Section 8.1 The Nature of Integral Problems and Ways to Approach Them
Think "integral" any time that you are confronted with a situation where you know how fast Quantity A varies with respect to Quantity B and you want to know an amount of Quantity A that has accumulated with respect to Quantity B.
Example 1. A rock is thrown vertically from a height of 2 meters with an initial velocity of 21 m/s. Acceleration due to gravity is -9.8 (m/s)/s. What is the rock's height after 2.8 seconds? What was the rock's height at each number of seconds after being thrown?
As was the case in Section
7.3 (Optimization), the most time-consuming part of answering such questions is to construct an appropriate model of relationships among quantities. Once you have a rate of change function that models how fast one quantity varies with respect to another, then all that remains is to integrate the rate of change function over an appropriate interval.
It is worthwhile to note something about the problem in Example 1. Contrary to intuition, you must, in principle, answer the general
question before you can answer the specific question. You must answer the question of the rock's height at each moment in time to answer the question about its height after 2.8 seconds have elapsed.
Solution to Example 1. First, notice that the two relevant quantities in Example 1 are the rock's velocity at each moment in time and the elapsed time since being thrown. Then we can reason as follows:
The rock's height variations in small bits by moving at some velocity for some amount of time.
The rock's velocity $v(t)$ at each value of t (at each moment in time since thrown) is its initial velocity plus added velocity due to gravity (recall that velocity is a signed quantity--direction matters).
The rock's initial velocity is 21 m/s. Gravity adds -$9.8t$ m/s to the initial velocity as t varies. Thus, $v(t)=21-9.8t$.
The rock's height at t seconds from being thrown varies by approximately $v(t)dt$ meters over as t varies by $dt$ seconds.
The rock's accumulated height at t seconds since launch is the sum of those bits of change from $u=0$ seconds to $u=t$ seconds.
We use "$u$" because some variable needs to vary in time between 0 seconds and t seconds to generate the bits of change in height that accumulate into the rock's height at t seconds. We could use any variable other than t.
Let $s$ be the function that gives the rock's height at each moment in time since being thrown. Then $s(t)=2+\int_0^t v(u)du$. This integral says that we are summing bits of variations in height from $u=0$ to $u=t$ seconds. We add 2 because the rock started at an initial height of 2 meters and the integral calculates net change in height from the starting height.
Notice that we answered the general question in order to answer the specific question. Our answer to the general question answers all specific questions.
Define $s$ in GC. Type $s(2.8)$. Then graph $y=s(x),x \gt 0$ (see Figure 8.1.1, left).
We could have defined added velocity as an integral, since it builds up as t varies (see Figure 8.1.2, right).
Figures 8.1.1 and 8.1.2. Solutions to Example 1, entered into GC.
In both solutions to Example 1, we defined the distance function in open form, as an integral. Defining an accumulation function in open form is always good. It outlines our solutions conceptually.
The definition of $v$ in Figure 8.1.2 reminds us that the rock's velocity at any time t seconds is the initial velocity plus the (negative) velocity that accrues due to gravity.
Velocity that accrues due to gravity accrues in bits of velocity that are due to the rock accelerating at the rate of -9.8 (m/s)/s over an interval of $du$ seconds as $u$ varies from $u=0$ to $u=t$.
So the net change in velocity at t seconds is $-\int_0^t 9.8du$.
The definition of $s$ reminds us that the rock's height at any time t is the initial height plus the height that accrues in bits of height that are $v(u)du$ meters as the rock travels at a velocity of $v(u)$ as $u$ varies by $du$.
The accrued height due to the rock's velocity being $v(u)$ over time intervals of length $du$ as $u$ varies from $u=0$ to $u=t$ seconds is therefore $$s(t)=2+\int_0^t v(u)du.$$
However, some questions can be difficult to answer precisely based on the open-form-integral approach:
After how many seconds does the rock
hit the ground?
We could approximate an answer by using GC's tracing feature on the graph of $y=s(x)$ to see that the rock hits the ground approximately 4.379 seconds after being released.
To answer this question precisely, we need to define $s$ symbolically. Using the FTC, we see that $$\begin{align}s(t)&=2+\int_0^t v(u)du\\[1ex] &=2+\int_0^t (21-9.8u)du\\[1ex] &=2+\left. \left(21u-\frac{9.8}{2}u^2\right)\right|_{u=0}^{u=t}\\[1ex]
&=2+21t-\frac{9.8}{2}t^2.\end{align}$$
Using the quadratic formula to solve for t in $$2+21t-\frac{9.8}{2}t^2=0,$$we get $$t=\frac{-21-\sqrt{21^2-4\cdot \frac{-9.8}{2} \cdot 2}}{2 \cdot \frac{-9.8}{2}}\approx 4.37911.$$
According to our model, the ball hit the ground approximately 4.37911 seconds after being thrown.
Reflection 8.1.1. Do the graphs in Figures 8.1.1 and 8.1.2 show the ball's path after being thrown? If not, what do they show?
Two Modes of Approaching Integral Problems
The discussion in Example 1 illustrated strengths and weaknesses of two modes of approaching problems that involve integrals.
The first mode is to model the situation using a function defined as an integral and to use this function to have GC calculate specific values or to generate a graph.
The second mode is to use the FTC to create a closed form definition of the same function that the integral defines and to use that function to answer questions about properties of the situation.
In Chapter 8 we will concentrate on the first mode of approaching problems. Modeling a situation using integrals allows us to focus on the conceptual nature of the situation's quantities and relationships among them.
In Chapter 9 (Integration Techniques), we will concentrate on the second mode, using the FTC to find closed form definitions of the same functions that we defined in Chapter 8 using integrals. Finding closed form definitions of integral functions will allow us to investigate structural properties of situations.
Exercise Set 8.1
Download the GC file for Section 8.1 Example 1 (a rock thrown vertically from a height of 2 meters with initial velocity of 21 m/s). Use the functions in this file to:
Give a numerical estimate for s(1.9):
Give a numerical estimate for v(1.9):
Represent s(1.9) in terms of the definition of $s$:
Represent v(1.9) in terms of the definition of $v$:
$V(T)$ is the volume (in $\text{cm}^3$) of a container when the temperature is T degrees Celsius. $H(T)$ is the rate at which the volume varies with respect to temperature, in $\text{cm}^3$ per degree Celsius.
$V(T)$ can be represented as the _____ of $H(T)$ with respect to _____.
Represent the small bit of volume that accumulates as the temperature varies slightly from 56 degrees C:
Use $H(T)$ to represent the exact net accumulation of volume as the temperature varies from 56 degrees C to 51 degrees C.
Use $V(T)$ to represent the exact net accumulation of volume as the temperature varies from 56 degrees C to 51 degrees C.
Suppose that the container's volume was 1023 $\text{cm}^3$ when the temperature was 51 degrees C.
Represent the container's total volume in $\text{cm}^3$ when the temperature is t degrees C.
Rewrite your answer to "i" using function notation instead of the number 1023. In what way is this representation more general than the one you wrote for "i"?
$q(t)$ is the mass, in grams, of a yeast culture t hours after it started growing. $r(t)$ is the rate, in $\mathrm{g/sec}$, at which the yeast mass is increasing with respect to time.
$q(t)$ can be represented by the _____ of $r(t)$ with respect to _____.
Represent the small bit of mass that accumulates as elapsed time varies slightly from 0.75 hours:
Use $q(t)$ to represent the exact net accumulation of mass as elapsed time varies from 1.2 hours to 2.1 hours.
Use $q(t)$ to represent the exact net accumulation of mass when we imagine elapsed time varying from 3.2 hours to 2.1 hours.
Suppose that the culture's mass was 102.7 grams when 1.72 hours had elapsed.
Represent the culture's initial mass (when $t=0$).
Represent the culture's total mass in grams after t hours have elapsed.