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Chapter 3: Problem 19

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. \(P(t)=\quad \frac{t^{3}}{3}-t^{2}+t\)

Short Answer

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

Step by step solution

01

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 \( t \): \[ 2t - 10 = 0 \] \[ 2t = 10 \] \[ t = 5 \]
02

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 \( t \): \[ 3t^2 - 3 = 0 \] \[ 3(t^2 - 1) = 0 \] \[ t^2 = 1 \] \[ t = \pm 1 \]
03

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 \( t \): \[ 10t - 1 = 0 \] \[ 10t = 1 \] \[ t = 0.1 \]
04

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: \[ (t-3)(t-1) = 0 \] \[ t = 3 \text{ or } t = 1 \]
05

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: \[ t^2 = 4 \] \[ t = \pm 2 \]
06

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 \( t > 0 \))
07

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 \( t > 0 \))
08

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:\[ (t - 1)^2 = 0 \] \[ t = 1 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攅ssential 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.

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Most popular questions from this chapter

In a chemical reaction of the form$$2 \mathrm{~A}+\mathrm{B} \longrightarrow \mathrm{A}_{2} \mathrm{~B} $$, where the reaction does not involve intermediate compounds, the reaction rate is proportional to [A] \(^{2}\) [B] where [A] and [B] denote, respectively, the concentrations of the components A and B. Let \(a\) and \(b\) denote \([\mathrm{A}]\) and \([\mathrm{B}]\), respectively, and assume that \([\mathrm{B}]\) is much greater than \([\mathrm{A}]\) so that \((b(t) \gg a(t) .\) The rate at which \(a\) changes may be written $$a^{\prime}(t)=-k(a(t))^{2} b(t)=-K(a(t))^{2}$$ We have assumed that \(b(t)\) is (almost) constant because [A] is the limiting concentration of the reaction. Let $$a(t)=\frac{a_{0}}{1-a_{0} K t}$$ Show that \(\quad a(0)=a_{0}\). Use the Definition of Derivative 3.22 , $$a^{\prime}(t)=\lim _{b \rightarrow t} \frac{a(b)-a(t)}{b-t}$$ to compute \(a^{\prime}(t)\). Then compute \((a(t))^{2}\) and show that $$a^{\prime}(t)=-K(a(t))^{2}$$

Compute the derivatives of the following polynomials as in the previous exercises. Use only one rule for each step written, and write the name of the rule used for each step. a. \(P(t)=15 t^{2}-32 t^{6}\) c. \(P(t)=\frac{t^{4}}{4}+\frac{t^{3}}{3}\) b. \(P(t)=1+t+t^{2}+t^{3}\) d. \(P(t)=\left(1+t^{2}\right)^{2}\) e. \(P(t)=31 t^{52}-82 t^{241}+\pi t^{314}\) f. \(P(t)=2^{5}+17 t^{5}\) g. \(P(t)=\sqrt{2}-\frac{t^{7}}{427}+18 t^{35}\) h. \(P(t)=17^{3}-\frac{t^{23}}{690}+5 t^{705}\)

What is the optimum radius of the trachea when coughing? The objective is for the flow of air to create a strong force outward in the throat to clear it. For this problem you should perform the following experiment. Hold your hand about \(10 \mathrm{~cm}\) from your mouth and blow on it (a) with your lips compressed almost closed but with a small stream of air escaping, (b) with your mouth wide open, and (c) with your lips adjusted to create the largest force on your hand. With (a) your lips almost closed there is a high pressure causing rapid air flow but a small stream of air and little force. With (b) your mouth wide open there is a large stream of air but with little pressure so that air flow is slow. The largest force (c) is created with an intermediate opening of your lips where there is a notable pressure and rapid flow of substantial volume of air. Let \(R\) be the normal radius of the trachea and \(r

Show that for \(s(t)=\frac{g}{2} t^{2}+v_{0} t+\gamma, s^{\prime}(t)=g t+v_{0}\).

a. Find a number, \(\delta>0,\) so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(|2 x-4|<0.01\). b. Find a number, \(\delta\), so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(\left|x^{2}-4\right|<0.01\). c. Find a number, \(\delta\), so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(\left|\frac{1}{x}-\frac{1}{2}\right|<0.01\). d. Find a number, \(\delta,\) so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(\left|x^{3}-8\right|<0.01\). e. Find a number, \(\delta\), so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(\left|\frac{x}{x+1}-\frac{2}{3}\right|<0.01\). f. Find a number, \(\delta>0\), so that if \(x\) is a number and \(0<|x-9|<\delta\) then \(|\sqrt{x}-3|<0.01\). g. Find a number, \(\delta\), so that if \(x\) is a number and \(0<|x-8|<\delta\) then \(|\sqrt[3]{x}-2|<0.01\). h. Find a number, \(\delta>0\), so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(\left|x^{4}-3 x-13\right|<0.01\)
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