Direction Fields — Question 7

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Question 7

For the equation y′=x2−y,y'=x^2-y, consider the supplied candidate through the origin w(x)=x2−2x+2−2e−x.w(x)=x^2-2x+2-2e^{-x}. You may use the calculus inequality et>1+te^t>1+t for t≠0t\ne 0.

Tasks

  1. Find the zero-slope isocline and describe the slope signs above and below it.

  2. Verify that ww solves the equation and that w(0)=w′(0)=0w(0)=w'(0)=0.

  3. Decide whether ww has a local extremum at 00. Determine its concavity on either side of 00 and classify the horizontal tangent there.

  4. Draw the field, the isocline, and ww near the origin in your solution. Explain why observing a horizontal field segment is insufficient to conclude that a solution has a maximum or minimum.

Original worksheet page 1: question and worked solution for 1-2-007
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Question 7 – Solution

Strategy. A zero first derivative identifies a stationary point, not its type. Inspect nearby derivative signs and concavity.

Step 1: Isocline and verification. Zero slopes occur on y=x2\boxed{y=x^2}. Slopes are positive below this parabola and negative above it. For the candidate, w′=2x−2+2e−x=x2−w,w(0)=w′(0)=0.w'=2x-2+2e^{-x}=x^2-w,\qquad w(0)=w'(0)=0. It is therefore a solution on ℝ\mathbb R with a horizontal tangent at the origin.

See the diagram in the original worksheet below.

Step 2: No local extremum. For x≠0x\ne 0, the supplied inequality gives e−x>1−xe^{-x}>1-x, so w′=2(x−1+e−x)>0.w'=2(x-1+e^{-x})>0. The derivative is positive on both sides of 00. Thus ww increases through the origin and has neither a local maximum nor a local minimum there. Moreover, x2−w=w′>0x^2-w=w'>0 for x≠0x\ne 0, so the curve lies below its zero-slope isocline on both sides.

Step 3: Concavity. Since w″=2−2e−xw''=2-2e^{-x}, the curve is concave down for x<0x<0 and concave up for x>0x>0. Hence (0,0) is an inflection point with a horizontal tangent.\boxed{(0,0)\text{ is an inflection point with a horizontal tangent.}} The field gives the tangent slope at a point; the surrounding behavior is needed to classify that point.

Original worksheet page 2: question and worked solution for 1-2-007

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