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Show that a. \(P(t)=5 t+3\) satisfies \(P(0)=3\) and \(P^{\prime}(t)=5\) b. \(\quad P(t)=8 t+2\) satisfies \(P(0)=2\) and \(P^{\prime}(t)=8\) c. \(P(t)=t^{2}+3 t+7 \quad\) satisfies \(P(0)=7\) and \(P^{\prime}(t)=2 t+3\) d. \(P(t)=-2 t^{2}+5 t+8\) satisfies \(P(0)=8\) and \(P^{\prime}(t)=-4 t+5\) e. \(\quad P(t)=(3 t+4)^{2}\) satisfies \(P(0)=16\) and \(P^{\prime}(t)=6 \sqrt{P(t)}\) f. \(P(t)=(5 t+1)^{2}\) satisfies \(P(0)=1\) and \(P^{\prime}(t)=10 \sqrt{P(t)}\) g. \(\quad P(t)=(1-2 t)^{-1}\) satisfies \(P(0)=1\) and \(P^{\prime}(t)=2(P(t))^{2}\) h. \(P(t)=\frac{5}{1-15 t}\) satisfies \(P(0)=5\) and \(P^{\prime}(t)=3(P(t))^{2}\) i. \(P(t)=(6 t+9)^{1 / 2} \quad\) satisfies \(P(0)=3\) and \(P^{\prime}(t)=3 / P(t)\) j. \(P(t)=(4 t+4)^{1 / 2} \quad\) satisfies \(P(0)=2\) and \(P^{\prime}(t)=2 / P(t)\) k. \(P(t)=(4 t+4)^{3 / 2} \quad\) satisfies \(P(0)=8 \quad\) and \(\quad P^{\prime}(t)=6 \sqrt[3]{P(t)}\) For parts \(\mathrm{g}-\mathrm{k}\), use the Definition of Derivative 3.2 .2 to compute \(P^{\prime}\).

Short Answer

Expert verified
All functions satisfy the conditions of their respective parts.

Step by step solution

01

Verify P(0) for Part a

Substitute \( t = 0 \) into \( P(t) = 5t + 3 \). \[ P(0) = 5(0) + 3 = 3 \]Thus \( P(0) = 3 \) is satisfied.
02

Calculate Derivative for Part a

Calculate the derivative of \( P(t) = 5t + 3 \).\[ P'(t) = \frac{d}{dt}(5t + 3) = 5 \]Thus \( P'(t) = 5 \) is satisfied.
03

Verify P(0) for Part b

Substitute \( t = 0 \) into \( P(t) = 8t + 2 \).\[ P(0) = 8(0) + 2 = 2 \]Thus \( P(0) = 2 \) is satisfied.
04

Calculate Derivative for Part b

Calculate the derivative of \( P(t) = 8t + 2 \).\[ P'(t) = \frac{d}{dt}(8t + 2) = 8 \]Thus \( P'(t) = 8 \) is satisfied.
05

Verify P(0) for Part c

Substitute \( t = 0 \) into \( P(t) = t^2 + 3t + 7 \).\[ P(0) = (0)^2 + 3(0) + 7 = 7 \]Thus \( P(0) = 7 \) is satisfied.
06

Calculate Derivative for Part c

Calculate the derivative of \( P(t) = t^2 + 3t + 7 \).\[ P'(t) = \frac{d}{dt}(t^2 + 3t + 7) = 2t + 3 \]Thus \( P'(t) = 2t + 3 \) is satisfied.
07

Verify P(0) for Part d

Substitute \( t = 0 \) into \( P(t) = -2t^2 + 5t + 8 \).\[ P(0) = -2(0)^2 + 5(0) + 8 = 8 \]Thus \( P(0) = 8 \) is satisfied.
08

Calculate Derivative for Part d

Calculate the derivative of \( P(t) = -2t^2 + 5t + 8 \).\[ P'(t) = \frac{d}{dt}(-2t^2 + 5t + 8) = -4t + 5 \]Thus \( P'(t) = -4t + 5 \) is satisfied.
09

Verify P(0) for Part e

Substitute \( t = 0 \) into \( P(t) = (3t + 4)^2 \).\[ P(0) = (3(0) + 4)^2 = 16 \]Thus \( P(0) = 16 \) is satisfied.
10

Calculate Derivative for Part e

Use the definition of the derivative to compute \( P'(t) \) for \( P(t) = (3t + 4)^2 \).\[ P'(t) = 2(3t + 4) \cdot 3 = 6(3t + 4) = 6\sqrt{P(t)} \]Thus \( P'(t) = 6\sqrt{P(t)} \) is satisfied.
11

Verify P(0) for Part f

Substitute \( t = 0 \) into \( P(t) = (5t + 1)^2 \).\[ P(0) = (5(0) + 1)^2 = 1 \]Thus \( P(0) = 1 \) is satisfied.
12

Calculate Derivative for Part f

Use the chain rule to compute \( P'(t) \) for \( P(t) = (5t + 1)^2 \).\[ P'(t) = 2(5t + 1) \cdot 5 = 10(5t + 1) = 10\sqrt{P(t)} \]Thus \( P'(t) = 10\sqrt{P(t)} \) is satisfied.
13

Verify P(0) for Part g

Substitute \( t = 0 \) into \( P(t) = (1-2t)^{-1} \).\[ P(0) = (1 - 2(0))^{-1} = 1 \]Thus \( P(0) = 1 \) is satisfied.
14

Calculate Derivative for Part g

Use the definition of the derivative to compute \( P'(t) \) for \( P(t) = (1 - 2t)^{-1} \).\[ P'(t) = -1(-2)(1-2t)^{-2} = 2(1-2t)^{-2} = 2(P(t))^2 \]Thus \( P'(t) = 2(P(t))^2 \) is satisfied.
15

Verify P(0) for Part h

Substitute \( t = 0 \) into \( P(t) = \frac{5}{1-15t} \).\[ P(0) = \frac{5}{1 - 15(0)} = 5 \]Thus \( P(0) = 5 \) is satisfied.
16

Calculate Derivative for Part h

Use the definition of the derivative to compute \( P'(t) \) for \( P(t) = \frac{5}{1-15t} \).\[ P'(t) = -15 \frac{5}{(1-15t)^2} = 3(P(t))^2 \]Thus \( P'(t) = 3(P(t))^2 \) is satisfied.
17

Verify P(0) for Part i

Substitute \( t = 0 \) into \( P(t) = (6t + 9)^{1/2} \).\[ P(0) = (6(0) + 9)^{1/2} = 3 \]Thus \( P(0) = 3 \) is satisfied.
18

Calculate Derivative for Part i

Use the chain rule to compute \( P'(t) \) for \( P(t) = (6t + 9)^{1/2} \).\[ P'(t) = \frac{1}{2}(6t + 9)^{-1/2} \cdot 6 = \frac{3}{\sqrt{6t + 9}} = \frac{3}{P(t)} \]Thus \( P'(t) = \frac{3}{P(t)} \) is satisfied.
19

Verify P(0) for Part j

Substitute \( t = 0 \) into \( P(t) = (4t + 4)^{1/2} \).\[ P(0) = (4(0) + 4)^{1/2} = 2 \]Thus \( P(0) = 2 \) is satisfied.
20

Calculate Derivative for Part j

Use the chain rule to compute \( P'(t) \) for \( P(t) = (4t + 4)^{1/2} \).\[ P'(t) = \frac{1}{2}(4t + 4)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4t + 4}} = \frac{2}{P(t)} \]Thus \( P'(t) = \frac{2}{P(t)} \) is satisfied.
21

Verify P(0) for Part k

Substitute \( t = 0 \) into \( P(t) = (4t + 4)^{3/2} \).\[ P(0) = (4(0) + 4)^{3/2} = 8 \]Thus \( P(0) = 8 \) is satisfied.
22

Calculate Derivative for Part k

Use the chain rule to compute \( P'(t) \) for \( P(t) = (4t + 4)^{3/2} \).\[ P'(t) = \frac{3}{2}(4t + 4)^{1/2} \cdot 4 = 6 \cdot \sqrt{4t + 4} = 6 \sqrt[3]{P(t)} \]Thus \( P'(t) = 6 \sqrt[3]{P(t)} \) is satisfied.

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Key Concepts

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

Polynomial Functions
Polynomial functions are mathematical expressions that consist of variables raised to whole number powers, combined with coefficients. They are one of the fundamental building blocks in algebra and calculus. For instance, the equation \( P(t) = 5t + 3 \) is a linear polynomial, where the variable \( t \) is raised to the power of 1. On the other hand, \( P(t) = t^2 + 3t + 7 \) is a quadratic polynomial, with the highest power being 2.
  • Linear Polynomials: These take the form \( ax + b \), where \( a \) and \( b \) are constants.
  • Quadratic Polynomials: These can be represented as \( ax^2 + bx + c \).
  • Higher Degree Polynomials: These involve powers greater than 2, like cubic \( ax^3 + bx^2 + cx + d \).

Understanding the structure of polynomial functions is vital for applying differentiation techniques, solving equations, and analyzing graphs.
Chain Rule
The chain rule is a fundamental technique in calculus, used to find the derivative of composite functions. A composite function is a function formed by combining two or more functions. If you have a function \( f(g(t)) \), you use the chain rule to differentiate it.
In application, consider \( P(t) = (5t + 1)^2 \). Here, the outer function can be described as \( u^2 \) with \( u = 5t + 1 \). To differentiate this composite, you first derive the outer function, treating \( u \) as a variable, which gives \( 2u \), and then multiply it by the derivative of the inner function \( u \), which is \( 5 \). So,
  • Outer function: \( d(u^2)/du = 2u \)
  • Inner function: \( du/dt = 5 \)
  • Derivative: \( P'(t) = 2u(du/dt) = 10(5t + 1) \)

This results in \( P'(t) = 10(5t + 1) \), demonstrating the utility of the chain rule.
Differentiation Techniques
Differentiation is a primary concept in calculus referring to the process of finding the derivative of a function, which essentially calculates the rate of change of the function. There are several rules for differentiation depending on the function type:
  • Simple Polynomials: Utilize the power rule, where if \( f(t) = t^n \), then \( f'(t) = nt^{n-1} \).
  • Constant Factors: Constants are treated as such, making the derivative of \( at \) equal to \( a \).
  • Chain Rule: For composite functions, as explained, apply the derivative to the outer function and multiply by the derivative of the inner function.

For example, the derivative of \( -2t^2 + 5t + 8 \) involves using the power rule:
  • \( rac{d}{dt}(-2t^2) = -4t \)
  • \( rac{d}{dt}(5t) = 5 \)
  • \( rac{d}{dt}(8) = 0 \)

Combining these results, the derivative is \( P'(t) = -4t + 5 \). By practicing these techniques, one develops the skill to differentiate a wide variety of mathematical functions.
Limits and Continuity
Limits and continuity are foundational concepts in calculus that underpin the process of finding derivatives and understanding the behavior of functions. A limit describes the value that a function approaches as the input (or \( t \) in many cases) approaches some value. Meanwhile, continuity ensures that a function has no interruptions in its domain.
In derivative calculation, limits are used to formally define what a derivative is. For example, the definition of a derivative is:
\[ \frac{d}{dt} P(t) = \lim_{h \to 0} \frac{P(t+h) - P(t)}{h} \]
This definition forms the basis for all differentiation but can be complex to work with for most practical calculations, which is why methods like the power rule, the product rule, or the chain rule are more commonly used in practice. A function is continuous if it doesn't have jumps, holes, or vertical asymptotes in its graph, making it easier to differentiate. In the practice exercises, these concepts help ensure correct computation of derivatives, as they validate that the function behaves predictably and smoothly as \( t \) approaches any given value.

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Most popular questions from this chapter

Data from Purdue University \(^{7}\) for the decrease of the titration marker phenolphthalein (Hln) in the presence of excess base are shown in Table 3.4.5. The data show the concentration of phenolphthalein that was initially at \(0.005 \mathrm{M}\) in a solution with \(0.61 \mathrm{M} \mathrm{OH}^{-}\) ion. a. Graph the data. b. Estimate the rate of change of the concentration of phenolphthalein (Hln) for each of the times shown. c. Draw a graph of the rate of reaction versus concentration of phenolphthalein. $$\begin{array}{|l|rrrrrrrr|}\hline \text { Time (sec) } & 0 & 10.5 & 22.3 & 35.7 & 51.1 & 69.3 & 91.6 & 120.4 & 160.9 \\ \text { Hln (M) } & 0.005 & 0.0045 & 0.0040 & 0.0035 & 0.0030 & 0.0025 & 0.0020 & 0.0015 & 0.0010 \\ \hline\end{array}$$

Probably baseball statistics should be discussed in British units rather than metric units. Professional pitchers throw fast balls in the range of \(90+\) miles per hour. Suppose the pop fly ball leaves the bat traveling 60 miles per hour \((88\) feet \(/ \mathrm{sec})\), in which case the height of the ball in feet will be \(s(t)=-16 t^{2}+88 t\) feet above the bat, \(t\) seconds after the batter hits the ball. How high will the ball go, and how long will the catcher have to position to catch it? How fast is the ball falling when the catcher catches it?

Show that for any quadratic function, \(Q(t)=a+b t+c t^{2}(a, b\) and \(c\) are constants), and any interval, \([u, v]\), the average rate of change of \(Q\) on \([u, v]\) is equal to the rate of change of \(Q\) at the midpoint, \((u+v) / 2,\) of \([u, v]\).

Add the equations, $$\begin{aligned} H_{2}-H_{1} &=\frac{-849}{2}\left(t_{2}^{2}-t_{1}^{2}\right)+126\left(t_{2}-t_{1}\right) \\\ H_{3}-H_{2} &=\frac{-849}{2}\left(t_{3}^{2}-t_{2}^{2}\right)+126\left(t_{3}-t_{2}\right) \end{aligned}$$ $$H_{n}-H_{n-1}=\frac{-849}{2}\left(t_{n}^{2}-t_{n-1}^{2}\right)+126\left(t_{n}-t_{n-1}\right) $$to obtain $$H_{n}-H_{1}=\frac{-849}{2}\left(t_{n}^{2}-t_{1}^{2}\right)+126\left(t_{n}-t_{1}\right)$$

Technology. Suppose plasma penicillin concentration in a patient following injection of 1 gram of penicillin is observed to be $$P(t)=2002^{-0.03 t}$$ where \(t\) is time in minutes and \(P(t)\) is \(\mu \mathrm{g} / \mathrm{ml}\) of penicillin. Use the following steps to approximate the rate at which the penicillin level is changing at time \(t=5\) minutes and at \(t=0\) minutes. a. \(\mathrm{t}=5\) minutes. Draw the graph of \(P(t)\) vs \(t\) for \(4.9 \leq t \leq 5.1 .\) (The graph should appear to be a straight line on this short interval.) b. Complete the table on the left, computing the average rates of change of penicillin level. $$\begin{array}{l|l}b & \frac{P(b)-P(5)}{b-5} \\\\\hline 4.9 & -3.7521 \\\4.95 & \\\4.99 & \\\4.995 & \\\& \\\5.005 & \\\5.01 & \\\5.05 & \\\5.1 & -3.744\end{array}$$ $$\begin{array}{l|l}b & \frac{P(b)-P(0)}{b-0} \\\\\hline-0.1 & \text { OMIT } \\\\-0.05 & \text { OMIT } \\ -0.01 & \text { OMIT } \\\\-0.005 & \text { OMIT } \\\0.005 & \\\0.01 & \\\0.05 & \\\0.1 & -4.155\end{array}$$ c. What is your best estimate of the rate of change of penicillin level at the time \(t=5\) minutes? Include units in your answer. d. \(\mathbf{t}=\mathbf{0}\) minutes. Complete the second table above. The OMIT entries in the second table refer to the fact that the level of penicillin, \(P(t),\) may not be given by the formula for negative values of time, \(t .\) What is your best estimate of the rate of change of penicillin level at the time \(t=0\) minutes?

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