Problem 19 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Find values of \(t\) for which \(P^{\prime}(t)=0\) for: a. \(P(t)=t^{2}-10 t+35\) c. \(P(t)=5 t^{2}-t+1\) b. \(P(t)=t^{3}-3 t+8\) d. \(P(t)=t^{3}-6 t^{2}+9 t+7\) e. \(P(t)=7 t^{4}-56 t^{2}+8\) f. \(P(t)=t+\frac{1}{t} \quad t>0\) g. \(P(t)=\frac{t}{2}+\frac{2}{t}\) h. \(P(t)=\quad \frac{t^{3}}{3}-t^{2}+t\)

Step-by-Step Solution

Answer

a. t=5; b. t=±1; c. t=0.1; d. t=1,3; e. t=0,±2; f. t=1; g. t=2; h. t=1.

Step 1: Derive the Function for a

For the function \[ P(t) = t^2 - 10t + 35 \] Take the derivative to find \( P'(t) \): \[ P'(t) = \frac{d}{dt}(t^2) - \frac{d}{dt}(10t) + \frac{d}{dt}(35) \] \[ P'(t) = 2t - 10 \] Set \( P'(t) = 0 \) and solve for \( t \): \[ 2t - 10 = 0 \] \[ 2t = 10 \] \[ t = 5 \]

Step 2: Derive the Function for b

For the function \[ P(t) = t^3 - 3t + 8 \] Take the derivative: \[ P'(t) = \frac{d}{dt}(t^3) - \frac{d}{dt}(3t) + \frac{d}{dt}(8) \] \[ P'(t) = 3t^2 - 3 \] Set \( P'(t) = 0 \) and solve for \( t \): \[ 3t^2 - 3 = 0 \] \[ 3(t^2 - 1) = 0 \] \[ t^2 = 1 \] \[ t = \pm 1 \]

Step 3: Derive the Function for c

For the function \[ P(t) = 5t^2 - t + 1 \] Take the derivative: \[ P'(t) = \frac{d}{dt}(5t^2) - \frac{d}{dt}(t) + \frac{d}{dt}(1) \] \[ P'(t) = 10t - 1 \] Set \( P'(t) = 0 \) and solve for \( t \): \[ 10t - 1 = 0 \] \[ 10t = 1 \] \[ t = 0.1 \]

Step 4: Derive the Function for d

For the function \[ P(t) = t^3 - 6t^2 + 9t + 7 \] Take the derivative: \[ P'(t) = \frac{d}{dt}(t^3) - \frac{d}{dt}(6t^2) + \frac{d}{dt}(9t) + \frac{d}{dt}(7) \] \[ P'(t) = 3t^2 - 12t + 9 \] Set \( P'(t) = 0 \) and solve for \( t \): \[ 3t^2 - 12t + 9 = 0 \] Divide by 3: \[ t^2 - 4t + 3 = 0 \] Use factoring: \[ (t-3)(t-1) = 0 \] \[ t = 3 \text{ or } t = 1 \]

Step 5: Derive the Function for e

For the function \[ P(t) = 7t^4 - 56t^2 + 8 \] Take the derivative: \[ P'(t) = \frac{d}{dt}(7t^4) - \frac{d}{dt}(56t^2) \] \[ P'(t) = 28t^3 - 112t \] Set \( P'(t) = 0 \) and solve for \( t \): \[ 28t^3 - 112t = 0 \] Factor out \[ t(28t^2 - 112) = 0 \] \[ t = 0 \text{ or } 28(t^2 - 4) = 0 \] Solve further: \[ t^2 = 4 \] \[ t = \pm 2 \]

Step 6: Derive the Function for f

For the function \[ P(t) = t + \frac{1}{t} \] where \( t > 0 \)Take the derivative: \[ P'(t) = \frac{d}{dt}(t) + \frac{d}{dt}\left(\frac{1}{t}\right) \] \[ P'(t) = 1 - \frac{1}{t^2} \] Set \( P'(t) = 0 \) and solve for \( t \): \[ 1 - \frac{1}{t^2} = 0 \] \[ \frac{1}{t^2} = 1 \] \[ t^2 = 1 \] \[ t = 1 \] (since \( t > 0 \))

Step 7: Derive the Function for g

For the function \[ P(t) = \frac{t}{2} + \frac{2}{t} \] Take the derivative: \[ P'(t) = \frac{d}{dt}\left(\frac{t}{2}\right) + \frac{d}{dt}\left(\frac{2}{t}\right) \] \[ P'(t) = \frac{1}{2} - \frac{2}{t^2} \] Set \( P'(t) = 0 \) and solve for \( t \): \[ \frac{1}{2} - \frac{2}{t^2} = 0 \] \[ \frac{2}{t^2} = \frac{1}{2} \] \[ t^2 = 4 \] \[ t = 2 \] (since \( t > 0 \))

Step 8: Derive the Function for h

For the function \[ P(t) = \frac{t^3}{3} - t^2 + t \] Take the derivative: \[ P'(t) = \frac{d}{dt}\left(\frac{t^3}{3}\right) - \frac{d}{dt}(t^2) + \frac{d}{dt}(t) \] \[ P'(t) = t^2 - 2t + 1 \] Set \( P'(t) = 0 \) and solve for \( t \): \[ t^2 - 2t + 1 = 0 \] Use factoring:\[ (t - 1)^2 = 0 \] \[ t = 1 \]

Key Concepts

Quadratic Equations

A quadratic equation is a polynomial equation of degree 2, and it is usually written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. When dealing with quadratic equations, you'll often find yourself solving for \( x \) in a way that makes the equation equal to zero. This is because quadratic equations often describe parabolic shapes and are crucial in finding important points like vertices and intercepts. For example, consider \( P(t) = t^2 - 10t + 35 \). This is a quadratic function where the derivative, \( P'(t) = 2t - 10 \), helps us find the rate of change of the function. We solve \( 2t - 10 = 0 \) to find the critical points where the rate of change equals zero, giving us \( t = 5 \). This critical point is where the graph might reach a maximum or minimum, depending on the direction of the parabola.

Factoring

Factoring is a mathematical process used to break down complex equations into simpler multipliers that can be multiplied together to give the original equation. This process is especially useful in solving quadratic equations as it allows us to find the roots, or the values of \( x \) that make the equation zero. To factor a quadratic equation like \( t^2 - 4t + 3 = 0 \), we look for two numbers that multiply to give us the constant term \( c \) (which is 3 in this case) and add up to the linear coefficient \( b \) (which is -4). After identifying these numbers, we can express the equation as \( (t - 3)(t - 1) = 0 \). Solving these factors gives us the roots \( t = 3 \) and \( t = 1 \). Factoring is a critical step that simplifies quadratic equations into solvable parts, providing clear intersection points on the graph.

Differentiation

Differentiation is a fundamental mathematical process used to find the derivative of a function. The derivative represents the rate at which the function value changes as the input changes, essentially capturing the function's gradient or slope at any particular point. In calculus, differentiation is essential for finding slopes of curves, and rates of change, and for solving optimization problems. Consider the function \( P(t) = t^3 - 3t + 8 \). Differentiating gives us \( P'(t) = 3t^2 - 3 \), which shows how \( P(t) \) changes as \( t \) changes. Solving \( 3t^2 - 3 = 0 \) yields \( t^2 = 1 \), thus \( t = \pm 1 \). Differentiation is crucial for identifying such critical points, where the slope of the tangent to the curve is zero, indicating a potential maximum, minimum, or point of inflection on the graph.

Critical Points

Critical points are specific locations on the graph of a function where the derivative is zero or undefined. These points are central to understanding the behavior of the function because they can indicate maximums, minimums, or points of inflection—essential features of the graph. By studying these points, mathematicians and teachers can interpret significant changes in direction on the graph of the function.For the function \( P(t) = 7t^4 - 56t^2 + 8 \), the derivative is \( P'(t) = 28t^3 - 112t \). Solving \( 28t^3 - 112t = 0 \) involves factoring out commont terms as \( t(28t^2 - 112) = 0 \), leading to critical points at \( t = 0 \) or \( t = \pm 2 \). Evaluating the function at these points helps in determining whether they represent peaks, valleys, or saddle points in the graph. Recognizing and analyzing critical points is key for a comprehensive exploration of the function's overall shape and the transition points.