Unit 7 · Topic 7.4 Beta

Reasoning with Slope Fields

Trace a solution through a point by following the segments across the field.

Practice 2: Connecting RepresentationsPractice 3: Justification

Question set for this topic

Part 1 · Hook

Why this matters

A slope field is a map of every solution at once. Drop a pencil on a point and let the segments steer it, and you trace the one curve through that point: where it rises, where it bends, where it settles. Rows of flat segments act like rails: solution curves approach them or run away from them, but never cross them.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If f′f' changes from negative to positive at x=ax = a, then ff has

  1. a relative minimum at x=ax = a
  2. a relative maximum at x=ax = a
  3. a point of inflection at x=ax = a
  4. no critical point at x=ax = a
Show the answer

Decreasing then increasing: a relative minimum, by the First Derivative Test.

  • Correct: a relative minimum at x=ax = a:
  • a relative maximum at x=ax = a:
  • a point of inflection at x=ax = a:
  • no critical point at x=ax = a:

2. If f″>0f'' > 0 on an interval, the graph of ff there is

  1. concave up
  2. concave down
  3. increasing
  4. decreasing
Show the answer

A positive second derivative means f′f' is increasing: concave up.

  • Correct: concave up:
  • concave down:
  • increasing:
  • decreasing:

Part 4 · See it

See it first

The slope field of dy/dx = (3 − y)/2 drawn every half unit. The dashed green line y = 3 has slope 0 everywhere. A crimson curve through (0, 1) rises and flattens toward y = 3; a blue curve through (0, 5.5) falls and flattens toward y = 3. Neither crosses the line.
The slope field of dydx=3−y2\frac{dy}{dx} = \frac{3 - y}{2}. The line y=3y = 3 is an equilibrium solution; the curves through (0,1)(0, 1) and (0,5.5)(0, 5.5) both approach it and never cross it. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A solution curve passes through a given pointit runs along the segment there and keeps following the segments in both directions
  2. dydx=0\frac{dy}{dx} = 0 for every xx on a horizontal line y=cy = cy=cy = c is an equilibrium solution, a row of flat segments
  3. Solution curves of these equations do not cross one anothera curve that starts below an equilibrium line stays below it
  4. d2ydx2\frac{d^2y}{dx^2} is found by differentiating dydx\frac{dy}{dx} with the chain rule and substituting dydx\frac{dy}{dx}its sign at a point tells the concavity of the solution there

Part 6 · Key ideas

Key ideas

  • Sketch a solution through a point by following the segments both ways, all the way across the field; never cut across the segments.
  • An equilibrium solution y=cy = c makes dydx=0\frac{dy}{dx} = 0 for every xx. Solve "right side =0= 0" for yy.
  • The sign of dydx\frac{dy}{dx} says where a solution rises or falls; long-run behavior shows whether curves approach an equilibrium or move away.
  • Concavity: d2ydx2=ddx(dydx)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right). Every yy is a function of xx, so ddx(y)=dydx\frac{d}{dx}(y) = \frac{dy}{dx}; then substitute the equation.

Part 7 · Misconception

A common mistake

The wrong idea: For dydx=y−12\frac{dy}{dx} = \frac{y - 1}{2}, d2ydx2=0\frac{d^2y}{dx^2} = 0, because there is no xx in the formula.

What actually happens: yy depends on xx, so the chain rule applies: d2ydx2\frac{d^2y}{dx^2}=12dydx= \frac{1}{2}\frac{dy}{dx}=y−14= \frac{y - 1}{4}.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The slope field of dydx=y−12\frac{dy}{dx} = \frac{y - 1}{2}

−2−1123−2−11234xy

1. Which is an equilibrium solution of the differential equation?

  1. y=1y = 1
  2. y=0y = 0
  3. x=1x = 1
  4. y=2y = 2
Show the answer

Set the right side to 0: y−12=0\frac{y - 1}{2} = 0 gives y=1y = 1.

  • Correct: y=1y = 1: On y=1y = 1, y−12=0\frac{y - 1}{2} = 0 for every xx: a row of flat segments.
  • y=0y = 0: At y=0y = 0 the slope is −12-\frac{1}{2}, not 0.
  • x=1x = 1: An equilibrium solution is a horizontal line y = c; on x = 1 the slopes vary.
  • y=2y = 2: At y=2y = 2 the slope is 12\frac{1}{2}.

2. Describe the solution curve through (0,2)(0, 2) as xx increases.

  1. It falls and approaches y=1y = 1.
  2. It levels off at y=2y = 2.
  3. It rises and approaches y=3y = 3.
  4. It rises ever more steeply, moving away from y=1y = 1.
Show the answer

For y>1y > 1, dydx=y−12>0\frac{dy}{dx} = \frac{y - 1}{2} > 0 and grows as yy grows, so the curve climbs faster and faster.

  • It falls and approaches y=1y = 1.: Above y=1y = 1, dydx>0\frac{dy}{dx} > 0: the curve rises.
  • It levels off at y=2y = 2.: At y=2y = 2 the slope is 12\frac{1}{2}, not 0.
  • It rises and approaches y=3y = 3.: Nothing is flat at y=3y = 3; the slope there is 1.
  • Correct: It rises ever more steeply, moving away from y=1y = 1.: Above y=1y = 1 the slopes are positive and grow with yy.

3. Find d2ydx2\frac{d^2y}{dx^2} at the point (0,3)(0, 3) for dydx=y−12\frac{dy}{dx} = \frac{y - 1}{2}.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

dydx=3−12=1\frac{dy}{dx} = \frac{3 - 1}{2} = 1, so d2ydx2=12(1)=12\frac{d^2y}{dx^2} = \frac{1}{2}(1) = \frac{1}{2}.

  • Answer: 12\frac{1}{2} ≈ 0.500

4. Which are equilibrium solutions of dydx=x2(y−3)(y+1)\frac{dy}{dx} = x^2(y - 3)(y + 1)? Select all that apply.

  1. y=3y = 3
  2. y=−1y = -1
  3. x=0x = 0
  4. y=0y = 0
Show the answer

An equilibrium solution is a constant y=cy = c with dydx=0\frac{dy}{dx} = 0 for every xx: y=3y = 3 and y=−1y = -1.

  • Correct: y=3y = 3: The factor y−3y - 3 makes the slope 0 for every xx.
  • Correct: y=−1y = -1: The factor y+1y + 1 makes the slope 0 for every xx.
  • x=0x = 0: The slope is 0 along x=0x = 0, but a vertical line is not a function yy of xx.
  • y=0y = 0: At y=0y = 0, dydx=−3x2\frac{dy}{dx} = -3x^2, not 0 for every xx.

5. For dydx=y2−x\frac{dy}{dx} = y^2 - x, find d2ydx2\frac{d^2y}{dx^2} at the point (1,2)(1, 2).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

d2ydx2=2ydydx−1\frac{d^2y}{dx^2} = 2y\frac{dy}{dx} - 1. At (1,2)(1, 2): dydx=4−1=3\frac{dy}{dx} = 4 - 1 = 3, so d2ydx2\frac{d^2y}{dx^2}=2(2)(3)−1= 2(2)(3) - 1=11= 11.

  • Answer: 11

6. A solution curve of dydx=y−x2\frac{dy}{dx} = y - x^2 passes through (0,1)(0, 1). Near x=0x = 0 the curve is

  1. increasing and concave down
  2. decreasing and concave up
  3. decreasing and concave down
  4. increasing and concave up
Show the answer

dydx=1−0=1\frac{dy}{dx} = 1 - 0 = 1; d2ydx2\frac{d^2y}{dx^2}=dydx−2x= \frac{dy}{dx} - 2x=1−0= 1 - 0=1= 1. Both positive.

  • increasing and concave down: d2ydx2=dydx−2x=1\frac{d^2y}{dx^2} = \frac{dy}{dx} - 2x = 1 at (0,1)(0, 1), positive.
  • decreasing and concave up: dydx=1−0=1>0\frac{dy}{dx} = 1 - 0 = 1 > 0.
  • decreasing and concave down: Both derivatives are positive at (0,1)(0, 1).
  • Correct: increasing and concave up: dydx=1>0\frac{dy}{dx} = 1 > 0 and d2ydx2\frac{d^2y}{dx^2}=dydx−2x= \frac{dy}{dx} - 2x=1>0= 1 > 0.

Part 9 · Summary

Summary

Trace a solution through a point by following the segments across the field. Rows of flat segments are equilibrium solutions, which other solutions approach or leave but never cross. The sign of dydx\frac{dy}{dx} gives direction; d2ydx2\frac{d^2y}{dx^2}, found with the chain rule, gives concavity.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections