Investigating The Meaning Of Change And Quantities With A Theoretical Hydroponic system
Introduction to the Problem
As I work my way through Khan Academy’s Calculus 2 curriculumn, I intend to take the parts that inspire me and apply them to the field of agriculture by designing and solving a problem with the help of Gemini AI.
In the Khan Academy article, “Exploring accumulation of change”, there is a reminder to distinguish between the net change of a quantity and the value of an actual quantity.
Consider a farmer monitoring nitrate (NO3-) concentrations in their hydroponic system. Over the course of 24 hours, let $R(t)$ be the rate nitrate is added to the system and let $C(t)$ be the rate nitrate is consumed by plants and is lost to evaporation.
\[R(t) = 0.5 + 0.3sin(\frac{\pi t}{12})\:ppm/hr\] \[C(t) = 0.2 + 0.1cos(\frac{\pi t}{6})\:ppm/hr\]
With this information you can calculate the change in concentration of nitrate in the system, for any period of time in between $t=0$ and $t=24$, but you cannot know the quantity of nitrate in the system at any point in time without knowing the initial concentration of nitrate or the initial condition.
Assuming the initial concentration of NO3- in the system at $t=0$ was 150 ppm of NO3-, what is the actual NO3- concentration at noon ($t=12$)?
Subtracting functions
To calculate the net change of NO3- concentration in the system, we can employ algebra to substract the rate NO3- is leaving the system from the rate NO3- is entering the system.
Net rate of change:
\[N(t) = R(t) - C(t)\] \[= (0.5 + 0.3sin(\frac{\pi t}{12})) - (0.2 + 0.1cos(\frac{\pi t}{6}))\] \[= 0.3 + 0.3sin(\frac{\pi t}{12}) - 0.1cos(\frac{\pi t}{6})\]Using the Sum and Difference Rule to Integrate Terms in a Function Seperately
The sum and difference rule tells us the difference or sum of functions is the difference or sum of their individual derivatives. This also applies to integration. According to Open Stax’s Calculus 1 book, Section 4.10 Antiderivatives, if $F$ and $G$ are antiderivatives of $f$ and $g$, and $k$ is any real number then:
\[\int f(x) \pm g(x) dx = F(x) \pm G(x) + C\]The sum and difference rule allows us to break down one long integration problem
\[\int_0^{12} N(t)\, dt= \int_0^{12} 0.3 + 0.3sin(\frac{\pi t}{12}) - 0.1cos(\frac{\pi t}{6})\, dt\]into three sizeable chunks:
\[= \int_0^{12} 0.3\,dt+ \int_0^{12} 0.3sin(\frac{\pi t}{12})\,dt + \int_0^{12} 0.1cos(\frac{\pi t}{6})\, dt\]For now, we’ll surpress the bounds:
\[= \int 0.3\,dt+ \int 0.3sin(\frac{\pi t}{12})\,dt + \int0.1cos(\frac{\pi t}{6})\, dt\]Using the Constant Rule for Integration
To integrate the first term in our Net Change in NO3- Concentration function, we must use the constant rule for integration. If $k$ is a constant, then:
\[\int k\,dx = kx + C\]Therefore:
\[\int 0.3\, dt= 0.3t + C\]We can ignore the constant $C$ because we are evaluating a definite integral. Remember, we are temporarily surpessing the bounds of our integral, $t=0$ and $t=12$.
Focusing on the Derivation Component of U-Substition
To integrate the second term of the function, we must use $u$ substitution.
Let $u= \frac{\pi t}{12}$
For the function to be completely in terms of $u$, we must find the derivative of $u$.
$\frac{d}{dx}\left[\frac{\pi t}{12}\right] = du$
To make the constant more of apparent we can the rewrite the expression as the following:
$\frac{d}{dx}\left[(\frac{\pi}{12})t\right] = du$
The power rule states that for any function $f(x)=x^n$, the derivative is:
\[\frac{d}{dx}\left[x^n\right] = nx^{(n-1)}\]Therefore:
\[\frac{d}{dx}\left[(\frac{\pi}{12})t\right] = 1 * \frac{\pi}{12} t^{(1-1=0)}= \frac{\pi}{12}\]Our value for $du$ is $\frac{\pi}{12}$ and in differntial form it is written as
\[du= \frac{\pi}{12}dt\]which can be rearranged to
\[dt= \frac{12}{\pi}du\]Integrating the 2nd and Final Term
The second term, $\int 0.3sin(\frac{\pi t}{12})\,dt$, in terms of $u$ is the following:
\[\int 0.3 sin(u)\frac{12}{\pi}\,du\]We can bring the constants in front of the integral due to the rule of linearity:
\[0.3* \frac{12}{\pi}\int sin(u)\,du\]Knowledge of basic integration functions such as
\[\int sin(x)dx = -cos(x) +C\]allows us to arrive to
\[-\frac{3.6}{\pi} cos(u)\]and to finish the process of integration we resubstitute $\frac{\pi}{12}$ for $u$:
\[-\frac{3.6}{\pi}cos(\frac{\pi}{12})\]The process we used to integrate the second term of our Net Change in NO3- Concentration function, can be used to integrate the final term.
\[\int -0.1cos(\frac{\pi t}{6})\, dt = -\frac{0.6}{\pi}sin(\frac{\pi t}{6})\]Combining Antiderivatives and Evaluating the Integral
To evalulate the integral of our Net Change in NO3- Concentration function, we must combine our three antiderivatives.
\[\left[0.3t - \frac{3.6}{\pi}cos(\frac{\pi t}{12})-\frac{0.6}{\pi}sin(\frac{\pi t}{6})\right]_0^{12}\]Using a calculator and rounding to the nearest tenth place, at $t=0$ (the beginning of the day) the value is $-1.15$ ppm. At $t=12$ (the afternoon) the value is $4.75$ ppm. Substracting the value at $t=12$ from the value at $t=0$ will result in the Net Change in NO3- Concentration.
\[4.75 - (-1.15) = 5.9\: ppm\]Remembering the Initial Condition
The original question asks for the actual quantity of NO3- concentration in the hydroponic system at noon. $5.9$ ppm is the quantity of change. We must add the initial condition, $150$ ppm, to the quantity of change to solve for the actual quantity. Our final answer is:
\[155.9\,ppm\]Conclusion
I did not orignally imagine this blogpost would be as expansive as it is, and still I skipped many oppurtunties to explain my thinking. My math courses thus far have not required us to explain what we understand, but only perform. If there are any errors in my work, I welcome feedback. Additionally, if there are any nuances or complexities to be added to our theoretical hydroponics system, please share.