/*! 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 91 The concentration (in milligrams... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 91

The concentration (in milligrams/cubic centimeter) of a certain drug in a patient's body \(t\) hr after injection is given by $$ C(t)=\frac{t^{2}}{2 t^{3}+1} \quad(0 \leq t \leq 4) $$ When is the concentration of the drug increasing, and when is it decreasing?

Short Answer

Expert verified
The concentration of the drug is increasing when \(0 < t < 1\) hours and decreasing when \(1 < t \leq 4\) hours.

Step by step solution

01

Find the derivative of the concentration function

Given the function: $$ C(t)=\frac{t^{2}}{2 t^{3}+1} \quad(0 \leq t \leq 4) $$ We need to find its derivative, C'(t), with respect to time t. Using the quotient rule, where \(u(t) = t^{2}\) and \(v(t) = 2t^{3} + 1\), we get: $$ C'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} $$ Now we must find \(u'(t)\) and \(v'(t)\) by taking the derivatives of u(t) and v(t) with respect to t: $$ u'(t) = \frac{d(t^2)}{dt} = 2t $$ $$ v'(t) = \frac{d(2t^3 + 1)}{dt} = 6t^2 $$ Now, plug these derivatives back into the quotient rule formula for C'(t): $$ C'(t) = \frac{(2t)(2t^3 + 1) - (t^2)(6t^2)}{ (2t^3 + 1)^2} $$
02

Simplify the derivative

Now, we simplify the expression for C'(t): $$ C'(t) = \frac{4t^4 + 2t - 6t^4}{(2t^3 + 1)^2} $$ $$ C'(t) = \frac{-2t^4 + 2t}{(2t^3 + 1)^2} $$ We can factor out the common factor (2t) from the numerator: $$ C'(t) = \frac{2t(-t^3 + 1)}{(2t^3 + 1)^2} $$
03

Analyze the derivative to determine when the concentration is increasing or decreasing

Now, we find the intervals where C'(t) > 0 (increasing) and C'(t) < 0 (decreasing): $$ C'(t) > 0 \implies 2t(-t^3 + 1) > 0 $$ The function is increasing when t > 0 and -t^3 + 1 > 0, which implies t < 1 by solving for t in the inequality. So, the concentration is increasing when \(0 < t < 1\). Similarly, $$ C'(t) < 0 \implies 2t(-t^3 + 1) < 0 $$ The function is decreasing when t > 0 and -t^3 + 1 < 0, which implies t > 1 by solving for t in the inequality. So, the concentration is decreasing when \(1 < t \leq 4\). Therefore, the concentration of the drug is increasing when \(0 < t < 1\) hours and decreasing when \(1 < t \leq 4\) hours.

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.

Calculus in Pharmacokinetics
Calculus plays an instrumental role in studying and predicting the concentration of drugs within the human body, a crucial aspect of pharmacokinetics. One might wonder why the math of change and motion, calculus, is so vital in determining the fate of a drug administered to a patient. The answer lies in our body's dynamic nature and how it processes substances over time.

Drug concentration, as described by mathematical models, can help predict how long a medication will be effective and determine the appropriate dosing schedules. The concentration of a drug over time typically follows a curve that requires differential calculus to analyze. As in the given exercise, by differentiating the concentration function with respect to time, we can determine when the drug levels are increasing or decreasing, indicating the periods of absorption and elimination respectively. This understanding is critical for clinicians to design an appropriate dosage regimen to ensure the drug is delivered at the right time, in the right amount, for optimal therapeutic effect.
Quotient Rule Differentiation
Differentiation is a fundamental concept in calculus used to find the rate at which something changes. When functions are divided by one another, the quotient rule comes into play for differentiation. It's an efficient way to tackle complex problems in which the division of two functions is involved, such as the concentration function in our original exercise.

The quotient rule states that for two differentiable functions, u(t) and v(t), the derivative of their quotient u(t)/v(t) is given by \
\[\frac{{d}}{{dt}}\left(\frac{{u}}{{v}}\right) = \frac{{u'(t) \cdot v(t) - u(t) \cdot v'(t)}}{{[v(t)]^2}}\]
The quotient rule is a reliable tool used in pharmacokinetics for assessing how drug concentrations change over time. By applying this rule, as shown in the exercise solution, we obtain the first derivative of the concentration function, which informs us about the rate and direction (increase or decrease) of change in the drug's concentration.
Analyzing Function's Increasing and Decreasing Intervals
Determining when a function is increasing or decreasing is essential in various fields, including economics, physics, and, as illustrated in our exercise, pharmacokinetics. This analysis is based on the sign of the function's first derivative.

When the derivative of a function is positive over an interval, it indicates that the function is increasing on that interval. Conversely, if the derivative is negative, the function is decreasing. By setting the derivative equal to zero, we can find critical points, which could represent peaks, valleys, or inflection points in the function. In the context of the drug concentration problem, finding where the function C'(t) changes its sign reveals when the concentration of the drug stops increasing and starts decreasing, and vice versa. This fine detail provides valuable insight into the drug's absorption and metabolism, which are critical for ensuring the safety and efficacy of pharmacological treatments.

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

A tank initially contains 10 gal of brine with \(2 \mathrm{lb}\) of salt. Brine with \(1.5 \mathrm{lb}\) of salt per gallon enters the tank at the rate of \(3 \mathrm{gal} / \mathrm{min}\), and the well-stirred mixture leaves the tank at the rate of \(4 \mathrm{gal} / \mathrm{min}\). It can be shown that the amount of salt in the tank after \(t\) min is \(x \mathrm{lb}\) where \(x=f(t)=1.5(10-t)-0.0013(10-t)^{4} \quad(0 \leq t \leq 10)\) What is the maximum amount of salt present in the tank at any time?

A developing country's gross domestic product (GDP) from 2000 to 2008 is approximated by the function $$ G(t)=-0.2 t^{3}+2.4 t^{2}+60 \quad(0 \leq t \leq 8) $$ where \(G(t)\) is measured in billions of dollars, with \(t=0\) corresponding to 2000 . Sketch the graph of the function \(G\) and interpret your results.

The total world population is forecast to be \(P(t)=0.00074 t^{3}-0.0704 t^{2}+0.89 t+6.04 \quad(0 \leq t \leq 10)\) in year \(t\), where \(t\) is measured in decades with \(t=0\) corresponding to 2000 and \(P(t)\) is measured in billions. a. Show that the world population is forecast to peak around 2071 . Hint: Use the quadratic formula. b. What will the population peak at?

It has been conjectured that a fish swimming a distance of \(L \mathrm{ft}\) at a speed of \(v \mathrm{ft} / \mathrm{sec}\) relative to the water and against a current flowing at the rate of \(u \mathrm{ft} / \mathrm{sec}(u

An apple orchard has an average yield of 36 bushels of apples/tree if tree density is 22 trees/acre. For each unit increase in tree density, the yield decreases by 2 bushels/tree. How many trees should be planted in order to maximize the yield?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

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.