From a Rate Graph to Net Change
A rate graph shows how quickly a quantity changes as the input changes. The graph’s height alone does not tell you the total change over an interval. To find that change, you must account for both the rate’s size and how long it acts.
For example, if \(r(t)\) is a rate measured in liters per minute, then the area between its graph and the horizontal axis over a time interval has units of liters. That area represents a change in the quantity. Regions above the axis contribute positively; regions below it contribute negatively.
This tutorial focuses on reading that signed area from a graph. The graph may be made of straight segments or curves, but the main task is the same: identify the regions, determine their signs, calculate their areas, and add their contributions.
Read the Axes Before Finding Areas
First inspect the graph’s labels and scale. The horizontal axis might measure time in minutes, while the vertical axis measures a rate in liters per minute. A horizontal distance of \(2\) then means \(2\) minutes, and a vertical height of \(3\) means \(3\) liters per minute. Their product has units of liters.
Next, mark the interval endpoints and any places where the graph meets or crosses the horizontal axis. These points divide the graph into regions whose signs and shapes are easier to identify. A region above the axis has positive signed area; a region below has negative signed area. If a graph crosses the axis within the interval, the two parts must not be treated as one unsigned area.
Common shapes can be calculated with familiar geometry. A triangle has area \(\frac12 bh\), a rectangle has area \(bh\), and a trapezoid with parallel sides of lengths \(u\) and \(v\) and width \(w\) has area \(\frac12(u+v)w\). For a rate graph, these geometric areas become signed according to whether they lie above or below the axis.
A Reliable Graph-Reading Process
Identify the starting and ending input values, and note the units on both axes.
Mark zeros, crossings, corners, and interval endpoints so each region has a clear shape and sign.
Use the appropriate geometric formula, keeping horizontal width and vertical height in the graph’s stated units.
Count regions above the axis as positive and regions below it as negative; the sum is the net change.
For a straight-line segment, the signed contribution over an interval can be calculated as a trapezoid using the two endpoint heights. The heights are signed: an endpoint below the axis has a negative value. This convention handles a segment that crosses the axis without changing the rule.
Why Signed Geometric Area Gives Net Change
The connection between area and change is not an arbitrary graph-reading trick. A rate measures change per unit of input. Over a very short interval, the change is approximately the rate multiplied by the interval width. Adding these small contributions across the interval produces the accumulated change represented by the definite integral.
When the graph is curved, the same signed-area interpretation applies. The exact area may require a known geometric formula or an integral calculation; do not assume a curved region is a triangle or trapezoid just because it looks similar in a sketch.
Worked Examples
Worked Example: Adding Triangles and a Rectangle
A rate \(r(t)\), measured in gallons per minute, has a graph formed by straight segments through \((0,0)\), \((2,4)\), \((5,4)\), and \((7,0)\). Find the net change in the amount of water from \(t=0\) to \(t=7\) minutes.
Step 1: Identify the regions and their signs. The whole graph lies on or above the horizontal axis, so all three regions contribute positively.
Step 2: Find the first triangle. Its base is \(2-0=2\) minutes and its height is \(4\) gallons per minute: $$ A_1=\frac12(2)(4)=4\text{ gallons}. $$
Step 3: Find the rectangle. Its width is \(5-2=3\) minutes and its height is \(4\) gallons per minute: $$ A_2=(3)(4)=12\text{ gallons}. $$
Step 4: Find the final triangle. Its base is \(7-5=2\) minutes and its height is \(4\) gallons per minute: $$ A_3=\frac12(2)(4)=4\text{ gallons}. $$
Step 5: Add the signed areas. $$ \text{Net change}=A_1+A_2+A_3=4+12+4=20\text{ gallons}. $$ The result is consistent with the graph: every contribution is positive, and the total is \(20\) gallons. The units also check: minutes multiplied by gallons per minute gives gallons.
Worked Example: Splitting at a Zero Crossing
A rate \(r(t)\), measured in units per hour, is piecewise linear through \((0,3)\), \((2,3)\), \((4,-1)\), \((6,-1)\), and \((8,0)\). Find the net change from \(t=0\) to \(t=8\) hours.
Step 1: Locate the zero crossing. Between \(t=2\) and \(t=4\), the graph falls linearly by \(3-(-1)=4\) units of rate over \(2\) hours. It falls \(2\) units of rate per hour, so starting from \(3\), it reaches zero after \(\frac{3}{2}\) hours. The crossing is at \(t=2+\frac32=3.5\).
Step 2: Calculate the positive regions. From \(0\) to \(2\), the graph makes a rectangle: $$ A_1=(2)(3)=6. $$ From \(2\) to \(3.5\), it makes a triangle above the axis: $$ A_2=\frac12(3.5-2)(3)=\frac12(1.5)(3)=2.25. $$
Step 3: Calculate the negative regions. From \(3.5\) to \(4\), the graph makes a triangle below the axis: $$ A_3=-\frac12(4-3.5)(1)=-\frac12(0.5)(1)=-0.25. $$ From \(4\) to \(6\), it makes a rectangle below the axis: $$ A_4=-(6-4)(1)=-2. $$ From \(6\) to \(8\), it makes a triangle below the axis: $$ A_5=-\frac12(8-6)(1)=-1. $$
Step 4: Add the signed contributions. $$ \text{Net change}=6+2.25-0.25-2-1=5\text{ units}. $$ As a check, the accumulated signed contributions at successive breakpoints are \(6\), \(8.25\), \(8\), \(6\), and finally \(5\). In particular, the negative-rate portions reduce the net change rather than adding to it.
Worked Example: A Curved Region and a Negative Rectangle
On \(0\le t\le4\), a rate graph is the upper semicircle of radius \(2\), centered on the horizontal axis. On \(4\le t\le6\), the rate is constant at \(-2\) units per hour. Find the net change over \(0\le t\le6\).
Step 1: Find the semicircle’s contribution. The semicircle lies above the axis, so its signed area is positive. A full circle of radius \(2\) has area \(\pi(2)^2=4\pi\). Half of that area is $$ A_1=\frac12(4\pi)=2\pi\text{ units}. $$
Step 2: Find the rectangle’s contribution. The interval width is \(6-4=2\) hours, and the rate is \(-2\) units per hour: $$ A_2=(2)(-2)=-4\text{ units}. $$
Step 3: Add and check the signs. $$ \text{Net change}=2\pi-4\text{ units}\approx2.283\text{ units}. $$ Here the approximation is rounded to three decimal places. The positive semicircle contributes about \(6.283\) units, and the negative rectangle subtracts \(4\), leaving about \(2.283\) units. This agrees with the exact result \(2\pi-4\).
Using Net Change with an Initial Value
If the quantity has a known value at the beginning of the interval, add the graph’s net change to that initial value to find its value at the end. For example, if a quantity starts at \(Q(0)=10\) units and its rate graph gives a net change of \(5\) units on the interval, then \(Q(8)=10+5=15\) units. The rate graph supplies the change, not the starting amount.
Common Pitfall: Making Every Area Positive
A frequent error is to calculate the geometric size of every region and add all those sizes as positive numbers, even when some regions are below the axis. For the second worked example, that incorrect method gives $$ 6+2.25+0.25+2+1=11.5. $$ But the regions after the zero crossing lie below the axis, so their signed contributions are \(-0.25\), \(-2\), and \(-1\), not positive. The corrected calculation is $$ 6+2.25-0.25-2-1=5. $$ When the question asks for net change, the sign of each region is essential.
Another useful check is to consider the direction of change. Where the rate is positive, the quantity increases; where the rate is negative, it decreases. If a calculation treats a below-axis region as increasing the net change, its sign is wrong.
Check Your Understanding
Use signed areas and show how you determine each region’s sign.
- A rate graph is a triangle above the axis with base \(6\) minutes and height \(5\) units per minute. What is the net change?
- A rate is constant at \(-3\) units per hour for \(4\) hours. What is the net change?
- A straight segment goes from rate \(4\) to rate \(-2\) over an interval of width \(3\). Find its signed contribution using the trapezoid formula.
- A graph has a positive region of area \(9\) units and a negative region of geometric area \(4\) units. What is the net change?
- A quantity starts at \(12\) units and the rate graph gives a net change of \(-5\) units. What is its ending value?
