/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 The temperature change \(T\) (in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 6

The temperature change \(T\) (in Fahrenheit degrees), in a patient, that is generated by a dose \(q\) (in milliliters), of a drug, is given by the function \(T=f(q)\). a. What does it mean to say \(f(50)=0.75 ?\) Write a complete sentence to explain, using correct units. b. A person's sensitivity, \(s,\) to the drug is defined by the function \(s(q)=f^{\prime}(q) .\) What are the units of sensitivity? c. Suppose that \(f^{\prime}(50)=-0.02 .\) Write a complete sentence to explain the meaning of this value. Include in your response the information given in (a).

Short Answer

Expert verified
At a 50 ml dose, the temperature increases by 0.75掳F. Sensitivity has units of Fahrenheit degrees per milliliter. A 50 ml dose decreases temperature at a rate of 0.02掳F per ml.

Step by step solution

01

Understanding the Function Value - Step 1

The expression \( f(50) = 0.75 \) means that when the dose of the drug, \( q \), is 50 milliliters, the patient's temperature change \( T \) is 0.75 Fahrenheit degrees. In other words, administering a dose of 50 milliliters of the drug results in an increase of 0.75 degrees Fahrenheit in the patient's temperature.
02

Sensitivity Units - Step 2

To determine the units of sensitivity \( s(q) \), which is defined as \( f'(q) \), note that \( f'(q) \) represents the rate of change of temperature with respect to the dose. Since \( T \) is measured in Fahrenheit degrees and \( q \) is measured in milliliters, the units of \( f'(q) \) (and hence sensitivity \( s(q) \)) are \( \text{Fahrenheit degrees per milliliter} \).
03

Interpreting the Derivative Value - Step 3

The expression \( f'(50) = -0.02 \) indicates that at a dose of 50 milliliters, the patient's temperature decreases at a rate of 0.02 Fahrenheit degrees per milliliter. This means that increasing the dose from 50 milliliters by one milliliter will decrease the patient's temperature by 0.02 degrees Fahrenheit. Combining this with the information given in (a), even though the patient鈥檚 temperature increases to 0.75 degrees Fahrenheit at a 50 milliliter dose, any further increase in the dose will cause the temperature to start decreasing.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Function Interpretation
In this problem, we encountered the function \(T = f(q)\), which describes the change in a patient's temperature \(T\) (measured in Fahrenheit degrees) depending on the dose \(q\) (measured in milliliters) of a drug. When we say \(f(50) = 0.75\), it means that administering a dose of 50 milliliters of the drug causes the patient's temperature to increase by 0.75 degrees Fahrenheit. This interpretation is crucial because it tells us precisely how a particular dose affects temperature, which can be vital for determining the correct dose for patients.
Sensitivity Analysis
Sensitivity in this context is defined by the function \(s(q) = f'(q)\). Sensitivity analysis is used to determine how sensitive the patient's temperature is to changes in the drug dose. The unit of sensitivity is derived from the derivative \(f'(q)\).

Since \(f(q)\) represents temperature in Fahrenheit degrees and \(q\) represents the dose in milliliters, the derivative \(f'(q)\) tells us the rate of change in temperature per unit change in the dose. Therefore, the unit of sensitivity \(s(q)\) is Fahrenheit degrees per milliliter. This means we are looking at how much a small change in the drug dose will impact the patient's temperature.

Knowing the sensitivity can help doctors make better decisions about adjusting drug doses to achieve the desired temperature change in patients.
Derivative Interpretation
The derivative \(f'(q)\) shows us how the temperature change rate varies with the dose. When \(f'(50) = -0.02\), it means that at a dose of 50 milliliters, the patient's temperature is decreasing at a rate of 0.02 Fahrenheit degrees per milliliter. This is a negative value, indicating a decrease in temperature.

Combining this with our previous information that \(f(50) = 0.75\), we understand that while a 50 milliliter dose initially increases the patient's temperature by 0.75 degrees Fahrenheit, any increase in dose beyond 50 milliliters will start to decrease the temperature. This highlights the importance of not just knowing the effect of a specific dose but also understanding how small changes in dose can impact the patient's temperature.

By interpreting derivatives, healthcare professionals can better manage doses to achieve the optimal therapeutic effect, avoiding potential adverse effects from incorrect dosing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

According to the U.S. census, the population of the city of Grand Rapids, MI, was 181,843 in \(1980 ; 189,126\) in 1990 ; and 197,800 in 2000 . a. Between 1980 and 2000 , by how many people did the population of Grand Rapids grow? b. In an average year between 1980 and 2000 , by how many people did the population of Grand Rapids grow? c. Just like we can find the average velocity of a moving body by computing change in position over change in time, we can compute the average rate of change of any function \(f\). In particular, the average rate of change of a function \(f\) over an interval \([a, b]\) is the quotient $$\frac{f(b)-f(a)}{b-a}$$ What does the quantity \(\frac{f(b)-f(a)}{b-a}\) measure on the graph of \(y=f(x)\) over the interval \([a, b] ?\) d. Let \(P(t)\) represent the population of Grand Rapids at time \(t,\) where \(t\) is measured in years from January \(1,1980 .\) What is the average rate of change of \(P\) on the interval \(t=0\) to \(t=20 ?\) What are the units on this quantity? e. If we assume the population of Grand Rapids is growing at a rate of approximately \(4 \%\) per decade, we can model the population function with the formula $$P(t)=181843(1.04)^{t / 10}$$ Use this formula to compute the average rate of change of the population on the intervals \([5,10],[5,9],[5,8],[5,7],\) and [5,6] f. How fast do you think the population of Grand Rapids was changing on January 1 , 1985 ? Said differently, at what rate do you think people were being added to the population of Grand Rapids as of January \(1,1985 ?\) How many additional people should the city have expected in the following year? Why?

Suppose that an accelerating car goes from 0 mph to 61.4 mph in five seconds. Its velocity is given in the following table, converted from miles per hour to feet per second, so that all time measurements are in seconds. (Note: \(1 \mathrm{mph}\) is \(22 / 15 \mathrm{ft} / \mathrm{sec} .)\) Find the average acceleration of the car over each of the first two seconds. $$\begin{array}{|l|l|l|l|l|l|l|} \hline t(\mathrm{~s}) & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline v(t)(\mathrm{ft} / \mathrm{s}) & 0.00 & 30.68 & 53.18 & 69.55 & 81.82 & 90.00 \\ \hline \end{array}$$ average acceleration over the first second \(=\square\) help (units) average acceleration over the second second =\(\square\) help (units)

A potato is placed in an oven, and the potato's temperature \(F\) (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. Time \(t\) is measured in minutes. $$\begin{array}{ll} \hline t & F(t) \\ \hline 0 & 70 \\ \hline 15 & 180.5 \\ \hline 30 & 251 \\ \hline 45 & 296 \\ \hline 60 & 324.5 \\ \hline 75 & 342.8 \\ \hline 90 & 354.5 \\ \hline \end{array}$$ a. Use a central difference to estimate \(F^{\prime}(60)\). Use this estimate as needed in subsequent questions. b. Find the local linearization \(y=L(t)\) to the function \(y=F(t)\) at the point where \(a=60\). c. Determine an estimate for \(F(63)\) by employing the local linearization. d. Do you think your estimate in (c) is too large or too small? Why?

A cup of coffee has its temperature \(F\) (in degrees Fahrenheit) at time \(t\) given by the function \(F(t)=75+110 e^{-0.05 t}\), where time is measured in minutes. a. Use a central difference with \(h=0.01\) to estimate the value of \(F^{\prime}(10)\). b. What are the units on the value of \(F^{\prime}(10)\) that you computed in (a)? What is the practical meaning of the value of \(F^{\prime}(10) ?\) c. Which do you expect to be greater: \(F^{\prime}(10)\) or \(F^{\prime}(20) ?\) Why? d. Write a sentence that describes the behavior of the function \(y=F^{\prime}(t)\) on the time interval \(0 \leq t \leq 30\). How do you think its graph will look? Why?

The cost, \(C\) (in dollars) to produce \(g\) gallons of ice cream can be expressed as \(C=f(g)\). (a) In the expression \(f(300)=350\), what are the units of \(300 ?\) [Choose: ? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] what are the units of \(350 ?\) [Choose: ? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] (b) In the expression \(f^{\prime}(300)=1.2,\) what are the units of 300? [Choose:? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] what are the units of 1.2? [Choose: ? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] (Be sure that you can carefully put into words the meanings of each of these statement in terms of ice cream and money.)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.