First — talk for 2 minHas anyone in your family farmed, gardened, or grown anything — even on a windowsill? What was the soil like? If no one, Mr. Bennett will share about a community garden.
What is the soil under your feet doing?
Most of the carbon on a healthy farm is not in the corn or the cows — it is buried under your boots. Soil organic carbon is the leftover stems, roots, and microbes built up over centuries of life and death. Healthy farm soils can hold dozens of tons of it per hectare. Plowed-and-replanted soils, every year, lose a little of it back to the air as carbon dioxide.
A regenerative farmer in South Jersey is transitioning a 1-hectare field from conventional corn-and-soybean to silvopasture — cattle grazing under widely-spaced hardwood trees. At year 0, the field has about 40 tons of carbon per hectare. Studies project the soil will gain carbon at a continuous rate of about 3.4% per year for the first decade or two.
"Continuous" is the new word. Soil microbes do not pause and wait for January 1 to add the year's growth. They are working every second — root tip, raindrop, fungus, beetle. So instead of writing \(C(t) = 40 \cdot (1.034)^{t}\), soil scientists write \(C(t) = 40 \cdot e^{0.034\,t}\). Same starting carbon. Same 3.4%. Different timing rule. Does that little change in timing rule actually move any tons of carbon?
Worked Example #1 — Build the continuous-rate model (silvopasture)
Problem\(C(t)\) is soil organic carbon (tons per hectare), \(t\) = years since transition. Write a continuous-rate model.
DataYear 0 carbon: 40 t/ha. Continuous growth rate: 3.4% per year. Use the form \(P_{0} \cdot e^{rt}\), where \(r\) is the continuous rate per year and \(P_{0}\) is the starting amount (the value when \(t = 0\)).
Word Bank
starting amount
multiply
decimal
continuous rate
equation
subtract
substitute
exponent
Number Bank
40
0.034
3.4
0.34
1.034
0
| # | Reasoning (fill in) | Math |
|---|
| 1 |
The \(P_{0}\) is the soil carbon at \(t = 0\), before any growth. |
\(P_{0} = \) |
| 2 |
Convert 3.4% to a by dividing by 100. This is the \(r\). |
\(r = \) |
| 3 |
Write the using \(P_{0}\), \(e\), \(r\), and \(t\). |
\(C(t) = \) \(\cdot\, e^{\,r\,t}\) |
| 4 |
Plug in the value of \(r\) so the formula is ready to use. |
\(C(t) = 40 \cdot e^{\,\,}\)\(\,t\) |
| 5 |
To predict \(t = 10\), 10 in for \(t\) and use the \(e^{x}\) key. |
\(C(10) = 40 \cdot e^{0.34} \approx \) t/ha |
Page 1 of 3 · Continuous Growth and Soil Carbon · Algebra 2 · Unit 5 · Lesson 13
Worked Example #2 — Identify \(a\) and \(r\) from the equation (no-till + cover crops)
ProblemA US Midwest field under no-till + cover crops is modeled by \(C(t) = 35 \cdot e^{0.012\,t}\). Find the starting carbon and the continuous growth rate as a percent per year.
DataThe form is \(C(t) = a \cdot e^{rt}\). Here \(a\) is the value at \(t = 0\) (before any growth) and \(r\) is the continuous rate per year, written as a decimal.
Word Bank
starting amount
exponent
add
percent
multiply
continuous rate
decimal
Number Bank
35
0.012
1.2
12
1.012
0.12
| # | Reasoning (fill in) | Math |
|---|
| 1 |
The coefficient (the value when \(t = 0\)) is the \(a\). |
\(a = \) t/ha |
| 2 |
The number multiplying \(t\) inside the is the continuous rate \(r\) (as a decimal). |
\(r = \) |
| 3 |
To turn the decimal into a , by 100. |
\(0.012 \times 100 = \) % |
| 4 |
State the : the soil grows at this percent per year. |
1.2% continuous growth per year |
| 5 |
Watch out: \(e^{0.012} \approx 1.0121\) is the discrete annual factor, not the continuous rate. The continuous rate is \(r\) itself. |
Answer: \(a = 35\) t/ha; rate \(= 1.2\%/\)yr |
Three Practices, Three Rates — fill in from your work
Silvopasture (Drawdown): \(C_0 = 50\) t/ha, continuous rate 4.5% per year. \(C(t) = \) \(\cdot\, e^{\,}\)\(\,t\).
"4 per 1000" UN target: \(C_0 = 60\) t/ha, continuous rate 0.4% per year. \(C(t) = \) \(\cdot\, e^{\,}\)\(\,t\).
Conventional tillage: \(C(t) = 45 \cdot e^{-0.008\,t}\). Continuous rate \(=\) % per year. Growth or decay? (because the exponent is ).
Conclude — what \(r\) would the US need?
How fast would adoption have to grow?
Today only about 5% of US cropland uses regenerative practice. Suppose adoption grows continuously: \(p(t) = 0.05 \cdot e^{r\,t}\). To hit a 25-year target of \(p(25) = 0.30\) (30% of cropland), what continuous rate \(r\) would we need?
From the equation: \(0.05 \cdot e^{25 r} = 0.30\), so \(e^{25 r} = 6\). Using \(\ln\): \(25 r = \) \(\ln(6)\) \(\approx 1.7918\), so \(r \approx \) per year (about %).
Reflect: Cover-crop adoption grew at about 5.5% per year between 2012 and 2022. Pick one intervention from class (EQIP cost-share, CSP, Farmer-First Pilot, carbon markets, agroforestry grants, or state soil-health policy). Would it move \(r\) toward the value you found, or away? Why?
Page 2 of 3 · Continuous Growth and Soil Carbon · Algebra 2 · Unit 5 · Lesson 13
Tier 2 · General Academic Words
- coefficient — the number multiplying a variable or expression (in \(P_{0} \cdot e^{rt}\), \(P_{0}\) is the coefficient).
- continuous — happening at every moment, with no pauses or breaks.
- decay — getting smaller over time, the opposite of growth.
- decimal — a number written with a dot, like 0.034 or 1.2 (3.4% as a decimal is 0.034).
- discrete — happening in separate steps, not smoothly (like once-a-year payments).
- percent — a number out of 100 (3.4% means 3.4 out of every 100).
- rate — how fast something changes over time.
- regenerative — helps a system rebuild itself; healing instead of using up.
- soil — the top layer of dirt that plants grow in — alive with roots, microbes, and fungi.
- transition — the time when a farm switches from one practice to another.
Tier 3 · Soil, Climate, & Math Words
- soil organic carbon (SOC) — the carbon stored inside dirt as old roots, leaves, and microbes; usually measured in tons per hectare.
- regenerative agriculture — a way of farming that builds healthier soil each year instead of using it up.
- cover crop — a plant (like clover or rye) grown to protect and feed the soil between main crops.
- no-till — a way of farming that does not plow the soil; seeds are planted into the ground as is.
- silvopasture — a farm system with pastured animals grazing under widely-spaced trees on the same land.
- ton per hectare (t/ha) — the standard unit for soil carbon: 1 ton of carbon spread across one hectare (about 2.5 acres).
- \(e\) (Euler's number) — a special math constant, \(e \approx 2.71828\); it shows up whenever something grows continuously.
- continuous growth — growth where the rate is applied at every instant, not just once a year.
- continuous growth rate (\(r\)) — the rate \(r\) inside the exponent of \(P_{0} \cdot e^{rt}\); the percent per unit of \(t\), applied continuously.
- natural log (\(\ln\)) — the inverse of \(e^{x}\); use it to solve for the exponent (\(\ln(e^{k}) = k\)).
- \(P_{0}\) / \(C_{0}\) — the starting amount at \(t = 0\), before any growth or decay.
- EQIP — a USDA program that pays farmers to adopt soil-friendly practices like cover crops.
- "4 per 1000" — a UN goal of growing global soil carbon by 0.4% per year — small per acre, huge worldwide.
Page 3 of 3 · Continuous Growth and Soil Carbon · Algebra 2 · Unit 5 · Lesson 13