Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Suppose the rate of glucose production in a corn plant is proportional to sunlight intensity and can be approximated by $$ R(t)=K(t+7)^{2}(t-7)^{2}=K\left(t^{4}-98 t^{2}+2401\right) \quad-7 \leq t \leq 7 $$ Time is measured so that sunrise is at -7 hours, the sun is at its zenith at 0 hours and sets at 7 hours. The quantity \(Q(x)\) of glucose produced during the period \([-7, x]\) is $$ Q(x)=\int_{-7}^{x} R(t) d t=\int_{-7}^{x} K\left(t^{4}-98 t^{2}+2401\right) d t=K \int_{-7}^{x}\left(t^{4}-98 t^{2}+2401\right) d t $$ By the Fundamental Theorem of Calculus, $$ Q^{\prime}(x)=K\left(x^{4}-98 x^{2}+2401\right) $$ a. Sketch the graph of \(R(t)\). At what time is the sun most intense? b. Show that if \(U(x)=\frac{x^{5}}{5}\) then \(U^{\prime}(x)=x^{4}\). c. Find an example of a function, \(V(x)\) such that \(V^{\prime}(x)=-98 x^{2}\). d. Find an example of a function, \(W(x)\) such that \(W^{\prime}(x)=2401\). e. Let \(G(x)=K\left(\frac{x^{5}}{5}-\frac{98}{3} x^{3}+2401 x\right)\). Show that \(G^{\prime}(x)=Q^{\prime}(x)\). f. Conclude that there is a number, \(C\), such that $$ Q(x)=G(x)+C=K\left[\frac{x^{5}}{5}-\frac{98}{3} x^{3}+2401 x\right]+C $$ a. Sketch the graph of \(R(t)\). At what time is the sun most intense? b. Show that if \(U(x)=\frac{x^{5}}{5}\) then \(U^{\prime}(x)=x^{4}\). c. Find an example of a function, \(V(x)\) such that \(V^{\prime}(x)=-98 x^{2}\). d. Find an example of a function, \(W(x)\) such that \(W^{\prime}(x)=2401\). e. Let \(G(x)=K\left(\frac{x^{5}}{5}-\frac{98}{3} x^{3}+2401 x\right)\). Show that \(G^{\prime}(x)=Q^{\prime}(x)\). f. Conclude that there is a number, \(C,\) such that $$ Q(x)=G(x)+C=K\left[\frac{x^{5}}{5}-\frac{98}{3} x^{3}+2401 x\right]+C $$ g. Why is \(Q(-7)=0\). h. Evaluate \(C\). i. Compute \(Q(7),\) the amount of glucose produced during the day.

Step-by-Step Solution

Answer

Sunlight intensity peaks at noon (t=0). C=0 and compute Q(7) based on the function given.

Step 1: Graph of R(t) and Most Intense Sunlight

To sketch the graph of \(R(t)\), multiply out the polynomial: \( R(t) = K[(t+7)^2(t-7)^2] = K[t^4 - 98t^2 + 2401] \). This quartic polynomial is symmetric about \(t=0\) and has its maximum magnitude at \(t=0\) (zenith). Hence, the sun is most intense at noon, \(t=0\).

Step 2: Differentiate U(x)

To show that \(U(x)=\frac{x^5}{5}\), differentiate: \( U'(x) = \frac{d}{dx} \left( \frac{x^5}{5} \right) = x^4 \), which matches the requirement.

Step 3: Find V(x) such that V'(x)=-98x^2

Integrate \(-98x^2\) to find \(V(x)\): \(V(x) = \int -98x^2 \, dx = -\frac{98}{3}x^3 + C\). Choose \(C = 0\) for simplicity.

Step 4: Find W(x) such that W'(x)=2401

Integrate the constant \(2401\) with respect to \(x\): \(W(x) = \int 2401 \, dx = 2401x + C\). Choose \(C = 0\) for simplicity.

Step 5: Show G'(x) = Q'(x)

Differentiate \(G(x) = K\left(\frac{x^5}{5} - \frac{98}{3}x^3 + 2401x \right)\). This yields: \(G'(x) = K(x^4 - 98x^2 + 2401) = Q'(x)\).

Step 6: Conclude the Form of Q(x)

\( Q(x) = G(x) + C = K\left[ \frac{x^5}{5} - \frac{98}{3} x^3 + 2401 x \right] + C \), where \(C\) is a constant of integration.

Step 7: Reason Q(-7)=0

The integral \(Q(x) = \int_{-7}^x R(t) \, dt\) computes the area from \(-7\) to \(x\). \(Q(-7) = 0\) because the interval is a single point, meaning there is no area under the curve when \(x=-7\).

Step 8: Evaluate C

From \(Q(-7) = G(-7) + C = 0\), substitute \(-7\) into \(G(x)\):\[ G(-7) = K\left(\frac{(-7)^5}{5} - \frac{98}{3}(-7)^3 + 2401(-7) \right) \]. This simplifies to 0; hence, \(C = 0\).

Step 9: Compute Q(7)

Evaluate \(Q(7)\) using \(Q(x) = G(x) + C\): \[ Q(7) = K\left(\frac{7^5}{5} - \frac{98}{3} \, 7^3 + 2401 \, 7 \right) + 0 \]. Compute the value inside the brackets and multiply by \(K\).

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving variables, coefficients, and exponents. In this exercise, the rate of glucose production, \( R(t) \), is represented by a polynomial of degree 4, which is also known as a quartic polynomial. Such functions are useful because they can model a wide range of natural phenomena, including the changing rate of glucose production in plants as light intensity varies throughout the day.
This polynomial is represented as \( R(t) = K[(t+7)^2(t-7)^2] = K[t^4 - 98t^2 + 2401] \). Here, \( t \) plays the role of 'time', and the function is expanded to reveal its more familiar polynomial form. This method helps in simplifying complex expressions, making it easier to interpret and analyze their behavior over time.

Differentiation

Differentiation is a fundamental concept in calculus used to determine the rate at which a function is changing at any point. It involves calculating the derivative of a function and can tell us how the function behaves dynamically - in this case, how quickly glucose production accelerates or decelerates with each passing hour.
In this problem, we differentiate functions like \( U(x) = \frac{x^5}{5} \) to find \( U'(x) = x^4 \) and \( V(x) \) so that \( V'(x) = -98x^2 \). Differentiation indicates the polynomial's direction at any given point, helping analyze where maxima, minima, or inflection points occur. Here, differentiating helps not just to understand the curve’s behavior, but also to verify the parental relationship between \( G(x) \) and \( Q'(x) \), as \( G'(x) = Q'(x) \).

Integration

Integration is the inverse process of differentiation. It sums up all the infinitesimal changes of a function, thus calculating the total accumulation such as area under a curve. Here, we use integration to determine the total glucose produced over time, expressed in \( Q(x) = \int_{-7}^{x} R(t) \, dt \).
By integrating polynomial terms individually, like \(-98x^2\) and the constant \(2401\), we advance from rates of change to quantities. Integration helps bridge the gap between differential coefficients and the comprehensive function, unveiling cumulative outputs over time intervals. The integral \( Q(x) \) essentially tells us the amount of glucose produced between sunrise and any time \( x \) during daylight.

Area under a curve

In calculus, the area under a curve on a graph represents the integral of a function over a given interval. Calculating this area provides insight into the total accumulated quantity, such as glucose produced over time.
For this exercise, \( Q(x) = \int_{-7}^{x} R(t) \, dt \) calculates the glucose produced from sunrise \( t = -7 \) to any time \( x \). When the time does not exceed its starting point \( t = -7 \), \( Q(-7) = 0 \), as there is no area (or production) yet.
To find maximum daily production, we compute the integral up to \( t = 7 \), obtaining \( Q(7) \), which expresses the total glucose synthesis throughout the entire day under the curve of \( R(t) \). The entire process exemplifies how integration can sift significant information out of rate functions by finding corresponding areas below the curve.