Calculus 2 Quiz: Reasoning With Slope Fields
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Reasoning With Slope FieldsQuestion 1 of 20

Consider the differential equation dydx=2x−y\frac{dy}{dx} = 2x-y with initial condition y(0)=1y(0)=1. If Euler's method with a step size of h=0.5h=0.5 is used to approximate y(1)y(1), the process is geometrically equivalent to:

finding the value at x=1x=1 on a single tangent line drawn from the point (0,1)(0,1).
following the tangent line from (0,1)(0,1) to find a point at x=0.5x=0.5, then following a new tangent line from that point to find a point at x=1x=1.
averaging the slopes calculated at x=0x=0 and x=1x=1 and using that average slope to move from the initial point.
constructing a quadratic approximation to the solution curve at (0,1)(0,1) and evaluating it at x=1x=1.
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Calculus 2 Quiz

Calculus 2 Quiz: Reasoning With Slope Fields

Practice Reasoning With 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.

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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 2.

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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

Consider the differential equation dydx=2x−y\frac{dy}{dx} = 2x-y with initial condition y(0)=1y(0)=1. If Euler's method with a step size of h=0.5h=0.5 is used to approximate y(1)y(1), the process is geometrically equivalent to:

  1. finding the value at x=1x=1 on a single tangent line drawn from the point (0,1)(0,1).
  2. following the tangent line from (0,1)(0,1) to find a point at x=0.5x=0.5, then following a new tangent line from that point to find a point at x=1x=1. (correct answer)
  3. averaging the slopes calculated at x=0x=0 and x=1x=1 and using that average slope to move from the initial point.
  4. constructing a quadratic approximation to the solution curve at (0,1)(0,1) and evaluating it at x=1x=1.
Explanation: Euler's method is a numerical procedure for approximating solutions of differential equations. Geometrically, it involves taking discrete steps. From a known point on the solution curve, we calculate the slope using the differential equation. We then move a small step (of size hh) along the tangent line to find the next approximate point. This process is then repeated from the new point. With a step size of 0.5 to get from x=0x=0 to x=1x=1, two steps are required. This corresponds to following one tangent line from x=0x=0 to x=0.5x=0.5, then recalculating the slope and following a new tangent line from x=0.5x=0.5 to x=1x=1.

Question 2

A slope field for dydx=f(x,y)\frac{dy}{dx} = f(x,y) has the property that along any vertical line x=cx = c, the slopes increase as yy increases. Additionally, along any horizontal line y=ky = k, the slopes decrease as xx increases. Which condition must be satisfied by f(x,y)f(x,y)?

  1. ∂f∂y>0\frac{\partial f}{\partial y} > 0 and ∂f∂x>0\frac{\partial f}{\partial x} > 0
  2. ∂f∂y<0\frac{\partial f}{\partial y} < 0 and ∂f∂x>0\frac{\partial f}{\partial x} > 0
  3. ∂f∂y>0\frac{\partial f}{\partial y} > 0 and ∂f∂x<0\frac{\partial f}{\partial x} < 0 (correct answer)
  4. ∂f∂y<0\frac{\partial f}{\partial y} < 0 and ∂f∂x<0\frac{\partial f}{\partial x} < 0
Explanation: When you encounter slope field problems involving how slopes change in different directions, you're working with partial derivatives of the function f(x,y)f(x,y) that defines dydx\frac{dy}{dx}. The key insight is connecting the given information to partial derivatives. Along a vertical line x=cx = c, only yy changes while xx stays constant. If slopes increase as yy increases, this means ∂f∂y>0\frac{\partial f}{\partial y} > 0. Think of this as: "How does the slope change when we move up vertically?" Since slopes get larger, the partial derivative with respect to yy is positive. Similarly, along a horizontal line y=ky = k, only xx changes while yy stays constant. If slopes decrease as xx increases, this means ∂f∂x<0\frac{\partial f}{\partial x} < 0. Moving rightward makes slopes smaller, so the partial derivative with respect to xx is negative. Therefore, we need ∂f∂y>0\frac{\partial f}{\partial y} > 0 and ∂f∂x<0\frac{\partial f}{\partial x} < 0, which is option C. Option A incorrectly suggests ∂f∂x>0\frac{\partial f}{\partial x} > 0, meaning slopes would increase (not decrease) as xx increases. Option B incorrectly has ∂f∂y<0\frac{\partial f}{\partial y} < 0, meaning slopes would decrease (not increase) as yy increases. Option D gets both signs wrong, reversing the behavior in both directions. Study tip: For slope field questions, always ask "What happens to the slope when I move in this direction?" Then translate that directly to the sign of the corresponding partial derivative. Increasing slopes mean positive partials, decreasing slopes mean negative partials.

Question 3

Consider the differential equation dydx=y(4−y)\frac{dy}{dx} = y(4-y). The slope field for this equation would show that solution curves passing through the region 0<y<20 < y < 2 are:

  1. increasing and concave up. (correct answer)
  2. increasing and concave down.
  3. decreasing and concave up.
  4. decreasing and concave down.
Explanation: First, determine the sign of the first derivative. For 0<y<20 < y < 2, both yy and 4−y4-y are positive, so dydx>0\frac{dy}{dx} > 0. Thus, the solution is increasing. Next, determine concavity by finding the second derivative: d2ydx2=ddx(4y−y2)=(4−2y)dydx=(4−2y)y(4−y)\frac{d^2y}{dx^2} = \frac{d}{dx}(4y - y^2) = (4 - 2y) \frac{dy}{dx} = (4-2y)y(4-y). In the region 0<y<20 < y < 2, the term (4−2y)(4-2y) is positive, yy is positive, and (4−y)(4-y) is positive. Therefore, d2ydx2>0\frac{d^2y}{dx^2} > 0, and the solution curve is concave up.

Question 4

A student observes a slope field and notes that all line segments on any given horizontal line are parallel to each other. Which of the following differential equations could represent the observed slope field?

  1. dydx=x2+1\frac{dy}{dx} = x^2 + 1
  2. dydx=y2−1\frac{dy}{dx} = y^2 - 1 (correct answer)
  3. dydx=xy\frac{dy}{dx} = xy
  4. dydx=x/y\frac{dy}{dx} = x/y
Explanation: If all line segments on any given horizontal line (where yy is constant) are parallel, it means the slope dydx\frac{dy}{dx} does not change with xx. This implies that the differential equation depends only on yy. Such an equation is called autonomous. Among the choices, only dydx=y2−1\frac{dy}{dx} = y^2 - 1 is a function of yy alone.

Question 5

The slope field for dydx=x−y2\frac{dy}{dx} = x-y^2 has line segments with a slope of -1. Which of the following equations describes the curve, known as an isocline, on which these segments lie?

  1. y=x+1y = \sqrt{x+1}
  2. y=x+1y = x+1
  3. x=y2−1x = y^2 - 1 (correct answer)
  4. x=−y2−1x = -y^2 - 1
Explanation: Isoclines are curves where the slope is constant. To find the isocline for a slope of -1, we set the differential equation equal to -1: x−y2=−1x - y^2 = -1. To describe this curve, we can solve for xx, which gives x=y2−1x = y^2 - 1. This equation represents a parabola opening to the right with its vertex at (−1,0)(-1, 0). Choice A only represents the top half of this parabola.

Question 6

The slope field of a differential equation dydx=f(x,y)\frac{dy}{dx} = f(x, y) is observed to be symmetric with respect to the y-axis, but it is not symmetric with respect to the x-axis. Which of the following could be the differential equation?

  1. dydx=x2y2\frac{dy}{dx} = x^2 y^2
  2. dydx=x3y\frac{dy}{dx} = x^3 y
  3. dydx=ycos⁡(x)\frac{dy}{dx} = y \cos(x) (correct answer)
  4. dydx=x2+y2\frac{dy}{dx} = x^2 + y^2
Explanation: Symmetry with respect to the y-axis means that the slope at (−x,y)(-x, y) is the same as at (x,y)(x, y), so f(−x,y)=f(x,y)f(-x, y) = f(x, y). This means ff must be an even function of xx. No symmetry with respect to the x-axis means that the slope at (x,−y)(x, -y) is not necessarily the same as at (x,y)(x, y), so f(x,−y)≠f(x,y)f(x, -y) \neq f(x, y) in general. Let's check the options: A) f(−x,y)=(−x)2y2=x2y2=f(x,y)f(-x,y) = (-x)^2y^2 = x^2y^2 = f(x,y) and f(x,−y)=x2(−y)2=x2y2=f(x,y)f(x,-y)=x^2(-y)^2=x^2y^2=f(x,y). Symmetric wrt both. B) f(−x,y)=(−x)3y=−x3y≠f(x,y)f(-x,y)=(-x)^3y = -x^3y \neq f(x,y). Not symmetric wrt y-axis. C) f(−x,y)=ycos⁡(−x)=ycos⁡(x)=f(x,y)f(-x,y)=y\cos(-x)=y\cos(x)=f(x,y) (symmetric wrt y-axis). f(x,−y)=−ycos⁡(x)≠f(x,y)f(x,-y)=-y\cos(x) \neq f(x,y) (not symmetric wrt x-axis). This matches. D) f(−x,y)=(−x)2+y2=x2+y2=f(x,y)f(-x,y)=(-x)^2+y^2=x^2+y^2=f(x,y) and f(x,−y)=x2+(−y)2=x2+y2=f(x,y)f(x,-y)=x^2+(-y)^2=x^2+y^2=f(x,y). Symmetric wrt both.

Question 7

Let S1S_1 be the slope field for dydx=f(x,y)\frac{dy}{dx} = f(x,y) and S2S_2 be the slope field for dydx=−1f(x,y)\frac{dy}{dx} = -\frac{1}{f(x,y)}. Geometrically, how are the line segments at any corresponding point (x,y)(x,y) in S1S_1 and S2S_2 related, assuming f(x,y)≠0f(x,y) \neq 0?

  1. They are parallel.
  2. They are perpendicular. (correct answer)
  3. They are reflections of each other across the x-axis.
  4. They are reflections of each other across the line y=xy=x.
Explanation: The slope of the line segments in S1S_1 at a point (x,y)(x,y) is m1=f(x,y)m_1 = f(x,y). The slope of the line segments in S2S_2 at the same point is m2=−1/f(x,y)m_2 = -1/f(x,y). Two lines with slopes m1m_1 and m2m_2 are perpendicular if and only if their product m1m2=−1m_1 m_2 = -1. Here, m1m2=f(x,y)⋅(−1f(x,y))=−1m_1 m_2 = f(x,y) \cdot \left(-\frac{1}{f(x,y)}\right) = -1. Therefore, the line segments at any given point are perpendicular. The solution curves of the second equation are the orthogonal trajectories of the solution curves of the first.

Question 8

The slope field for dydx=cos⁡(y)\frac{dy}{dx} = \cos(y) has horizontal line segments at certain y-values, corresponding to equilibrium solutions. Which statement accurately describes the stability of these solutions?

  1. All equilibrium solutions are stable.
  2. All equilibrium solutions are unstable.
  3. The equilibrium solutions are alternately stable and unstable. (correct answer)
  4. The equilibrium solutions are all semi-stable.
Explanation: Equilibrium solutions occur when dydx=0\frac{dy}{dx} = 0, so cos⁡(y)=0\cos(y) = 0. This happens at y=π2+nπy = \frac{\pi}{2} + n\pi for any integer nn. Let f(y)=cos⁡(y)f(y) = \cos(y). We can test stability by checking the sign of f(y)f(y) around the equilibria. Near y=π/2y=\pi/2, if y<π/2y < \pi/2, cos⁡(y)>0\cos(y)>0, and if y>π/2y > \pi/2, cos⁡(y)<0\cos(y)<0. Slopes point towards y=π/2y=\pi/2, so it's stable. Near y=3π/2y=3\pi/2, if y<3π/2y < 3\pi/2, cos⁡(y)<0\cos(y)<0, and if y>3π/2y > 3\pi/2, cos⁡(y)>0\cos(y)>0. Slopes point away from y=3π/2y=3\pi/2, so it's unstable. This pattern of stable then unstable repeats every π\pi, so the solutions are alternately stable and unstable.

Question 9

Consider the slope field for dydx=x2−y\frac{dy}{dx} = x^2 - y. A solution curve that passes through the point (1,0)(1, 0)...

  1. will have a local minimum at the point where it crosses the parabola y=x2y=x^2.
  2. will have a local maximum at the point where it crosses the parabola y=x2y=x^2. (correct answer)
  3. will approach the parabola y=x2y=x^2 as an asymptote.
  4. will decrease without bound.
Explanation: The slope of the solution curve is zero when dydx=x2−y=0\frac{dy}{dx} = x^2 - y = 0, which occurs along the parabola y=x2y=x^2. At the initial point (1,0)(1, 0), we are below the parabola since 0<120 < 1^2. The slope at this point is 12−0=11^2 - 0 = 1, which is positive. The solution is increasing. Since the curve starts below the parabola and is increasing, it must eventually cross the parabola y=x2y=x^2 at some point where x>1x>1. Just before crossing, y<x2y < x^2, so dydx>0\frac{dy}{dx} > 0 (increasing). Just after crossing, y>x2y > x^2, so dydx<0\frac{dy}{dx} < 0 (decreasing). A function that changes from increasing to decreasing has a local maximum.

Question 10

The slope field for dydx=f(x,y)\frac{dy}{dx} = f(x, y) shows that near the origin, the slopes are very close to 1. As xx and yy increase, the slopes appear to approach 0. Which of the following equations is most consistent with this description?

  1. dydx=e−x+e−y\frac{dy}{dx} = e^{-x} + e^{-y}
  2. dydx=cos⁡(x)cos⁡(y)\frac{dy}{dx} = \cos(x) \cos(y)
  3. dydx=1−xy\frac{dy}{dx} = 1 - xy
  4. dydx=11+x2+y2\frac{dy}{dx} = \frac{1}{1+x^2+y^2} (correct answer)
Explanation: When analyzing slope fields, you need to evaluate how the given differential equation behaves at specific points, particularly near the origin and as variables increase. Let's check each option at the origin (0,0) and see how slopes change as x and y grow larger. Option D, dydx=11+x2+y2\frac{dy}{dx} = \frac{1}{1+x^2+y^2}, gives us exactly what we need. At the origin: 11+02+02=1\frac{1}{1+0^2+0^2} = 1. As x and y increase, the denominator 1+x2+y21+x^2+y^2 grows much larger, making the entire fraction approach 0. This perfectly matches the described behavior. Option A, dydx=e−x+e−y\frac{dy}{dx} = e^{-x} + e^{-y}, gives e0+e0=2e^0 + e^0 = 2 at the origin, not 1. While it does approach 0 as variables increase, the initial value is wrong. Option B, dydx=cos⁡(x)cos⁡(y)\frac{dy}{dx} = \cos(x)\cos(y), equals cos⁡(0)cos⁡(0)=1\cos(0)\cos(0) = 1 at the origin, which is correct. However, as x and y increase, the cosine functions oscillate between -1 and 1, so the slope oscillates rather than consistently approaching 0. Option C, dydx=1−xy\frac{dy}{dx} = 1 - xy, gives the right value at the origin (1−0=11 - 0 = 1), but as x and y increase, xy grows rapidly, making the slope become increasingly negative rather than approaching 0. When interpreting slope fields, always test your differential equation at key points mentioned in the problem. Look for functions that smoothly transition between the described behaviors rather than oscillating or changing sign unexpectedly.

Question 11

For the differential equation dydx=y2−x\frac{dy}{dx} = y^2 - x, a student uses Euler's method with step size hh starting at (x0,y0)(x_0, y_0) where y02<x0y_0^2 < x_0. The first step yields the point (x1,y1)(x_1, y_1). Which statement must be true?

  1. y1<y0y_1 < y_0 (correct answer)
  2. y1>y0y_1 > y_0
  3. y1=y0y_1 = y_0
  4. The relationship between y1y_1 and y0y_0 depends on the value of hh.
Explanation: Euler's method formula is y1=y0+h⋅f(x0,y0)y_1 = y_0 + h \cdot f(x_0, y_0). Here, f(x,y)=y2−xf(x,y) = y^2 - x. So, y1=y0+h(y02−x0)y_1 = y_0 + h(y_0^2 - x_0). The problem states that y02<x0y_0^2 < x_0, which means y02−x0<0y_0^2 - x_0 < 0. Since the step size hh must be positive, the term h(y02−x0)h(y_0^2 - x_0) is negative. Therefore, y1=y0+(a negative number)y_1 = y_0 + (\text{a negative number}), which implies y1<y0y_1 < y_0. The geometric interpretation is that at (x0,y0)(x_0, y_0), the slope field indicates a negative slope, so the first step of Euler's method will move to a point with a smaller y-value.

Question 12

If a solution curve in the slope field of dydx=f(y)\frac{dy}{dx} = f(y) is translated horizontally, the resulting curve is also a solution curve. This property is a consequence of the differential equation being:

  1. linear.
  2. separable.
  3. autonomous. (correct answer)
  4. exact.
Explanation: The property described is time-invariance, which is characteristic of autonomous differential equations. An autonomous equation has the form dydx=f(y)\frac{dy}{dx} = f(y), where the rate of change depends only on the current state yy, not on the independent variable xx (often representing time). Geometrically, this means the slope field is the same for all xx. If you have a solution curve y(x)y(x), shifting it horizontally to y(x−c)y(x-c) results in a new curve. The slope of the new curve at xx is the same as the slope of the old curve at x−cx-c. Since the slope only depends on yy, and y(x−c)y(x-c) takes on the same y-values as y(x)y(x), the new curve will also satisfy the differential equation.

Question 13

The slope field for a certain differential equation shows that all tangent lines are horizontal along the parabola y=x2y = x^2 and that tangent lines are vertical on the line y=−1y = -1. Which of the following differential equations could correspond to this slope field?

  1. dydx=y−x2y+1\frac{dy}{dx} = \frac{y-x^2}{y+1} (correct answer)
  2. dydx=y+1y−x2\frac{dy}{dx} = \frac{y+1}{y-x^2}
  3. dydx=(y−x2)(y+1)\frac{dy}{dx} = (y-x^2)(y+1)
  4. dydx=y−x2y−1\frac{dy}{dx} = \frac{y-x^2}{y-1}
Explanation: Horizontal tangents occur where the slope dydx=0\frac{dy}{dx} = 0. This requires the numerator of the fraction to be zero. The condition is y=x2y = x^2, so the numerator should be proportional to y−x2y - x^2. Vertical tangents occur where the slope is undefined, which means the denominator of the fraction is zero. The condition is y=−1y = -1, so the denominator should be proportional to y+1y+1. Combining these gives dydx=ky−x2y+1\frac{dy}{dx} = k \frac{y-x^2}{y+1}. Choice A is the only one that matches this form.

Question 14

In a particular slope field, it is noted that all line segments lying on any line of the form y=x+Cy = x + C for any constant CC are parallel. Which of the following differential equations could produce this slope field?

  1. dydx=y−x+1\frac{dy}{dx} = y - x + 1 (correct answer)
  2. dydx=y+x+1\frac{dy}{dx} = y + x + 1
  3. dydx=xy\frac{dy}{dx} = xy
  4. dydx=y/x\frac{dy}{dx} = y/x
Explanation: The condition that slopes are constant on lines of the form y=x+Cy = x + C means the slope depends only on the value of y−xy-x. Rewriting the line equation as y−x=Cy - x = C makes this clear. We are looking for a differential equation of the form dydx=f(y−x)\frac{dy}{dx} = f(y-x). Choice A, dydx=(y−x)+1\frac{dy}{dx} = (y-x) + 1, is a function of y−xy-x. Choice B is a function of y+xy+x, so its slopes would be constant on lines of the form y+x=Cy+x=C.

Question 15

In the slope field for an autonomous differential equation dydx=f(y)\frac{dy}{dx} = f(y), slopes are positive for y>3y > 3, negative for 0<y<30 < y < 3, and zero for y=0y=0 and y=3y=3. What is the long-term behavior of a solution curve passing through the point (1,2)(1, 2)?

  1. The solution y(x)y(x) approaches 3 as x→∞x \to \infty.
  2. The solution y(x)y(x) approaches 0 as x→∞x \to \infty. (correct answer)
  3. The solution y(x)y(x) is periodic.
  4. The solution y(x)y(x) decreases without bound as x→∞x \to \infty.
Explanation: The point (1,2)(1, 2) is in the region 0<y<30 < y < 3, where the slopes are negative. This means that for a solution curve passing through this region, y(x)y(x) is a decreasing function. As xx increases, y(x)y(x) must decrease from its initial value of 2. The next lower equilibrium solution (where the slope is zero) is at y=0y=0. Therefore, the solution will approach y=0y=0 as x→∞x \to \infty. The equilibrium at y=3y=3 is unstable, and the equilibrium at y=0y=0 is stable.

Question 16

The slope field for an autonomous differential equation dydx=f(y)\frac{dy}{dx} = f(y) shows horizontal tangent lines at y=1y=1 and y=4y=4. Slopes are positive for y>4y > 4, negative for 1<y<41 < y < 4, and positive for y<1y < 1. How would the equilibrium solutions be classified?

  1. y=1y=1 is stable and y=4y=4 is unstable. (correct answer)
  2. y=1y=1 is unstable and y=4y=4 is stable.
  3. Both y=1y=1 and y=4y=4 are stable.
  4. Both y=1y=1 and y=4y=4 are unstable.
Explanation: For the equilibrium at y=1y=1: for y<1y < 1, slopes are positive, so solutions move up toward y=1y=1. For 1<y<41 < y < 4, slopes are negative, so solutions move down toward y=1y=1. Since solutions on both sides of y=1y=1 approach it, y=1y=1 is a stable equilibrium. For the equilibrium at y=4y=4: for 1<y<41 < y < 4, slopes are negative, so solutions move down away from y=4y=4. For y>4y > 4, slopes are positive, so solutions move up away from y=4y=4. Since solutions on both sides move away, y=4y=4 is an unstable equilibrium.

Question 17

Consider the slope field for dydx=ky\frac{dy}{dx} = ky, where kk is a constant. How does the steepness of the line segments change as one moves from left to right along any specific solution curve other than y=0y=0?

  1. The segments become steeper if k>0k>0 and less steep if k<0k<0. (correct answer)
  2. The segments become less steep if k>0k>0 and steeper if k<0k<0.
  3. The steepness of the segments does not change.
  4. The change in steepness depends on the specific solution curve.
Explanation: The solution to dydx=ky\frac{dy}{dx} = ky is y=Cekxy = Ce^{kx}. The slope at any point is kyky. Let's consider the case k>0k>0 and a solution curve with C>0C>0. As we move from left to right, xx increases, so y=Cekxy = Ce^{kx} increases. Since the slope is kyky, and both kk and yy are positive and increasing, the slope increases, meaning the segments get steeper. If k<0k<0, as xx increases, y=Cekxy = Ce^{kx} decreases toward 0. The magnitude of the slope, ∣ky∣|ky|, also decreases, so the segments become less steep. The same logic applies if C<0C<0. Thus, segments become steeper for k>0k>0 and less steep for k<0k<0.

Question 18

A slope field has positive slopes in Quadrants I and III, and negative slopes in Quadrants II and IV. Furthermore, the slopes are zero along the entire x-axis and y-axis (except possibly at the origin). Which of the following differential equations best matches this description?

  1. dydx=xy\frac{dy}{dx} = xy (correct answer)
  2. dydx=x/y\frac{dy}{dx} = x/y
  3. dydx=x−y\frac{dy}{dx} = x-y
  4. dydx=x+y\frac{dy}{dx} = x+y
Explanation: The sign pattern of slopes in the quadrants fits both dydx=xy\frac{dy}{dx} = xy and dydx=x/y\frac{dy}{dx} = x/y. In QI(+,+), both give +. In QII(-,+), both give -. In QIII(-,-), both give +. In QIV(+,-), both give -. We must use the second condition. Slopes are zero on the x-axis (where y=0y=0) and y-axis (where x=0x=0). For dydx=xy\frac{dy}{dx} = xy, if x=0x=0 or y=0y=0, the slope is 0. This matches. For dydx=x/y\frac{dy}{dx} = x/y, if y=0y=0, the slope is undefined (vertical tangent), which contradicts the description.

Question 19

A slope field for the differential equation dydx=f(x,y)\frac{dy}{dx} = f(x,y) shows that all slopes are positive in the first quadrant and negative in the second quadrant. Additionally, the slopes approach zero as yy approaches infinity for any fixed x>0x > 0. Which of the following could represent f(x,y)f(x,y)?

  1. f(x,y)=xy+1f(x,y) = \frac{x}{y+1} (correct answer)
  2. f(x,y)=x−yy+2f(x,y) = \frac{x-y}{y+2}
  3. f(x,y)=xyy2+1f(x,y) = \frac{xy}{y^2+1}
  4. f(x,y)=x+yx+1f(x,y) = \frac{x+y}{x+1}
Explanation: For f(x,y)=xy+1f(x,y) = \frac{x}{y+1}: In the first quadrant (x>0,y>0x > 0, y > 0), we have x>0x > 0 and y+1>0y+1 > 0, so f(x,y)>0f(x,y) > 0. In the second quadrant (x<0,y>0x < 0, y > 0), we have x<0x < 0 and y+1>0y+1 > 0, so f(x,y)<0f(x,y) < 0. As y→∞y \to \infty for fixed x>0x > 0, f(x,y)=xy+1→0f(x,y) = \frac{x}{y+1} \to 0. All conditions are satisfied. Choice B fails because slopes can be positive in the second quadrant when x−y<0x-y < 0 but y+2>0y+2 > 0 doesn't guarantee negative slopes. Choice C fails because it can be negative in the first quadrant when the denominator behavior doesn't match the limit condition properly. Choice D fails because slopes are positive in the second quadrant when x+y>0x+y > 0.

Question 20

A slope field for a differential equation has the property that slopes are steeper (more positive or more negative) farther from the xx-axis. Along the line y=2y = 2, all slopes are positive and equal to 6. Along the line y=−1y = -1, all slopes are negative and equal to -3. What is the most likely form of the differential equation?

  1. dydx=3y\frac{dy}{dx} = 3y (correct answer)
  2. dydx=y2+2y\frac{dy}{dx} = y^2 + 2y
  3. dydx=2y+2\frac{dy}{dx} = 2y + 2
  4. dydx=y2−y\frac{dy}{dx} = y^2 - y
Explanation: For dydx=3y\frac{dy}{dx} = 3y: When y=2y = 2, slope = 3(2)=63(2) = 6 ✓. When y=−1y = -1, slope = 3(−1)=−33(-1) = -3 ✓. The slopes are proportional to yy, so they get steeper farther from the x-axis ✓. Choice B: At y=2y = 2, slope = 4+4=8≠64 + 4 = 8 \neq 6. Choice C: At y=2y = 2, slope = 4+2=64 + 2 = 6 ✓, but at y=−1y = -1, slope = −2+2=0≠−3-2 + 2 = 0 \neq -3. Choice D: At y=2y = 2, slope = 4−2=2≠64 - 2 = 2 \neq 6.