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

Question

Find \(p_{0}, p_{1}\), and \(p_{2}\) for which the polynomial, \(p(t)=p_{0}+p_{1} t+p_{2} t^{2}\), satisfies a. \(p(0)=5, \quad p^{\prime}(0)=-2,\) and \(p^{\prime \prime}(0)=\frac{1}{3}\). b. \(p(0)=1, \quad p^{\prime}(0)=0, \quad\) and \(\quad p^{\prime \prime}(0)=-\frac{1}{2}\). c. \(p(0)=0, \quad p^{\prime}(0)=1, \quad\) and \(\quad p^{\prime \prime}(0)=0\). d. \(p(0)=1, \quad p^{\prime}(0)=0, \quad\) and \(\quad p^{\prime \prime}(0)=-1\). e. \(p(0)=1, \quad p^{\prime}(0)=1, \quad\) and \(\quad p^{\prime \prime}(0)=1\). f. \(p(0)=17, p^{\prime}(0)=-15,\) and \(p^{\prime \prime}(0)=12\).

Step-by-Step Solution

Answer

Each condition set gives specific values for \( p_0, p_1, \) and \( p_2 \). For example, in (a), they are 5, -2, and \( \frac{1}{6} \).

Step 1: Understand polynomial and derivatives

The polynomial is given as \( p(t) = p_0 + p_1 t + p_2 t^2 \). The first derivative is \( p'(t) = p_1 + 2p_2 t \) and the second derivative is \( p''(t) = 2p_2 \).

Step 2: Use the conditions for each case

Substitute \( t = 0 \) in the polynomial and its derivatives to find \( p_0, p_1, \) and \( p_2 \).

Step 3: Solve for condition (a)

Given \( p(0) = 5 \), \( p'(0) = -2 \), and \( p''(0) = \frac{1}{3} \):- \( p_0 = 5 \) (from \( p(0) = 5 \))- \( p_1 = -2 \) (from \( p'(0) = -2 \))- \( 2p_2 = \frac{1}{3} \Rightarrow p_2 = \frac{1}{6} \) (from \( p''(0) = \frac{1}{3} \))

Step 4: Solve for condition (b)

Given \( p(0) = 1 \), \( p'(0) = 0 \), \( p''(0) = -\frac{1}{2} \):- \( p_0 = 1 \) (from \( p(0) = 1 \))- \( p_1 = 0 \) (from \( p'(0) = 0 \))- \( 2p_2 = -\frac{1}{2} \Rightarrow p_2 = -\frac{1}{4} \) (from \( p''(0) = -\frac{1}{2} \))

Step 5: Solve for condition (c)

Given \( p(0) = 0 \), \( p'(0) = 1 \), \( p''(0) = 0 \):- \( p_0 = 0 \) (from \( p(0) = 0 \))- \( p_1 = 1 \) (from \( p'(0) = 1 \))- \( 2p_2 = 0 \Rightarrow p_2 = 0 \) (from \( p''(0) = 0 \))

Step 6: Solve for condition (d)

Given \( p(0) = 1 \), \( p'(0) = 0 \), \( p''(0) = -1 \):- \( p_0 = 1 \) (from \( p(0) = 1 \))- \( p_1 = 0 \) (from \( p'(0) = 0 \))- \( 2p_2 = -1 \Rightarrow p_2 = -\frac{1}{2} \) (from \( p''(0) = -1 \))

Step 7: Solve for condition (e)

Given \( p(0) = 1 \), \( p'(0) = 1 \), \( p''(0) = 1 \):- \( p_0 = 1 \) (from \( p(0) = 1 \))- \( p_1 = 1 \) (from \( p'(0) = 1 \))- \( 2p_2 = 1 \Rightarrow p_2 = \frac{1}{2} \) (from \( p''(0) = 1 \))

Step 8: Solve for condition (f)

Given \( p(0) = 17 \), \( p'(0) = -15 \), \( p''(0) = 12 \):- \( p_0 = 17 \) (from \( p(0) = 17 \))- \( p_1 = -15 \) (from \( p'(0) = -15 \))- \( 2p_2 = 12 \Rightarrow p_2 = 6 \) (from \( p''(0) = 12 \))

Key Concepts

Understanding Derivatives

Derivatives are fundamental in calculus and describe how a function changes as its input changes. When analyzing polynomial functions like the one in this exercise, understanding derivatives is crucial as they indicate the rate of change. The first derivative of a polynomial function provides information on the slope of the tangent line to the curve at any point. For the polynomial \( p(t) = p_0 + p_1 t + p_2 t^2 \), the first derivative is \( p'(t) = p_1 + 2p_2 t \). This formula demonstrates that the slope of the curve changes linearly with \( t \). The value of \( p'(0) \) gives a quick insight into the initial rate of change at \( t = 0 \).
Derivatives are also used to optimize functions, find maximum and minimum points, and solve real-world problems where rates of change are essential. In the given exercise, finding the values of the derivative at specific points helps determine the coefficients of the polynomial, ensuring it behaves as described by the conditions set.

Exploring Quadratic Polynomials

Quadratic polynomials, expressed in the form \( ax^2 + bx + c \), have unique properties that make them useful in various fields such as physics, economics, and engineering. These polynomials are characterized by their curved graph, known as a parabola. In this exercise, the quadratic polynomial is \( p(t) = p_0 + p_1 t + p_2 t^2 \).
Quadratics are notable for their symmetry and the fact that they have only one maximum or minimum point when plotted on a graph, known as the vertex. The position and nature of this vertex depend on the coefficients. In this form, \( p_2 \) affects the opening direction of the parabola (upward if \( p_2 > 0 \), downward if \( p_2 < 0 \)), \( p_1 \) affects the tilt of the parabola, while \( p_0 \) handles the vertical shift.
By determining \( p_0, p_1, \) and \( p_2 \) as in the exercise, one sets the specific characteristics of the quadratic to meet precise requirements, such as specific initial values and rates of change at certain points.

Importance of the Second Derivative

The second derivative of a function provides significant insight into the curvature and concavity of the graph. For the quadratic polynomial \( p(t) = p_0 + p_1 t + p_2 t^2 \), the second derivative is constant: \( p''(t) = 2p_2 \). The constancy arises because the highest degree term in the polynomial is quadratic.
The second derivative tells us about the concavity of the parabola. If \( p''(t) > 0 \), the graph is concave up, resembling a U shape; if \( p''(t) < 0 \), it is concave down, similar to an upside-down U. This attribute is critical when analyzing the behavior of the polynomial over its domain, as it helps predict the direction of the graph.
In practical terms, understanding the second derivative helps identify acceleration in physics, determine economic growth rates, and in optimization, it can distinguish local maxima from minima. In this exercise, setting \( p''(0) \) according to different scenarios helps tailor the polynomial to meet specific concavity and curvature conditions.