From Short Segments to a Smooth Curve
A slope field gives a visual rule for how a solution curve should move near each point. Each short segment represents the slope specified by a differential equation at that location. To sketch a solution through an initial point, begin there and trace a smooth curve that follows the direction and steepness of the nearby segments.
The curve should not be drawn by connecting the little segments as if they were a route made of straight pieces. The segments indicate tangent directions, not corners. As the curve moves, its location changes, so the slope usually changes too. A good sketch repeatedly checks the field and adjusts its direction accordingly.
How to Trace a Curve
At the initial point, find the segment’s slope. If it is positive, the curve rises as you move to the right; if it is negative, the curve falls as you move to the right. A slope near zero means the curve is nearly horizontal there, while a large slope in magnitude means it is steep.
After moving a short distance, look up the slope at the new location rather than extending the first tangent indefinitely. A curve can become steeper, flatten, or change from rising to falling. When tracing left from the initial point, remember that a positive slope still means the curve rises as \(x\) increases; consequently, moving left along that curve makes \(y\) decrease.
Place the curve exactly through the given point \((x_0,y_0)\).
Use the segment at the point to determine whether the curve rises, falls, or is locally horizontal, and judge how steep it should be.
Sketch a small smooth portion in the indicated direction. Do not travel so far that you assume the slope stays constant.
At the curve’s new location, follow the segment there. Repeat this process to trace the curve forward, backward, or both.
Check that the curve passes through the initial point, remains smooth, and agrees with the field’s direction and steepness all along its path.
Why the Segments Give the Right Local Direction
The reason to follow a segment is the definition of the derivative. If a solution passes through \((x_0,y_0)\), its derivative there is the limiting slope of nearby secant lines. The differential equation specifies that derivative as \(f(x_0,y_0)\). Thus, over a sufficiently short horizontal change, the curve’s change in height is approximately the tangent slope multiplied by that horizontal change.
Rearranging the secant quotient gives the local approximation below. It explains why a short move along a segment is a useful sketching step, but also why a long move can be misleading: the actual slope may change during that move.
This approximation is sometimes called a forward Euler step. It is a way to estimate a nearby point, not a guarantee that the estimated point lies on the exact solution curve. The approximation improves as the step is made sufficiently small, provided the solution and the slope field behave smoothly nearby.
Worked Example: Tracing a Curve That Is Still Falling
Worked Example: Tracing a Curve That Is Still Falling
Sketch the solution of \(y'=x-y\) through \((0,2)\), and use the field to judge whether it has turned by \(x=1\).
At the initial point, the slope is
So the curve starts by falling steeply as it moves right. A forward Euler estimate with step \(h=0.5\) gives
This is an estimate, not an exact solution value. If we continue from the estimated point \((0.5,1)\), the slope there is \(0.5-1=-0.5\), and the next estimate is
At this estimated point, the field slope is \(1-0.75=0.25\). It may look as though the curve has already turned somewhere before \(x=1\). That conclusion is not justified: these are coarse step estimates, not actual solution values or proof of where the derivative changes sign.
For a reliable check, consider \(y=x-1+3e^{-x}\). Its derivative and initial value are
Thus this function satisfies both the differential equation and the specified initial point. At \(x=0.5\), its height is \(-0.5+3e^{-0.5}\approx1.3196\), so its slope is \(0.5-1.3196\approx-0.8196\). At \(x=1\), its height is \(3/e\approx1.1036\), so its slope is \(1-3/e\approx-0.1036\). It is still falling at both locations. Its derivative is zero when \(1-3e^{-x}=0\), or \(x=\ln 3\approx1.099\); that is where this curve reaches its minimum.
Check: Substitution shows the proposed function has \(y'=x-y\) and \(y(0)=2\). The calculation confirms that the apparent early turn in the coarse estimates is not the actual turn. A sketch should follow the field’s changing slopes and should not treat an approximate step as an exact point.
Worked Example: Following a Curve Through a Horizontal Tangent
Worked Example: Following a Curve Through a Horizontal Tangent
Sketch the solution of \(y'=1-x\) through \((0,0)\). Here the slope depends only on the horizontal coordinate. At \(x=0\), the slope is \(1-0=1\), so the curve rises to the right. At \(x=0.5\), the slope is \(1-0.5=0.5\), so it is still rising, but less steeply. At \(x=1\), the slope is \(1-1=0\), so the curve has a horizontal tangent there. For \(x=1.5\), the slope is \(1-1.5=-0.5\), so the curve falls to the right of that location.
These field readings suggest a smooth curve that rises, flattens at \(x=1\), and then falls. For a check, the function \(y=x-x^2/2\) passes through the initial point and has derivative
Its values at the sample locations are \(y(0)=0\), \(y(0.5)=0.5-0.125=0.375\), \(y(1)=1-0.5=0.5\), and \(y(1.5)=1.5-1.125=0.375\). These values agree with the rising-and-falling shape indicated by the field.
Check: Differentiating the proposed curve gives exactly the field slope at every \(x\), and substituting \(x=0\) gives \(y=0\). The slope changes from positive to negative at \(x=1\), so the horizontal tangent there is a local high point for this curve.
Worked Example: Letting the Curve Become Steeper
Worked Example: Letting the Curve Become Steeper
Sketch the solution of \(y'=y\) through \((0,1)\). Since the slope equals the current height, the initial slope is \(1\). The curve starts by rising. To see how its steepness changes, use Euler estimates with step \(h=0.25\).
From \((0,1)\), the slope is \(1\), so the first estimate is \(y(0.25)\approx1+(1)(0.25)=1.25\). At the estimated point \((0.25,1.25)\), the slope is \(1.25\), giving \(y(0.5)\approx1.25+(1.25)(0.25)=1.5625\). At \((0.5,1.5625)\), the slope is \(1.5625\), giving \(y(0.75)\approx1.5625+(1.5625)(0.25)=1.953125\).
The successive estimates increase, and the slopes used for the steps are \(1\), \(1.25\), and \(1.5625\). The sketch should therefore rise and become steeper rather than continue as a straight line with the initial slope.
As a check on that shape, \(y=e^x\) satisfies \(y'=e^x=y\), and \(y(0)=1\). This confirms that it is a solution through the initial point and that its slope increases as its height increases.
Check: The estimated points are obtained by multiplying each current slope by \(0.25\) and adding that change to the current estimated height. They are approximations; the exact solution check verifies the equation and initial point without confusing those estimates with exact values.
What a Sketch Can—and Cannot—Tell You
A carefully traced curve lets you see how a solution moves locally: whether it rises or falls, how steep it is, and where the field indicates a horizontal tangent. If slopes change sign along the curve, that may indicate a change from rising to falling or vice versa. The curve should be smooth when the solution is differentiable, but its steepness need not stay constant.
A slope field alone does not make every hand sketch exact. The spacing of the grid, the length of the displayed segments, and the size of your tracing steps all affect visual estimates. In particular, a rough step can jump across a region where the slope changes sign. A curve may appear to turn too early or too late if you infer a turning point only from estimated points. When the location matters, use shorter steps and check the slopes on both sides; if a known solution is available, substitute it into the equation and check the initial value.
Another important distinction is between a tangent segment and the curve itself. The segment at one point gives the curve’s local direction at that point; it does not say that the curve must remain parallel to that segment at nearby points. Rechecking the field as you trace is what keeps the sketch faithful to the differential equation.
Check Your Understanding
Use the ideas in this tutorial to describe or check each sketch.
- At an initial point, the field segment has slope \(-2\). In which direction does the curve move as \(x\) increases, and how does it move as you trace left?
- Why is a forward Euler step an estimate rather than an exact solution value?
- For \(y'=1-x\), what does the field indicate at \(x=1\), and what additional slope information supports a turning point there?
- For \(y'=y\) at \((0,1)\), what is wrong with continuing the initial tangent as a straight line?
- Why can a pair of coarse step estimates suggest a turning point at a location different from the actual one?
