Problem 54
Question
Chemical Mixtures Two chemicals react to form another chemical. Suppose that the amount of the chemical formed in time \(t\) (in hours) is given by $$ x(t)=\frac{15\left[1-\left(\frac{2}{3}\right)^{3 t}\right]}{1-\frac{1}{4}\left(\frac{2}{3}\right)^{3 t}} $$ where \(x(t)\) is measured in pounds. a. Plot the graph of \(x\) using the viewing window \([0,10] \times[0,16] .\) b. Find the rate at which the chemical is formed when \(t=1\). c. How many pounds of the chemical are formed eventually?
Step-by-Step Solution
Verified Answer
a. To plot the graph of the function, use graphing software or a calculator with the viewing window \([0, 10] \times [0, 16]\).
b. The rate at which the chemical is formed at \(t=1\) is approximately \(3.76\) pounds per hour.
c. The chemical formed eventually amounts to \(15\) pounds.
1Step 1: Plot the function
The function to plot is \(x(t)=\frac{15\left[1-\left(\frac{2}{3}\right)^{3t}\right]}{1-\frac{1}{4}\left(\frac{2}{3}\right)^{3t}}\). Use graphing software or a graphing calculator to plot the function in the specified viewing window, which is \([0, 10] \times [0, 16]\).
For step 2, we will calculate the derivative of this function and evaluate it at \(t=1\).
2Step 2: Calculate the rate of formation at \(t=1\)
To find the rate of formation at \(t=1\), we have to calculate the first derivative of \(x(t)\) concerning \(t\). Then, evaluate the derivative at \(t=1\).
The derivative of the function \(x(t)\) can be found by applying the quotient rule, which states:
\( \frac{d}{dt}\left(\frac{f(t)}{g(t)}\right) = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2} \)
Differentiating both the numerator and denominator will give us the derivative of the function \(x(t)\). After calculating the derivative, you get:
\(x'(t) = \frac{-30(\frac{2}{3})^{3t}(1-2(\frac{2}{3})^{3t})}{[1-\frac{1}{4}(\frac{2}{3})^{3t}]^2}\)
Now, we evaluate the derivative at \(t=1\):
\(x'(1) = \frac{-30(\frac{2}{3})^{3}(1-2(\frac{2}{3})^{3})}{[1-\frac{1}{4}(\frac{2}{3})^{3}]^2} \approx 3.76 \)
The rate at which the chemical is formed when \(t=1\) is approximately \(3.76\) pounds per hour.
For step 3, we will determine the limit of the function as time goes to infinity.
3Step 3: Find the amount of chemical formed eventually
To find the amount of chemical formed eventually, we need to find the limit of the function \(x(t)\) as \(t\) approaches infinity:
\(\lim_{t \to \infty} x(t) =\lim_{t \to \infty} \frac{15\left[1-\left(\frac{2}{3}\right)^{3t}\right]}{1-\frac{1}{4}\left(\frac{2}{3}\right)^{3t}}\)
The exponential terms in the above function go to zero as \(t\) goes to infinity, which means:
\(\lim_{t \to \infty} x(t) = \frac{15(1-0)}{1-0} = 15\)
Thus, the chemical formed eventually is \(15\) pounds.
To summarize the solution:
a. The graph of the function can be plotted using graphing software or graphing calculator with the viewing window \([0, 10] \times [0, 16]\).
b. The rate at which the chemical is formed when \(t=1\) is approximately \(3.76\) pounds per hour.
c. The chemical formed eventually is \(15\) pounds.
Key Concepts
Quotient RuleLimitsDerivativesGraphical Analysis
Quotient Rule
In calculus, the quotient rule is a method for finding the derivative of a function that is the ratio of two differentiable functions. This is particularly useful when you have a function presented as a fraction, as in our original problem. The quotient rule can be stated as follows:
For two functions, say \( f(t) \) and \( g(t) \), the derivative of their quotient is:
\[ \frac{d}{dt}\left( \frac{f(t)}{g(t)} \right) = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2} \]
This rule helps us differentiate complex expressions like the one from the exercise for \( x(t) \), by letting us focus separately on the numerator and denominator, which are less complex. Always remember to first find the derivatives of the numerator \( f'(t) \) and the denominator \( g'(t) \), and put them into the quotient rule formula. Also, pay attention to simplifying the result wherever possible to make your further calculations easier.
For two functions, say \( f(t) \) and \( g(t) \), the derivative of their quotient is:
\[ \frac{d}{dt}\left( \frac{f(t)}{g(t)} \right) = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2} \]
This rule helps us differentiate complex expressions like the one from the exercise for \( x(t) \), by letting us focus separately on the numerator and denominator, which are less complex. Always remember to first find the derivatives of the numerator \( f'(t) \) and the denominator \( g'(t) \), and put them into the quotient rule formula. Also, pay attention to simplifying the result wherever possible to make your further calculations easier.
Limits
Limits in calculus play a crucial role, especially when dealing with expressions as they approach a certain value. In the context of the exercise, we needed to determine the amount of chemical formed eventually, which translates mathematically to finding the limit of \( x(t) \) as \( t \rightarrow \infty \).
When calculating limits, particularly for functions involving exponents like \( (\frac{2}{3})^{3t} \), it is important to recognize behaviour as \( t \) increases. These exponential components will tend to 0 because the base \( \frac{2}{3} \) is less than 1, causing the entire term to shrink as \( t \) grows.
To find this limit, simply evaluate the remaining constant terms once the shrinking terms tend to zero. In our case, it simplified the limit to \( \frac{15(1-0)}{1-0} = 15 \). This is how the limiting process helps us find the quantity of chemical formed at equilibrium or over an infinite timespan, ensuring no terms guide calculations to infinity unnecessarily.
When calculating limits, particularly for functions involving exponents like \( (\frac{2}{3})^{3t} \), it is important to recognize behaviour as \( t \) increases. These exponential components will tend to 0 because the base \( \frac{2}{3} \) is less than 1, causing the entire term to shrink as \( t \) grows.
To find this limit, simply evaluate the remaining constant terms once the shrinking terms tend to zero. In our case, it simplified the limit to \( \frac{15(1-0)}{1-0} = 15 \). This is how the limiting process helps us find the quantity of chemical formed at equilibrium or over an infinite timespan, ensuring no terms guide calculations to infinity unnecessarily.
Derivatives
Derivatives represent the rate of change or the slope of a function at any point. In the exercise, finding the derivative of \( x(t) \) was necessary to determine how quickly the chemical formed at \( t=1 \).
Let's look at the derivative process briefly. Begin by applying the quotient rule to the function \( x(t) \). Once the derivative formula is set, evaluate it at the desired point, \( t=1 \). The derivative gives valuable insight into how the production of chemical pounds scales with time, showing us around 3.76 pounds are formed per hour at \( t=1 \).
Understanding derivatives is critical not only for calculating rates like this but also for exploring any function's sensitivity to change. This concept forms the basis for optimization, physics applications, and deep understanding of any system's dynamics over time.
Let's look at the derivative process briefly. Begin by applying the quotient rule to the function \( x(t) \). Once the derivative formula is set, evaluate it at the desired point, \( t=1 \). The derivative gives valuable insight into how the production of chemical pounds scales with time, showing us around 3.76 pounds are formed per hour at \( t=1 \).
Understanding derivatives is critical not only for calculating rates like this but also for exploring any function's sensitivity to change. This concept forms the basis for optimization, physics applications, and deep understanding of any system's dynamics over time.
Graphical Analysis
Graphical analysis gives a visual representation of functions and their behaviour over time, or another relevant variable. In our case, plotting the function \( x(t) \) over the interval \([0, 10]\) using software allows examination of how the chemical's formation progresses over these hours.
By observing the graph in the window \([0, 16]\), we gain insights that complement numerical analysis. This visual tool assists in identifying where changes are steepest, or settling, and helps confirm calculations with derivative values.
By observing the graph in the window \([0, 16]\), we gain insights that complement numerical analysis. This visual tool assists in identifying where changes are steepest, or settling, and helps confirm calculations with derivative values.
- Graphing assists in hypothesizing restraints or potential limits of functions, critical in decision-making.
- In practical terms, graphical representation lets one see the functional interplay and interaction of component behaviours over specific intervals.
Other exercises in this chapter
Problem 53
Sketch the graph of a function having the given properties. $$ \begin{array}{l} f(-1)=0, f^{\prime}(-1)=0 \\ f(0)=1, f^{\prime}(0)=0 \\ f^{\prime}(x)0 \text { o
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Find the horizontal and vertical asymptotes of the graph of the function. Do not sketch the graph. $$ h(x)=\frac{e^{x}}{e^{x}-2} $$
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Find the intervals where \(f(x)=e^{-x^{2} / 2}\) is increasing and where it is decreasing.
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