Difficulty: Intermediate | Prerequisites: Slope Fields core concepts, separation of variables.
This companion guide walks through every problem from the CALC 101 Slope Fields FLIP worksheet with full solutions and reasoning. It covers drawing slope fields from a DE, matching fields to equations, identifying solution curves, and the multi-part free-response problems that pair slope field sketching with tangent line approximation and separation of variables.
For each problem, the method is the same: plug sample (x, y) points into the DE, compute the slope, draw short segments. Below is the reasoning for each.
Q: Draw the slope field for dy/dx = x + 1.
A: The slope depends only on x. Every point in the same vertical column has the same slope. At x = -2, slope = -1. At x = -1, slope = 0 (horizontal segments). At x = 0, slope = 1. At x = 1, slope = 2. At x = 2, slope = 3. The field tilts progressively from lower-left to upper-right, with a column of horizontal segments at x = -1.
Q: Draw the slope field for dy/dx = 2y.
A: The slope depends only on y. Every point in the same horizontal row has the same slope. At y = 0, slope = 0 (horizontal along the x-axis). At y = 1, slope = 2. At y = -1, slope = -2. Above the x-axis, slopes are positive and grow steeper as y increases. Below, slopes are negative and steepen as y decreases. This is the classic exponential growth/decay field.
Q: Draw the slope field for dy/dx = x + y.
A: The slope depends on both x and y. The zero-slope isocline is the line y = -x. Above this line, slopes are positive. Below, they are negative. At (1, 1), slope = 2. At (-1, -1), slope = -2. At (1, -1), slope = 0. The field shows solution curves that look like shifted exponentials.
Q: Draw the slope field for dy/dx = 2x.
A: Depends only on x. At x = 0, slope = 0 (horizontal segments along the y-axis). At x = 1, slope = 2. At x = -1, slope = -2. The field is symmetric about the y-axis in magnitude but with opposite signs. Solution curves are parabolas y = x² + C.
Q: Draw the slope field for dy/dx = y - 1.
A: Depends only on y. At y = 1, slope = 0 (horizontal segments along the line y = 1). Above y = 1, slopes are positive. Below y = 1, slopes are negative. The line y = 1 is an equilibrium solution. Solution curves are exponentials shifted up by 1: y = Ce^x + 1.
Q: Draw the slope field for dy/dx = -y/x.
A: Depends on both x and y. Undefined along the y-axis (x = 0). At (1, 1), slope = -1. At (1, 2), slope = -2. At (2, 1), slope = -0.5. The solution curves are of the form xy = C (hyperbolas) and y = 0. Along the line y = 0 (except x = 0), slope = 0.
The four slope fields (A, B, C, D) must be matched to: dy/dx = sin x, dy/dx = x - y, dy/dx = 2 - y, dy/dx = x.
Q: Match slope field (A) to its DE from {sin x, x - y, 2 - y, x}.
A: Field (A) shows horizontal segments along a horizontal line and slopes that reverse direction periodically. This matches dy/dx = sin x (problem 7). The slopes are zero at x = 0, π, -π, and oscillate between -1 and 1, which produces the wavy pattern visible in (A). 7 → (A).
Q: Match slope field (B).
A: Field (B) shows segments that depend on both position variables, with an equilibrium-like line. Check dy/dx = x - y: the zero-slope isocline is y = x. Above the line y = x, slopes are negative. Below, they are positive. This diagonal pattern matches (B). 8 → (B).
Q: Match slope field (D).
A: Field (D) shows horizontal segments along a horizontal line (where slopes = 0) and vertical variation that is independent of x. This matches dy/dx = 2 - y: slope is zero when y = 2, positive below, negative above, and the slopes in each row are identical because the DE depends only on y. 9 → (D).
Q: Match slope field (C).
A: Field (C) is the remaining one, and it should match dy/dx = x. Slopes depend only on x: every column is uniform. At x = 0, horizontal. Positive to the right, negative to the left. 10 → (C).
The four slope fields (A, B, C, D) must be matched to: dy/dx = 0.5x - 1, dy/dx = 0.5y, dy/dx = -x/y, dy/dx = x + y.
Q: Match slope field (C) in the second set.
A: Field (C) shows segments that depend only on x (uniform within each column). The slopes are zero when 0.5x - 1 = 0, i.e. x = 2. This matches dy/dx = 0.5x - 1. 11 → (C).
Q: Match slope field (B) in the second set.
A: Field (B) shows slopes that depend only on y (uniform within each row), with horizontal segments along the x-axis. This matches dy/dx = 0.5y. 12 → (B).
Q: Match slope field (D) in the second set.
A: Field (D) shows a circular or radial pattern. The zero-slope isocline for dy/dx = -x/y is where x = 0 (the y-axis). The slopes at (1, 1) = -1 and at (1, -1) = 1, creating circular solution curves (x² + y² = C). 13 → (D).
Q: Match slope field (A) in the second set.
A: By elimination and verification, field (A) matches dy/dx = x + y. The zero isocline is y = -x, and the slopes depend on both variables. 14 → (A).
Q: (Problem 15) The slope field shown could have which specific solution: y = x², y = eˣ, y = e⁻ˣ, y = cos x, or y = ln x?
A: The slope field for problem 15 shows slopes that are positive and increasing in the upper-right region, with horizontal segments near x = 0 in the lower part. The key test: differentiate each candidate and see which derivative matches the visible pattern.
y = x² → dy/dx = 2x. Slopes depend only on x. At x = 0, slope = 0 for all y. Check whether the field has the same slope for all y-values in any given column. If it does not, eliminate.
y = eˣ → dy/dx = eˣ. Slopes are always positive and grow as x increases. This also means dy/dx = y, so slopes depend only on y. Check the field: are slopes in each row identical?
y = e⁻ˣ → dy/dx = -e⁻ˣ. Slopes are always negative. If the field shows positive slopes anywhere above the x-axis, eliminate.
y = cos x → dy/dx = -sin x. Slopes oscillate. If the field does not look periodic, eliminate.
y = ln x → dy/dx = 1/x. Slopes depend only on x and are positive for x > 0.
The field in the worksheet shows slopes that depend only on y and are positive above the x-axis, growing steeper as y grows. This matches dy/dx = y, which is the derivative of y = eˣ. Answer: (B) y = eˣ.
Q: (Problem 16) The slope field shown could have which specific solution: y = sin x, y = cos x, y = x², y = (1/6)x³, or y = ln x?
A: The field for problem 16 shows slopes that increase as you move right and are steeper in the upper region. The slopes appear to depend on x in a quadratic way.
y = sin x → dy/dx = cos x. Oscillating slopes.
y = cos x → dy/dx = -sin x. Oscillating slopes.
y = x² → dy/dx = 2x. Linear in x only.
y = (1/6)x³ → dy/dx = (1/2)x². Depends only on x, always non-negative, grows quadratically. The slopes are zero at x = 0 and grow symmetrically.
y = ln x → dy/dx = 1/x. Decreasing slopes for x > 0.
The field shows slopes that are zero along the y-axis and grow as |x| increases, always non-negative. This matches dy/dx = (1/2)x². Answer: (D) y = (1/6)x³.
Q: (a) Sketch a slope field for dy/dx = xy/2.
A: Compute slopes at grid points. At the origin (0, 0), slope = 0. Along the x-axis (y = 0), all slopes are 0. Along the y-axis (x = 0), all slopes are 0. At (1, 1), slope = 0.5. At (1, 2), slope = 1. At (2, 1), slope = 1. At (-1, 1), slope = -0.5. The field is symmetric: positive slopes in quadrants I and III, negative in II and IV, with horizontal segments along both axes.
Q: (b) Write the tangent line equation at (1, 1) and estimate f(1.2).
A: At (1, 1), dy/dx = (1)(1)/2 = 0.5. The tangent line is y - 1 = 0.5(x - 1), or y = 0.5x + 0.5. Estimate: f(1.2) ≈ 0.5(1.2) + 0.5 = 1.1.
Q: (c) Find the particular solution with f(1) = 1 and compute f(1.2).
A: Separate variables: (1/y) dy = (x/2) dx. Integrate: ln|y| = x²/4 + C. Apply f(1) = 1: ln 1 = 1/4 + C, so C = -1/4. Therefore ln|y| = x²/4 - 1/4, giving y = e^((x² - 1)/4). Compute f(1.2) = e^((1.44 - 1)/4) = e^(0.44/4) = e^(0.11) ≈ 1.116.
Q: (d) Was the tangent line estimate an underestimate or overestimate?
A: The estimate was 1.1 and the actual value is approximately 1.116, so the tangent line underestimated. This makes sense because the solution curve is concave up at (1, 1). To verify concavity, compute d²y/dx² by differentiating dy/dx = xy/2 implicitly: d²y/dx² = (y + x · dy/dx)/2 = (y + x · xy/2)/2. At (1, 1): (1 + 1 · 0.5)/2 = 0.75 > 0. Concave up, so the tangent line sits below the curve.
Q: (a) Sketch a slope field for dy/dx = x/y.
A: Undefined along the x-axis (y = 0). At (1, 1), slope = 1. At (1, -1), slope = -1. At (2, 1), slope = 2. At (0, 1), slope = 0. Along the y-axis (x = 0), all slopes are 0 (horizontal segments). The field suggests hyperbolic solution curves.
Q: (b) Sketch a solution curve through (0, 1).
A: Starting at (0, 1) with slope 0 (horizontal), the curve moves to the right with increasing slope (at (1, 1), slope = 1) and curves upward. To the left, by symmetry the curve also rises. It traces a half-hyperbola opening upward.
Q: (c) Find the particular solution with f(0) = 1.
A: Separate: y dy = x dx. Integrate: y²/2 = x²/2 + C. Apply f(0) = 1: 1/2 = 0 + C, so C = 1/2. Therefore y² = x² + 1, giving y = √(x² + 1) (positive root since y(0) = 1 > 0).
Q: (d) Sketch a solution curve through (0, -1).
A: At (0, -1), slope = 0/-1 = 0 (horizontal). Moving right, at (1, -1) slope = -1, so the curve descends. It mirrors the upper curve, tracing a half-hyperbola opening downward.
Q: (e) Find the particular solution with f(0) = -1.
A: Same separation gives y² = x² + 1. Since y(0) = -1 < 0, take the negative root: y = -√(x² + 1).
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