/*! 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 3 Draw the graphs of $$y_{1}(t)=... [FREE SOLUTION] | 91Ó°ÊÓ

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Draw the graphs of $$y_{1}(t)=e^{t} \quad y_{2}(t)=t^{2} \quad y_{3}(t)=t^{3} \quad y_{4}(t)=t^{4} \quad-1 \leq t \leq 5$$ The graphs are close together near \(t=0\) and increase as \(t\) increases. Which one grows the most as \(t\) increases? Expand the domain and range to \(-1 \leq t \leq 10,0 \leq y \leq 25,000,\) and answer the same question.

Short Answer

Expert verified
On the expanded domain, \(y_4(t) = t^4\) grows the most.

Step by step solution

01

Understand the Functions

We need to graph four different functions: \(y_1(t) = e^t\), \(y_2(t) = t^2\), \(y_3(t) = t^3\), and \(y_4(t) = t^4\) within the domain \(-1 \leq t \leq 5\). Each function represents a different mathematical expression that describes how \(y\) changes with \(t\).
02

Analyze Graph Behavior in Given Domain

Plot each function to observe their behaviors in the domain \(-1 \leq t \leq 5\). All graphs start close to each other near \(t = 0\), but as \(t\) approaches 5, \(y_1(t) = e^t\) grows the fastest compared to \(y_2(t) = t^2\), \(y_3(t) = t^3\), and \(y_4(t) = t^4\). To confirm, calculate the values for a few points, like at \(t = 0, 2,\) and \(5\).
03

Expand Domain and Range

Now, expand the domain to \(-1 \leq t \leq 10\) and the range to \(0 \leq y \leq 25,000\). The functions grow larger, showing bigger variances in size as \(t\) approaches 10. Use this expanded domain to better visualize the rate of growth of each function, with \(y_1(t) = e^t\) potentially having more pronounced growth compared to polynomial functions.
04

Determine Which Function Grows the Most

Observe the graphs' behavior as \(t\) increases towards 10. Calculate \(y_1(10) = e^{10}\), \(y_2(10) = 10^2\), \(y_3(10) = 10^3\), and \(y_4(10) = 10^4\). Given the exponential nature of \(e^t\), \(y_1(t)\) is expected to grow faster initially, but \(y_4(t) = t^4\) ultimately surpasses it within the expanded domain.

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Key Concepts

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

Exponential Growth
Exponential growth is a fascinating mathematical concept commonly seen in processes that increase rapidly and continuously. One classical example is the function \( y_1(t) = e^t \), which describes exponential growth. At its core, exponential growth means that the growth rate of the value is proportional to its current size. Simply put, the larger the value, the faster it grows.
In the original exercise, as we examined the function \( e^t \) within a specific domain of \( -1 \leq t \leq 5 \), its growth rate was noticeable, especially as \( t \) approached higher values. When we expanded the domain to \( -1 \leq t \leq 10 \), the dramatic increase in \( y_1(t) \) was even more apparent.
- At \( t = 0 \), the function evaluates to \( e^0 = 1 \).- At \( t = 5 \), it evaluates to approximately \( e^5 \approx 148.41 \).- Moving further, at \( t = 10 \), \( e^{10} \approx 22,026.47 \).
This rapid increase showcases why exponential functions are often associated with swift and relentless expansion.
Polynomial Functions
Polynomial functions, on the other hand, grow based on both their degree and the coefficients of their terms. For instance, the exercises outline functions such as \( y_2(t) = t^2 \), \( y_3(t) = t^3 \), and \( y_4(t) = t^4 \). These functions represent quadratic, cubic, and quartic growth respectively. Here's why these distinctions matter:
- **Quadratic Function \( t^2 \):** This represents symmetrical growth around the origin. As \( t \) increases, \( t^2 \) grows steadily. At \( t = 5 \), it reaches \( 5^2 = 25 \), and at \( t = 10 \), it becomes \( 10^2 = 100 \).- **Cubic Function \( t^3 \):** This showcases asymmetrical growth with a more pronounced increase, gaining momentum faster than \( t^2 \). At \( t = 5 \), it evaluates to \( 5^3 = 125 \) and at \( t = 10 \), it gets up to \( 10^3 = 1,000 \).- **Quartic Function \( t^4 \):** With each increase in \( t \), this function's rate of growth accelerates significantly. At \( t = 5 \), \( t^4 = 625 \). By \( t = 10 \), it jumps to \( 10^4 = 10,000 \).
Despite its slower start compared to the exponential function, \( t^4 \) significantly rivals \( e^t \) by the time \( t \) reaches 10. Understanding these growth trends is key to mastering polynomial behaviors in calculus graphing.
Domain and Range Expansion
When dealing with graph functions, domain and range are crucial components that define where and how the function exists visually and numerically. The domain represents all possible input values (\( t \) in this case), while the range denotes all possible outcomes (\( y \) values).
In the given exercise, the initial domain was \(-1 \leq t \leq 5\), which allowed us to observe function behavior over a shorter interval. Expansion to \(-1 \leq t \leq 10\) introduced broader perspectives on the functions' growth capabilities.
- **Expanded Domain:** Enlarging the span of \( t \) helps in capturing more data on how functions perform across a larger set.- **Larger Range:** Adjusting the range to \(0 \leq y \leq 25,000\) accommodates greater outputs especially from polynomial growth like \( y_4 = t^4 \) and exponential growth from \( e^t \).
This expansion is particularly useful when comparing growth rates. It visually demonstrates how exponential functions, while initially strong, can be rivaled by high-degree polynomials. By broadening these parameters, one can better analyze and infer the functional trends over time.

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Most popular questions from this chapter

In Section 1.3 we found from the discrete model of light extinction, \(I_{d+1}=I_{d}-0.18 I_{d}\) that the solution \(\quad I_{d+1}=0.82 I_{d}\) Light decrease in water is continuous, however. Find a value of \(k\) for which the solution to the continuous model, \(I^{\prime}(x)=-k I(x),\) matches the data.

Suppose \(S_{2}\) is the set of numbers to which \(x\) belongs if and only if \(x\) is positive and \(x^{2}>2\) and \(S_{1}\) consists of all of the other numbers. 1\. Give an example of a number in \(S_{2}\). 2\. Give an example of a number in \(S_{1}\). 3\. Argue that every number in \(S_{1}\) is less than every number in \(S_{2}\). 4\. Which of the following two statements is true? (a) There is a number \(C\) which is the largest number in \(S_{1}\). (b) There is a number \(C\) which is the least number in \(S_{2}\). 5\. Identify the number \(C\) in the correct statement of the previous part.

Consider the kinetics of penicillin that is taken as a pill in the stomach. The diagram in Figure Ex. \(5.4 .9(\) a \()\) may help visualize the kinetics. We will find in Chapter 17 that a model of plasma concentration of antibiotic \(t\) hours after ingestion of an antibiotic pill yields an equation similar to $$C(t)=5 e^{-2 t}-5 e^{-3 t} \quad \mu \mathrm{g} / \mathrm{ml}$$ A graph of \(C\) is shown in Figure Ex. \(5.4 .9 .\) At what time will the concentration reach a maximum level, and what is the maximum concentration achieved? As we saw in Section 3.5 .2 and may be apparent from the graph in Figure Ex. 5.4 .9 , the highest concentration is associated with the point of the graph of \(C\) at which \(C^{\prime}=0 ;\) the tangent at the high point is horizontal. The question, then, is at what time \(t\) is \(C^{\prime}(t)=0\) and what is \(C(t)\) at that time?

Suppose penicillin concentration in the serum of a patient \(t\) minutes after a bolus injection of \(2 \mathrm{~g}\) is given by $$P(t)=200 \times 0.96^{t} \quad \mu \mathrm{g} / \mathrm{ml}$$ a. Approximate \(P(t)\) and \(P^{\prime}(t)\) for \(t=0,5,10,15,\) and 20 minutes. b. Plot a graph of \(P^{\prime}(t)\) vs \(P(t)\) using the five pairs of values you just computed.

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