d^2y}{dx^2}. We differentiate dydx\frac{dy}{dx} with respect to xx: d2ydx2=ddx(x2−y)=2x−dydx\frac{d^2y}{dx^2} = \frac{d}{dx}(x^2 - y) = 2x - \frac{dy}{dx}. Substitute the original equation for dydx\frac{dy}{dx}: d2ydx2=2x−(x2−y)=y−x2+2x\frac{d^2y}{dx^2} = 2x - (x^2 - y) = y - x^2 + 2x. Solution curves are concave down when d2ydx2<0\frac{d^2y}{dx^2} < 0, which means y−x2+2x<0y - x^2 + 2x < 0, or y<x2−2xy < x^2 - 2x. This inequality describes the region below the parabola y=x2−2xy = x^2 - 2x.

Question 7

The line segments of the slope field for the differential equation dydx=xyx2+y2−9\frac{dy}{dx} = \frac{xy}{x^2 + y^2 - 9} are vertical at certain points in the plane. These points form which of the following shapes?

  1. A pair of perpendicular lines, x=0x=0 and y=0y=0.
  2. A circle with center (0,0) and radius 9.
  3. A circle with center (0,0) and radius 3. (correct answer)
  4. A hyperbola defined by x2−y2=9x^2 - y^2 = 9.
Explanation: Vertical line segments correspond to an undefined slope. For a rational function, this occurs when the denominator is equal to zero. Setting the denominator to zero gives x2+y2−9=0x^2 + y^2 - 9 = 0, which can be rewritten as x2+y2=9x^2 + y^2 = 9. This is the equation of a circle centered at the origin with a radius of 9=3\sqrt{9} = 3.

Question 8

Consider the differential equation dydx=(y2−4)(x−1)\frac{dy}{dx} = (y^2 - 4)(x-1). A solution curve passes through the point (1,1)(1, 1). What is the nature of the solution curve at this point?

  1. The solution curve has a local maximum at x=1x=1. (correct answer)
  2. The solution curve has a local minimum at x=1x=1.
  3. The solution curve has a point of inflection at x=1x=1.
  4. The solution curve is constant in a neighborhood of x=1x=1.
Explanation: First, evaluate the slope at (1,1)(1,1): dydx=(12−4)(1−1)=(−3)(0)=0\frac{dy}{dx} = (1^2-4)(1-1) = (-3)(0) = 0. This indicates a critical point. To classify it, we examine the sign of dydx\frac{dy}{dx} around x=1x=1. For a point near (1,1)(1,1), yy is close to 1, so y2−4y^2-4 is negative. If x>1x > 1, then x−1>0x-1 > 0, so dydx=(neg)(pos)<0\frac{dy}{dx} = (\text{neg})(\text{pos}) < 0. If x<1x < 1, then x−1<0x-1 < 0, so dydx=(neg)(neg)>0\frac{dy}{dx} = (\text{neg})(\text{neg}) > 0. Since the derivative is positive to the left of x=1x=1 and negative to the right, the solution curve has a local maximum at x=1x=1.

Question 9

For the differential equation dydx=y2−4\frac{dy}{dx} = y^2 - 4, in which interval of yy-values are the solution curves both increasing and concave up?

  1. y>2y > 2 (correct answer)
  2. y<−2y < -2
  3. −2<y<0-2 < y < 0
  4. 0<y<20 < y < 2
Explanation: First, find where the solution is increasing: dydx>0  ⟹  y2−4>0  ⟹  y<−2\frac{dy}{dx} > 0 \implies y^2 - 4 > 0 \implies y < -2 or y>2y > 2. Second, find where the solution is concave up: d2ydx2>0\frac{d^2y}{dx^2} > 0. Using the chain rule, d2ydx2=ddx(y2−4)=2ydydx=2y(y2−4)\frac{d^2y}{dx^2} = \frac{d}{dx}(y^2-4) = 2y \frac{dy}{dx} = 2y(y^2-4). For this to be positive, yy and y2−4y^2-4 must have the same sign. Case 1: Both positive, y>0y>0 and (y<−2 or y>2)(y<-2 \text{ or } y>2), which gives y>2y>2. Case 2: Both negative, y<0y<0 and −2<y<2-2<y<2, which gives −2<y<0-2<y<0. So, concavity is up for y>2y>2 or −2<y<0-2<y<0. We need the intersection of the 'increasing' and 'concave up' regions. The only common region is y>2y>2.

Question 10

A solution curve for the differential equation dydx=1−xy\frac{dy}{dx} = 1 - xy passes through the origin (0,0)(0,0). Which of the following describes the behavior of the solution curve as it leaves the origin?

  1. It is initially horizontal and then its slope becomes negative.
  2. It has an initial slope of 1, and its slope continues to increase as it moves into the first quadrant.
  3. It has an initial slope of 1, and its slope decreases as it moves into the first quadrant. (correct answer)
  4. It is a vertical line passing through the origin.
Explanation: First, find the slope at the origin: dydx\frac{dy}{dx} at (0,0)(0,0) is 1−(0)(0)=11 - (0)(0) = 1. So the curve leaves the origin with a slope of 1, moving into the first quadrant where both xx and yy are positive. In the first quadrant, the term xyxy is positive. As xx and yy increase from 0, the product xyxy increases. Therefore, the slope, given by 1−xy1-xy, will decrease from its initial value of 1.

Question 11

A differential equation of the form dydx=f(x,y)\frac{dy}{dx} = f(x,y) is called homogeneous if f(tx,ty)=f(x,y)f(tx, ty) = f(x,y) for any t≠0t \neq 0. This property implies that the slope dydx\frac{dy}{dx} depends only on the ratio yx\frac{y}{x}. What geometric feature does the slope field of a homogeneous equation possess?

  1. Slopes are constant along any circle centered at the origin.
  2. Slopes are constant along any line passing through the origin. (correct answer)
  3. The slope field is symmetric with respect to both the x-axis and y-axis.
  4. All solution curves are lines passing through the origin.
Explanation: If the slope dydx\frac{dy}{dx} depends only on the ratio yx\frac{y}{x}, then for any line passing through the origin, given by y=mxy=mx, the ratio yx\frac{y}{x} is constant and equal to mm. Since the slope depends only on this ratio, the slope must be constant all along the line y=mxy=mx (excluding the origin). Therefore, the isoclines of a homogeneous equation are lines passing through the origin.

Question 12

Consider the differential equation dydx=2yx−x\frac{dy}{dx} = \frac{2y}{x} - x. What is the slope of any solution curve at a point where it intersects the parabola y=12x2y = \frac{1}{2}x^2?

  1. The slope is always -1.
  2. The slope is always 0. (correct answer)
  3. The slope is equal to the x-coordinate of the point.
  4. The slope is equal to the y-coordinate of the point.
Explanation: To find the slope of a solution curve at an intersection point with the parabola y=12x2y = \frac{1}{2}x^2, we substitute this expression for yy into the differential equation. dydx=2(12x2)x−x=x2x−x=x−x=0\frac{dy}{dx} = \frac{2(\frac{1}{2}x^2)}{x} - x = \frac{x^2}{x} - x = x - x = 0. This means that at any point where a solution curve intersects the parabola y=12x2y = \frac{1}{2}x^2, its slope is 0. This implies that all local extrema of the solution curves must lie on this parabola.

Question 13

The slope field for a differential equation has the property that all line segments along any vertical line are parallel to each other. Furthermore, the slopes are positive for x>1x > 1 and negative for x<1x < 1. Which of the following could be the differential equation?

  1. dydx=y−1\frac{dy}{dx} = y - 1
  2. dydx=x−1\frac{dy}{dx} = x - 1 (correct answer)
  3. dydx=(x−1)(y−1)\frac{dy}{dx} = (x-1)(y-1)
  4. dydx=1x−1\frac{dy}{dx} = \frac{1}{x-1}
Explanation: The property that slopes are constant along any vertical line means that the slope dydx\frac{dy}{dx} depends only on xx. This eliminates choices A and C, which depend on yy. The property that slopes are positive for x>1x > 1 and negative for x<1x < 1 means the expression must change sign at x=1x = 1. For choice B, x−1x-1 is positive when x>1x>1 and negative when x<1x<1, with a slope of 0 at x=1x=1. Choice D also changes sign correctly, but 1x−1\frac{1}{x-1} is undefined at x=1x=1, creating a discontinuity that would not appear in a typical slope field for a differential equation.

Question 14

The slope field for dydx=f(x,y)\frac{dy}{dx} = f(x,y) has positive slopes in quadrants I and III, and negative slopes in quadrants II and IV. Additionally, slopes are zero on the y-axis (for y≠0y \neq 0) and vertical on the x-axis (for x≠0x \neq 0). Which equation matches this description?

  1. dydx=xy\frac{dy}{dx} = xy
  2. dydx=yx\frac{dy}{dx} = \frac{y}{x}
  3. dydx=x+y\frac{dy}{dx} = x+y
  4. dydx=xy\frac{dy}{dx} = \frac{x}{y} (correct answer)
Explanation: The sign of xy\frac{x}{y} is positive when x,yx,y have the same sign (quadrants I and III) and negative when they have opposite signs (quadrants II and IV). Slopes are zero when the numerator is zero, so x=0x=0 (the y-axis). Slopes are vertical (undefined) when the denominator is zero, so y=0y=0 (the x-axis). This matches all the given conditions. Choice A has the correct sign behavior but no vertical slopes. Choice B has the correct sign behavior but the zero/vertical axes are swapped. Choice C does not have the correct sign behavior in all quadrants.

Question 15

The slope field for an autonomous differential equation dydx=f(y)\frac{dy}{dx} = f(y) indicates that solutions are strictly increasing for −1<y<3-1 < y < 3 and strictly decreasing elsewhere. Which of the following is a possible expression for f(y)f(y)?

  1. f(y)=(y−1)(y+3)f(y) = (y-1)(y+3)
  2. f(y)=(y+1)(y−3)f(y) = (y+1)(y-3)
  3. f(y)=(y+1)(3−y)f(y) = (y+1)(3-y) (correct answer)
  4. f(y)=y(y+1)(y−3)f(y) = y(y+1)(y-3)
Explanation: The information implies that f(y)>0f(y) > 0 on the interval (−1,3)(-1, 3) and f(y)<0f(y) < 0 for y<−1y < -1 or y>3y > 3. This means f(y)f(y) has roots at y=−1y=-1 and y=3y=3. This suggests a quadratic form k(y+1)(y−3)k(y+1)(y-3). Let's test the sign of (y+1)(y−3)(y+1)(y-3). In the interval (−1,3)(-1,3), picking y=0y=0, we get (1)(−3)=−3(1)(-3) = -3, which is negative. Since we need f(y)f(y) to be positive in this interval, we must multiply by a negative constant, e.g., k=−1k=-1. This gives f(y)=−(y+1)(y−3)=(y+1)(3−y)f(y) = -(y+1)(y-3) = (y+1)(3-y). This function is a downward-opening parabola with roots at -1 and 3, so it is positive between the roots and negative elsewhere, matching the description.

Question 16

For the differential equation dydx=xy\frac{dy}{dx} = \frac{x}{y}, which statement best describes the behavior of the slope field along the line y=xy = x?

  1. The slope segments have slope 1 and are parallel to the line y=xy = x (correct answer)
  2. The slope segments have slope 1 and are perpendicular to the line y=xy = x
  3. The slope segments have varying slopes that approach 1 as xx increases
  4. The slope segments are undefined at all points on the line y=xy = x
Explanation: Along the line y=xy = x, we substitute into the differential equation: dydx=xy=xx=1\frac{dy}{dx} = \frac{x}{y} = \frac{x}{x} = 1 (for x≠0x \neq 0). Since the slope of the line y=xy = x is also 1, the slope field segments are parallel to this line. Choice B is wrong because perpendicular lines would have slope -1. Choice C is incorrect because the slope is constantly 1, not varying. Choice D is wrong because the slope is well-defined except at the origin.

Question 17

For the differential equation dydx=cos⁡(x)⋅y\frac{dy}{dx} = \cos(x) \cdot y, at which point would the slope field show the steepest positive slope?

  1. (π,2)\left(\pi, 2\right)
  2. (π2,2)\left(\frac{\pi}{2}, 2\right)
  3. (0,2)\left(0, 2\right) (correct answer)
  4. (3π2,2)\left(\frac{3\pi}{2}, 2\right)
Explanation: When you encounter a slope field question, you're analyzing how steep the tangent lines are at different points based on the differential equation. The slope at any point (x,y)(x,y) is given directly by the right side of the equation: dydx=cos⁡(x)⋅y\frac{dy}{dx} = \cos(x) \cdot y. To find the steepest positive slope, you need to maximize the expression cos⁡(x)⋅y\cos(x) \cdot y. Since all answer choices have the same yy-value of 2, you're really looking for where cos⁡(x)\cos(x) is largest. The cosine function reaches its maximum value of 1 at x=0,2π,4π,x = 0, 2\pi, 4\pi, etc. At point (0,2)(0, 2), the slope equals cos⁡(0)⋅2=1⋅2=2\cos(0) \cdot 2 = 1 \cdot 2 = 2. This is the steepest positive slope among all the choices, making C correct. Let's check why the other options fall short. At point A (π,2)(\pi, 2), the slope is cos⁡(π)⋅2=(−1)⋅2=−2\cos(\pi) \cdot 2 = (-1) \cdot 2 = -2, which is negative, not positive. At point B (π2,2)(\frac{\pi}{2}, 2), the slope is cos⁡(π2)⋅2=0⋅2=0\cos(\frac{\pi}{2}) \cdot 2 = 0 \cdot 2 = 0, which is neither positive nor steep. At point D (3π2,2)(\frac{3\pi}{2}, 2), the slope is cos⁡(3π2)⋅2=0⋅2=0\cos(\frac{3\pi}{2}) \cdot 2 = 0 \cdot 2 = 0, again giving zero slope. Remember: for slope field problems, substitute the coordinates directly into the differential equation. The key is recognizing that maximizing cos⁡(x)⋅y\cos(x) \cdot y means finding where cosine is largest when yy values are equal.

Question 18

Consider the differential equation dydx=4x2+y22x2\frac{dy}{dx} = \frac{4x^2 + y^2}{2x^2}. What is the limiting behavior of the slopes of its slope field as ∣x∣→∞|x| \to \infty along the line y=2xy=2x?

  1. The slope approaches 0.
  2. The slope approaches 2.
  3. The slope approaches 4. (correct answer)
  4. The slope becomes infinitely large.
Explanation: To find the limiting behavior of the slope along the line y=2xy=2x, we substitute y=2xy=2x into the expression for dydx\frac{dy}{dx} and evaluate the limit as ∣x∣→∞|x| \to \infty. dydx=4x2+(2x)22x2=4x2+4x22x2=8x22x2=4\frac{dy}{dx} = \frac{4x^2 + (2x)^2}{2x^2} = \frac{4x^2 + 4x^2}{2x^2} = \frac{8x^2}{2x^2} = 4 for x≠0x \neq 0. Since the expression simplifies to the constant 4, the slope is 4 at every point on the line y=2xy=2x (except the origin). Therefore, the limiting slope is 4.

Question 19

Consider the autonomous differential equation dydx=y(3−y)\frac{dy}{dx} = y(3-y). Which statement best describes the line segments in the slope field for very large positive values of yy?

  1. They become nearly horizontal with a slope approaching zero.
  2. They become very steep with a large positive slope.
  3. They become very steep with a large negative slope. (correct answer)
  4. Their slope approaches -1.
Explanation: For very large positive values of yy, the term (3−y)(3-y) becomes a large negative number, while yy is a large positive number. Their product, y(3−y)y(3-y), will be a large negative number. For example, if y=100y=100, dydx=100(3−100)=−9700\frac{dy}{dx} = 100(3-100) = -9700. As y→∞y \to \infty, dydx≈−y2→−∞\frac{dy}{dx} \approx -y^2 \to -\infty. Therefore, the line segments become very steep with a large negative slope.

Question 20

Consider the slope field for the differential equation dydx=∣x∣−y\frac{dy}{dx} = |x| - y. The set of all points where the line segments are horizontal forms which shape?

  1. The line y=xy=x for x≥0x \ge 0 and y=−xy=-x for x<0x < 0. (correct answer)
  2. A parabola opening upwards with its vertex at the origin.
  3. The x-axis and the y-axis.
  4. The single point (0,0)(0,0).
Explanation: Horizontal line segments occur where the slope dydx\frac{dy}{dx} is zero. Setting the equation to zero gives ∣x∣−y=0|x| - y = 0, which is equivalent to y=∣x∣y = |x|. The graph of y=∣x∣y = |x| is a V-shape composed of two lines: y=xy=x for x≥0x \ge 0 and y=−xy=-x for x<0x < 0.