AP Calculus AB: Slope Fields (Drill 1)

Question 1. The slope field for a differential equation is described as follows: at every point \( (x, y) \), the slope equals \( x - y \). What is the slope of the tangent line to a solution curve passing through the point \( (3, 1) \)?

• A) \( -2 \)
• B) \( 2 \) ✓
• C) \( 3 \)
• D) \( 4 \)

Explanation: Choice B is correct. The differential equation is \( \frac{dy}{dx} = x - y \). At \( (3,1) \), the slope is \( 3 - 1 = 2 \). Choice A is incorrect because the student subtracted in the wrong order, computing \( y - x = 1 - 3 = -2 \). Choice C is incorrect because the student used only the \( x \)-coordinate and ignored \( y \). Choice D is incorrect because the student added \( x + y = 3 + 1 = 4 \) rather than computing \( x - y \).

Question 2. A slope field has the following properties: along the line \( y = x \), all slopes equal \( 0 \); for points where \( y > x \), slopes are negative; for points where \( y < x \), slopes are positive. Which differential equation matches this slope field?

• A) \( \dfrac{dy}{dx} = y \)
• B) \( \dfrac{dy}{dx} = x \)
• C) \( \dfrac{dy}{dx} = x - y \) ✓
• D) \( \dfrac{dy}{dx} = xy \)

Explanation: Choice C is correct. The description requires slope \( = 0 \) when \( x = y \), positive when \( y x \), exactly the behavior of \( x - y \). Choice A is incorrect because \( \frac{dy}{dx} = y \) is zero only along the x-axis (\( y=0 \)), not along the entire line \( y = x \). Choice B is incorrect because \( \frac{dy}{dx} = x \) is zero only along the y-axis (\( x=0 \)), not along \( y = x \). Choice D is incorrect because \( \frac{dy}{dx} = xy \) is zero on both coordinate axes and does not produce the described sign pattern.

Question 3. The slope field for \( \dfrac{dy}{dx} = y \) is given. A student sketches the solution curve through the point \( (0, 2) \). Which of the following best describes the long-run behavior of this curve as \( x \to \infty \)?

• A) The curve approaches the x-axis.
• B) The curve approaches a horizontal asymptote at \( y = 2 \).
• C) The curve increases without bound. ✓
• D) The curve oscillates above and below \( y = 2 \).

Explanation: Choice C is correct. The general solution to \( \frac{dy}{dx} = y \) is \( y = Ce^x \). With initial condition \( (0,2) \), the particular solution is \( y = 2e^x \), which grows without bound as \( x \to \infty \). Choice A is incorrect because for \( y = Ce^x \), any nonzero value of \( C \) leads to exponential growth in magnitude as \( x \to \infty \), not decay toward zero. Choice B is incorrect because \( Ce^x \) never levels off to a finite asymptote. Choice D is incorrect because oscillation characterizes sinusoidal solutions, not solutions to \( \frac{dy}{dx} = y \).

Question 4. Consider the slope field for \( \dfrac{dy}{dx} = y(y - 3) \). Which of the following correctly identifies all equilibrium solutions and describes the behavior of a solution curve through \( (0, 1) \)?

• A) Equilibrium solutions at \( y = 0 \) and \( y = 3 \); the solution through \( (0, 1) \) decreases toward \( y = 0 \). ✓
• B) Equilibrium solutions at \( y = 0 \) and \( y = 3 \); the solution through \( (0, 1) \) increases toward \( y = 3 \).
• C) Equilibrium solution only at \( y = 0 \); the solution through \( (0, 1) \) decreases toward \( y = 0 \).
• D) Equilibrium solution only at \( y = 3 \); the solution through \( (0, 1) \) increases without bound.

Explanation: Choice A is correct. Setting \( y(y-3) = 0 \) gives equilibria at \( y = 0 \) and \( y = 3 \). At \( (0,1) \), \( \frac{dy}{dx} = 1(1-3) = -2 < 0 \), so the solution is decreasing. Since it cannot cross the equilibrium at \( y = 0 \), it decreases toward that equilibrium. Choice B is incorrect because the slope at \( y = 1 \) is negative, not positive. Choice C is incorrect because it omits \( y = 3 \) as an equilibrium. Choice D is incorrect because it omits \( y = 0 \) as an equilibrium and misidentifies the direction of motion from \( y = 1 \).

Question 5. A slope field has all of the following properties: slopes are zero along the y-axis (\( x = 0 \)); slopes are positive for \( x > 0 \) and negative for \( x < 0 \); the slope at \( (2, 5) \) equals the slope at \( (2, -3) \); for a fixed \( x \), slope does not change as \( y \) varies. Which differential equation is consistent with all four properties?

• A) \( \dfrac{dy}{dx} = xy \)
• B) \( \dfrac{dy}{dx} = x^2 \)
• C) \( \dfrac{dy}{dx} = x \) ✓
• D) \( \dfrac{dy}{dx} = x + y \)

Explanation: Choice C is correct. Check \( \frac{dy}{dx} = x \): zero when \( x=0 \) ✓; positive for \( x>0 \), negative for \( x<0 \) ✓; slope at \( (2,5) \) and \( (2,-3) \) both equal \( 2 \) ✓; slope depends only on \( x \), not \( y \) ✓. Choice A is incorrect because \( xy \) depends on \( y \), so slopes at \( (2,5) \) and \( (2,-3) \) differ (\( 10 \) vs. \( -6 \)), violating property 3. Choice B is incorrect because \( x^2 \geq 0 \) always, so slopes are never negative, violating property 2. Choice D is incorrect because \( x+y \) depends on \( y \), violating properties 3 and 4.