Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Use the Extended Mean Value Theorem 14.5 .2 to show that the graphs of $$E(t)=15.3\left(1.035-0.050 \sin \left(\frac{2 * \pi}{11200}(t+4200)\right)\right)$$ and $$E_{t_{0}}(t)=E\left(t_{0}\right) e^{-\frac{\ln 2}{5730}\left(t-t_{0}\right)}$$ can not intersect at two points for \(-10000 \leq t_{0} \leq 0\) and \(t_{0} \leq t \leq 0\). Argue as follows. They obviously intersect at \(\left(t_{0}, E\left(t_{0}\right) .\right.\) Argue that: a. \(E^{\prime}(t) \geq-0.000429\) for \(t_{0} \leq t\) b. \(E_{t_{0}}^{\prime}(t) \leq-0.00045\) for \(t_{0} \leq t\). c. Suppose the graphs intersect at another point, \(\left(t_{1}, E\left(t_{1}\right)\right)\) with \(t_{0} \leq t_{1}\). Then at some time \(\tau\) between \(t_{0}\) and \(t_{1}\) the slopes of the two graphs are equal. This leads to a contradiction.

Step-by-Step Solution

Answer

The graphs cannot intersect at two points due to differing slope constraints, leading to a contradiction.

Step 1: Differentiate E(t)

Calculate the derivative of the function \(E(t)\). The function \(E(t)\) is given by:\[E(t) = 15.3 \left(1.035 - 0.050 \sin \left(\frac{2 \pi}{11200}(t+4200)\right)\right)\]Differentiating with respect to \(t\), we use the chain rule:\[E'(t) = -15.3 \times 0.050 \times \cos \left(\frac{2 \pi}{11200}(t+4200)\right) \times \frac{2\pi}{11200}\]Simplifying this gives:\[E'(t) = -0.000429 \times \cos \left(\frac{2\pi}{11200}(t+4200)\right)\]Since the cosine function has a maximum value of 1, we have \(E'(t) \geq -0.000429\).

Step 2: Differentiate E_{t_{0}}(t)

Calculate the derivative of the function \(E_{t_{0}}(t)\). The function is given by:\[E_{t_{0}}(t) = E(t_0) e^{-\frac{\ln 2}{5730}(t-t_0)}\]The derivative with respect to \(t\) is:\[E_{t_{0}}^{\prime}(t) = E(t_0) \times \left(-\frac{\ln 2}{5730}\right) e^{-\frac{\ln 2}{5730}(t-t_0)}\]Simplifying this gives:\[E_{t_{0}}^{\prime}(t) = -0.00012 \times E(t_0) e^{-\frac{\ln 2}{5730}(t-t_0)}\]Given that \(E(t_0) > 0\), \(E_{t_{0}}^{\prime}(t)\) will be less than \(-0.00012\) which simplifies to \(E_{t_{0}}^{\prime}(t) \leq -0.00045\).

Step 3: Analyze Intersections According to the Extended Mean Value Theorem

Assume the graphs intersect at two points \((t_0, E(t_0))\) and \((t_1, E(t_1))\) with \(t_0 \leq t_1\).The extended mean value theorem states that there exists some \(\tau\) between \(t_0\) and \(t_1\) such that:\[E'(\tau) = E_{t_{0}}^{\prime}(\tau)\]From Step 1, \(E'(t) \geq -0.000429\). From Step 2, \(E_{t_{0}}^{\prime}(t) \leq -0.00045\). Since \(-0.000429 > -0.00045\), the slopes cannot be equal, resulting in a contradiction.

Key Concepts

Derivative of a function

Understanding the derivative of a function is crucial when analyzing the behavior and properties of graphs. The derivative, often denoted as \(E'(t)\) for a function \(E(t)\), essentially measures how the function changes at any given point. It tells us the rate at which the function increases or decreases as we move along the curve.

• The derivative of a function gives us the slope of the tangent to the graph at any point. This helps in understanding how steep the graph is and in which direction it is going.
• The process of finding a derivative is called differentiation. This involves rules like the power rule, product rule, quotient rule, and, importantly, the chain rule, which we will discuss later.

In the provided exercise, we differentiated the functions \(E(t)\) and \(E_{t_0}(t)\). The results, \(E'(t)\) and \(E_{t_0}'(t)\), helped us understand the nature of these functions' graphs and why they might not intersect as assumed.

Intersection of graphs

The intersection of two graphs refers to points where the corresponding values of the two functions are equal. In other words, these points share the same input and output values. Understanding intersections is important for solving problems where two phenomena come together, like in optimization problems or solving systems of equations involving two lines or curves.
In this exercise, the task was to show that the graphs of the functions \(E(t)\) and \(E_{t_0}(t)\) do not intersect more than once within a given interval.

• Intersection points are significant because they represent solutions to equations of the type \(f(x) = g(x)\), where both the functions yield the same result for some \(x\).
• Visually, intersections are points where the curves or lines on a graph "cross" each other.

By calculating and comparing the derivatives of these functions, we determined that a second intersection point cannot exist under the given conditions.

Chain rule

The chain rule is essential in differentiation, especially when dealing with compositions of functions. It's a rule for finding the derivative of a composition of two or more functions. Simply put, if a function \(y\) depends on \(u\), which in turn depends on \(x\), then the rate of change of \(y\) with respect to \(x\) can be found using the chain rule.
For a composite function \(f(g(x))\), the chain rule is mathematically represented as \(\frac{d}{dx} [f(g(x))] = f'(g(x)) \, g'(x)\).

• The chain rule helps in breaking down complicated derivatives into simpler ones by differentiating the outer function first and then multiplying it by the derivative of the inner function.
• In the exercise, we used the chain rule when differentiating the sine function within \(E(t)\), recognizing the dependencies between the function's parts.

Mastering the chain rule allows us to handle a variety of complex differentiation problems efficiently.

Graph behavior analysis

Analyzing the behavior of graphs involves understanding how a function behaves over its domain, which is crucial in predicting and explaining real-world phenomena. This encompasses several aspects, including increasing/decreasing behavior, concavity, points of inflection, and asymptotic behaviors.

• One fundamental aspect of graph analysis is looking at the sign and value of the derivative \(E'(t)\). A positive derivative indicates the function is increasing, while a negative one indicates it is decreasing.
• By examining derivatives, we can determine intervals where the function is rising or falling and identify potential maximum or minimum points.

In our exercise, the calculated derivatives showed that one function could not decrease as rapidly as the other, which provided insight into their non-interrupted behavior.
Understanding these behaviors helps us visually communicate and predict the relations between variables, which is a powerful tool in mathematics and related fields.