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Chapter 10: Problem 3

Review concepts that are important in this section. In each exercise, sketch the graph of a function with the stated properties. Domain: \(0 \leq t \leq 5 ;(0,3)\) is on the graph; the slope is always negative, and the slope becomes less negative.

Short Answer

Expert verified
Domain: \(0 \leq t \leq 5\), passes through (0,3), always decreasing but at a decreasing rate.

Step by step solution

01

Identify the Domain

The domain of the function is given as: \[0 \leq t \leq 5\]. This means the function is defined for all values of \(t\) from 0 to 5.
02

Plot the Given Point

The point \((0, 3)\) is on the graph. This indicates that the function passes through \(t = 0\) and has a function value of \(3\). Plot this point on the graph.
03

Determine the Slope Behavior

The slope is always negative means the function is always decreasing. However, it is less negative as \(t\) increases which implies the function decreases at a decreasing rate.
04

Sketch the Graph

Start from point \((0, 3)\). Since the slope is negative and decreasing in magnitude, the function should curve downward initially, leveling off as \(t\) approaches 5.

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Key Concepts

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

Domain of a Function
When we talk about the domain of a function, we are identifying all the possible input values (usually denoted by the variable t) for which the function is defined and makes sense.
In this exercise, the domain is given as \[0 \leq t \leq 5 \].
This means that the function has input values ranging from t = 0 to t = 5, inclusive.

Knowing the domain helps us understand where to begin and end the sketch of the graph.
For this exercise, we only need to consider the behavior of the function between t = 0 and t = 5.
Always check the domain before plotting the function, because it tells you the 鈥渟cope鈥� of your graph.
Slope Behavior
The slope of a function represents the rate at which the function's output value changes with respect to the input value.
In this exercise, the slope is always negative, meaning the function is always decreasing.
However, it also states that the slope becomes less negative as t increases.
This means that while the function continues to decrease, it does so at a slowing rate.

Imagine a steep hill that gradually flattens out鈥攖his is similar to how the function behaves since its slope is becoming less steep.
This characteristic plays a crucial role in shaping the curve of the graph.
Function Plotting
To plot a function, we first identify key points and understand its slope behavior.
In this exercise, we start by plotting the given point (0, 3).
This means that when t = 0, the function value is 3.
Next, we consider that the function is always decreasing with a less negative slope as t increases.

Begin plotting by starting at the point (0, 3) and then draw the graph so that it trends downwards.
Make sure the slope is steep initially and becomes less steep as you approach t = 5.
Plotting functions correctly involves understanding these behaviors and carefully marking the points and slopes on your graph paper.
Decreasing Function
A function is described as decreasing if its output values get smaller as the input values increase.
In this exercise, the function is always decreasing since its slope is always negative.
However, the function's rate of decrease slows down over time, implying the graph flattens out as it moves to the right.

Visualize this concept by imagining a hill that gets less steep as you go down.
This is how the function will behave, starting from (0, 3) and decreasing smoothly until it levels out near t = 5.
Recognizing decreasing functions helps in sketching graphs accurately and understanding the overall trend of the function.

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Most popular questions from this chapter

A model that describes the relationship between the price and the weekly sales of a product might have a form such as $$ \frac{d y}{d p}=-\frac{1}{2}\left(\frac{y}{p+3}\right), $$ where \(y\) is the volume of sales and \(p\) is the price per unit. That is, at any time, the rate of decrease of sales with respect to price is directly proportional to the sales level and inversely proportional to the sales price plus a constant. Solve this differential equation. (Figure 6 shows several typical solutions.)

According to the National Kidney Foundation, in 1997 more than 260,000 Americans suffered from chronic kidney failure and needed an artificial kidney (dialysis) to stay alive. (Source: The National Kidney Foundation, www.kidney.org.) When the kidneys fail, toxic waste products such as creatinine and urea build up in the blood. One way to remove these wastes is to use a process known as peritoneal dialysis, in which the patient's peritonium, or lining of the abdomen, is used as a filter. When the abdominal cavity is filled with a certain dialysate solution, the waste products in the blood filter through the peritonium into the solution. After a waiting period of several hours, the dialysate solution is drained out of the body along with the waste products. In one dialysis session, the abdomen of a patient with an elevated concentration of creatinine in the blood equal to 110 grams per liter was filled with two liters of a dialysate (containing no creatinine). Let \(f(t)\) denote the concentration of creatinine in the dialysate at time \(t\). The rate of change of \(f(t)\) is proportional to the difference between 110 (the maximum concentration that can be attained in the dialysate) and \(f(t)\). Thus, \(f(t)\) satisfies the differential equation $$ y^{\prime}=k(110-y) . $$ (a) Suppose that, at the end of a 4-hour dialysis session, the concentration in the dialysate was 75 grams per liter and it was rising at the rate of 10 grams per liter per hour. Find \(k\). (b) What is the rate of change of the concentration at the beginning of the dialysis session? By comparing with the rate at the end of the session, can you give a (simplistic) justification for draining and replacing the dialysate with a fresh solution after 4 hours of dialysis? [Hint: You do not need to solve the differential equation.]

Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}-2 t y=-4 t $$

One or more initial conditions are given for each differential equation in the following exercises. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solutions. Include a \(y z\) -graph if one is not already provided. Always indicate the constant solutions on the \(t y\) -graph whether they are mentioned or not. \(y^{\prime}=-\frac{1}{2} y, y(0)=-2, y(0)=0, y(0)=2\)

Use Euler's method with \(n=4\) to approximate the solution \(f(t)\) to \(y^{\prime}=2 t-y+1, y(0)=5\) for \(0 \leq t \leq 2\) Estimate \(f(2)\).
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