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Chapter 2: Problem 20

In each of these cases, find the percentage rate of change of the function \(f(t)\) with respect to \(t\) at the given value of \(t\). a. \(f(t)=t^{2}-3 t+\sqrt{t}\) at \(t=4\) b. \(f(t)=\frac{t}{t-3}\) at \(t=4\)

Short Answer

Expert verified
The percentage rate of change for part (a) is approximately 87.5%. The percentage rate of change for part (b) is -75%.

Step by step solution

01

Find the derivative of the function for part (a)

The function is given as \(f(t) = t^2 - 3t + \sqrt{t}\). The derivative with respect to \(t\) is found by applying the rules of differentiation:\[\frac{d}{dt}(t^2 - 3t + \sqrt{t}) = 2t - 3 + \frac{1}{2\sqrt{t}}\].
02

Evaluate the derivative at the given point for part (a)

To find the rate of change at \(t = 4\), substitute \(t = 4\) into the derivative:\[2(4) - 3 + \frac{1}{2\sqrt{4}} = 8 - 3 + \frac{1}{4} = 5.25.\].
03

Convert the derivative to a percentage rate of change for part (a)

Use the formula for percentage rate of change: \[\text{Percentage rate of change} = \left(\frac{f'(t)}{f(t)}\right) \times 100\%.\] Evaluate \(f(4)\) first:\(f(4) = 4^2 - 3(4) + \sqrt{4} = 16 - 12 + 2 = 6\). Then:\[\text{Percentage rate of change at } t=4 = \left(\frac{5.25}{6}\right) \times 100\% \approx 87.5\%.\]
04

Find the derivative of the function for part (b)

The function for part (b) is given as \(f(t) = \frac{t}{t-3}\). The derivative with respect to \(t\) can be found using the quotient rule:\[\frac{d}{dt}\left(\frac{t}{t-3}\right) = \frac{(1)(t-3) - t(1)}{(t-3)^2} = \frac{t-3 - t}{(t-3)^2} = \frac{-3}{(t-3)^2}.\]
05

Evaluate the derivative at the given point for part (b)

To find the rate of change at \(t = 4\), substitute \(t = 4\) into the derivative:\[\frac{-3}{(4-3)^2} = \frac{-3}{1} = -3.\]
06

Convert the derivative to a percentage rate of change for part (b)

Use the formula for percentage rate of change: \[\text{Percentage rate of change} = \left(\frac{f'(t)}{f(t)}\right) \times 100\%.\] Evaluate \(f(4)\) first:\(f(4) = \frac{4}{4-3} = 4\). Then:\[\text{Percentage rate of change at } t=4 = \left(\frac{-3}{4}\right) \times 100\% = -75\%.\]

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Key Concepts

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

differentiation
Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which represents the rate at which the function changes with respect to a variable. The derivative is a measure of how a function's output value changes as the input value changes slightly.
To find the derivative of a function like \( f(t) = t^2 - 3t + \sqrt{t} \), you apply the differentiation rules:
  • The derivative of \( t^n \) is \( nt^{n-1} \).
  • The derivative of a constant times a function is the constant times the derivative of the function.
  • The derivative of the sum of functions is the sum of their derivatives.
Using these rules, we find: \[ \frac{d}{dt}(t^2 - 3t + \sqrt{t}) = 2t - 3 + \frac{1}{2\sqrt{t}}. \] This resulting function provides the rate of change of the original function \( f(t) \).
quotient rule
The quotient rule is a method for differentiating functions that are expressed as the ratio of two other functions.
If you have a function \( f(t) = \frac{g(t)}{h(t)} \), where both \( g \) and \( h \) are functions of \( t \), the quotient rule states: \[ \frac{d}{dt} \left( \frac{g(t)}{h(t)} \right) = \frac{g'(t)h(t) - g(t)h'(t)}{[h(t)]^2}. \]
For example, consider the function \( f(t) = \frac{t}{t-3} \). Here, \( g(t) = t \) and \( h(t) = t - 3 \).
Applying the quotient rule yields:
\[ \frac{d}{dt} \left( \frac{t}{t-3} \right) = \frac{(1)(t-3) - t(1)}{(t-3)^2} = \frac{-3}{(t-3)^2}. \] This helps to find the rate at which \( f(t) \) changes with respect to \( t \).
rate of change
The rate of change is a measure of how much a quantity changes relative to another quantity. In mathematics, it often refers to how a function changes over time or with respect to another variable.
The percentage rate of change provides this measure as a percentage, making it easier to understand the change in terms of proportions.
  • First, find the derivative of the function, which indicates its rate of change.
  • Next, evaluate the derivative at the given point.
  • Finally, apply the formula \[ \text{Percentage rate of change} = \left( \frac{f'(t)}{f(t)} \right) \times 100\%. \] to convert this rate into a percentage.
For instance, for the function \( f(t) = t^2 - 3t + \sqrt{t} \) at \( t = 4 \), the derivative is evaluated as \( 5.25 \) and the function value as \( 6 \). The percentage rate of change is then: \[ \left( \frac{5.25}{6} \right) \times 100\% = 87.5\%. \] Understanding these steps allows you to breakdown complex calculus problems with ease.

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Most popular questions from this chapter

The level of air pollution in a certain city is proportional to the square of the population. Use calculus to estimate the percentage by which the air pollution level will increase if the population increases by \(5 \%\).

When the price of a certain commodity is \(p\) dollars per unit, consumers demand \(x\) hundred units of the commodity, where $$ 75 x^{2}+17 p^{2}=5,300 $$ How fast is the demand \(x\) changing with respect to time when the price is \( 7\) and is decreasing at the rate of 75 cents per months? (That is, \(\frac{d p}{d t}=-0.75\).)

The formula \(D=36 m^{-1.14}\) is sometimes used to determine the ideal population density \(D\) (individuals per square kilometer) for a large animal of mass \(m\) kilograms (kg). a. What is the ideal population density for humans, assuming that a typical human weighs about \(70 \mathrm{~kg}\) ? b. The area of the United States is about \(9.2\) million square kilometers. What would the population of the United States have to be for the population density to be ideal? c. Consider an island of area \(3,000 \mathrm{~km}^{2}\). Two hundred animals of mass \(m=30 \mathrm{~kg}\) are brought to the island, and \(t\) years later, the population is given by $$ P(t)=0.43 t^{2}+13.37 t+200 $$ How long does it take for the ideal population density to be reached? At what rate is the population changing when the ideal density is attained?

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In its early phase, specifically the period 1984-1990, the AIDS epidemic could be modeled* by the cubic function $$ C(t)=-170.36 t^{3}+1,707.5 t^{2}+1,998.4 t+4,404.8 $$ for \(0 \leq t \leq 6\), where \(C\) is the number of reported cases \(t\) years after the base year \(1984 .\) a. Compute and interpret the derivative \(C^{\prime}(t)\). b. At what rate was the epidemic spreading in the year 1984 ? c. At what percentage rate was the epidemic spreading in 1984 ? In 1990 ?
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