From a Direction Field to a Particular Curve
A direction field gives local slope information for a differential equation. For an equation \(y'=F(x,y)\), each point \((x,y)\) is associated with the slope \(F(x,y)\). A solution curve is not just any path through the field: at every point on the curve, its tangent slope must agree with the slope prescribed there.
An initial condition \(y(x_0)=y_0\) specifies one point through which the curve must pass. Sketching the curve means starting at \((x_0,y_0)\), following the local directions, and repeatedly checking the direction at the curve’s current location. The curve should change direction smoothly as the field changes; it should not make a sharp corner or turn merely because a single direction mark is horizontal.
This definition gives two checks for a sketch. First, it must pass through the initial point. Second, its tangent must follow the field along the entire drawn curve—not just at the initial point.
Why the Curve Must Be Tangent to the Field
Suppose \(y(x)\) is a solution through \((x_0,y_0)\). By the definition of derivative, for a small change \(h\), the curve’s vertical change is approximately its derivative at \(x_0\) times \(h\). Since \(y'(x_0)=F(x_0,y_0)\), this gives a local straight-line guide for the curve:
Here \(o(h)\) is an error term whose size relative to \(h\) tends to zero as \(h\) tends to zero. The formula says that, very close to the initial point, the curve follows the tangent line with slope \(F(x_0,y_0)\). It does not say that the curve remains on that line. Away from the initial point, the field may prescribe a different slope, so the tangent direction must be updated.
This also explains why a pencil sketch should be smooth. If the curve has a corner, it does not have a single tangent slope at that point. Such a corner cannot represent a differentiable solution there.
A Practical Tracing Procedure
Place the curve exactly through \((x_0,y_0)\). Do not begin from a nearby grid point.
Use the direction mark at \((x_0,y_0)\), or calculate \(F(x_0,y_0)\). This determines the curve’s tangent there.
As the curve moves, inspect the direction marks at its new approximate location. Follow those marks rather than continuing with the original tangent.
Moving left reverses the horizontal direction of travel. Keep the slope convention \(dy/dx\); do not reverse the sign of the slope just because you are tracing left.
At several points on the drawn curve, compare the curve’s apparent tangent with the field. Correct any mismatch and remove unjustified corners or turns.
A slope of \(m\) means that for a small horizontal change \(\Delta x\), the corresponding vertical change is approximately \(m\Delta x\). Thus a positive slope means the graph rises as \(x\) increases, while a negative slope means it falls as \(x\) increases. If tracing toward smaller \(x\), a positive slope corresponds to moving downward as you move left. The slope itself has not changed sign.
Worked Example: Following Slopes That Become Steeper
Worked Example: Following Slopes That Become Steeper
Trace a solution of \(y'=x+y\) through \((0,1)\). Use short tangent-line steps as a sketching guide, and label them as estimates rather than exact solution points.
At the initial point, the slope is
The tangent line at \((0,1)\) is therefore \(y=1+x\). For a first short step to \(x=0.2\), that line estimates
At this estimated location, the field slope is
So the next part of the sketch should be steeper than the initial tangent. For example, using another horizontal step of \(0.2\), a tangent estimate from this location is
To trace to the left, first use the initial tangent to estimate the height at \(x=-0.2\):
The field slope at that estimated location is
That slope is still positive. Moving left along a positive-slope curve takes the sketch to a lower height, consistent with the estimate \(0.8<1\). The estimates \(1.2\), \(1.48\), and \(0.8\) are tangent-step guides, not exact values of the solution. Checking them back against \(F(x,y)\) confirms the directions: slopes increase along the rightward estimates, while the leftward estimate remains on a positive-slope part of the field.
Worked Example: A Horizontal Tangent and a Genuine Turn
Worked Example: A Horizontal Tangent and a Genuine Turn
Consider \(y'=x-y\) through \((0,0)\). At the initial point,
The curve has a horizontal tangent there. The tangent-line estimate is \(y\approx 0\) for a short step in either direction. At the estimated point \((0.2,0)\), the field slope is
At \((-0.2,0)\), it is
These are field slopes at estimated locations, not the exact slopes at points of the actual curve. To justify the local shape more firmly, differentiate the equation along a twice-differentiable solution. The chain rule gives
At \(x=0\), \(y'(0)=0\), so
Because \(y''\) is continuous near \(0\), it remains positive on a sufficiently small interval around \(0\). Therefore \(y'\) increases there. Since \(y'(0)=0\), the derivative is negative just to the left of \(0\) and positive just to the right. The curve decreases into \((0,0)\) and increases away from it. Thus the horizontal tangent is at a local minimum. This conclusion uses the change of slope through the point; the zero slope by itself would not establish a turn.
The check is consistent with the equation: the initial slope is zero, the slope changes from negative to positive locally, and the curve passes through the required point. The calculation does not determine the curve’s exact heights away from the initial point.
Worked Example: Tracking a Turn from a Slope Sign Change
Worked Example: Tracking a Turn from a Slope Sign Change
Trace a solution of \(y'=2-x\) through \((0,0)\). In this field, the slope depends on \(x\) and not on \(y\). At successive \(x\)-coordinates, the field slopes are
Start at \((0,0)\), where the curve rises with slope \(2\). A tangent estimate over \(\Delta x=0.5\) gives
At \(x=0.5\), the slope is \(1.5\), so another tangent estimate gives
The slope there is \(1\). Continuing with the same step size gives
and then
Each estimate uses the slope at the start of that step. Each is approximate: the field’s slope changes during the step. The slope is zero at \(x=2\) and negative at \(x=2.5\). Since \(y'=2-x\) is positive for \(x<2\) and negative for \(x>2\), a solution defined on both sides of \(x=2\) rises before \(x=2\), has a horizontal tangent there, and falls after it. This sign change justifies a local maximum. The estimated height \(2.5\) is not asserted to be the exact height at \(x=2\). Checking the field confirms that the sketch must flatten at \(x=2\) and then descend.
Horizontal Tangents Are Not Automatically Turns
A horizontal tangent means only that the derivative is zero at that point. To claim a local maximum or minimum, the curve’s slope must change in the appropriate way nearby. A slope that is zero but has the same sign on both sides does not produce a local turn. In a sketch, do not add a peak or valley just because one direction mark is horizontal.
There is also a useful quick check for proposed extrema. If a differentiable function has a local maximum or minimum at an interior point, its derivative there must be zero. This follows because the difference quotients from the two sides cannot have the same nonzero sign: at a local maximum the left-side quotients are nonnegative and the right-side quotients are nonpositive, while at a local minimum the inequalities reverse. If the derivative exists, the two one-sided limits agree, so their common value must be zero. The converse is not guaranteed: a zero derivative alone does not prove an extremum.
How Accurate Should a Sketch Be?
A direction-field sketch is a qualitative representation, not an exact formula for the solution. The tangent-step estimates in the examples show how to choose the next direction, but a large step can miss substantial changes in slope. Shorter steps generally make the trace more faithful to the field. When the field changes rapidly, check more often rather than drawing a long straight segment.
Keep the initial condition visible as an anchor, and distinguish the curve from the short direction marks around it. A curve should pass through the prescribed point and be tangent to the local marks, but it is not itself a sequence of disconnected line segments. Draw a smooth path that turns gradually as the prescribed slopes change. If the field gives no basis for a claimed feature—such as a sharp corner, an extra loop, or a turn—do not add it to the sketch.
Finally, a plausible sketch is not a proof that a solution exists or that it is the only one. The sketching task is to represent a curve consistent with the given local directions and initial point. Questions about existence or uniqueness require additional results; they are not settled merely by drawing one curve.
Check Your Understanding
Use the field directions and initial points to answer each question. Treat tangent-step values as estimates unless the equation makes them exact.
- For \(y'=x+y\) through \((1,0)\), calculate the initial slope. What does its sign tell you as \(x\) increases?
- For \(y'=3-x\) through \((0,1)\), calculate the slopes at \(x=0\), \(x=2\), and \(x=4\). What qualitative change should the sketch show?
- Why does a horizontal tangent at an initial point not, by itself, prove that the curve has a local maximum or minimum?
- When tracing leftward along a part of a curve with positive slope, does the slope become negative? Explain the change in height as \(x\) decreases.
- For \(y'=x-y\) through \((0,0)\), explain why the horizontal line through the initial point is not a solution on an interval.
