Active Calculus

Subsection 4.1.1 Area under the graph of the velocity function

In Preview Activity 4.1.1, we learned that when the velocity of a moving object's velocity is constant (and positive), the area under the velocity curve over an interval of time tells us the distance the object traveled.

The left-hand graph of Figure 4.1.2 shows the velocity of an object moving at 2 miles per hour over the time interval \([1,1.5]\text{.}\) The area \(A_1\) of the shaded region under \(y = v(t)\) on \([1,1.5]\) is

This result is simply the fact that distance equals rate times time, provided the rate is constant. Thus, if \(v(t)\) is constant on the interval \([a,b]\text{,}\) the distance traveled on \([a,b]\) is equal to the area \(A\) given by

where \(\Delta t\) is the change in \(t\) over the interval. (Since the velocity is constant, we can use any value of \(v(t)\) on the interval \([a,b]\text{,}\) we simply chose \(v(a)\text{,}\) the value at the interval's left endpoint.) For several examples where the velocity function is piecewise constant, see http://gvsu.edu/s/9T.¹

The situation is more complicated when the velocity function is not constant. But on relatively small intervals where \(v(t)\) does not vary much, we can use the area principle to estimate the distance traveled. The graph at right in Figure 4.1.2 shows a non-constant velocity function. On the interval \([1,1.5]\text{,}\) the velocity varies from \(v(1) = 2.5\) down to \(v(1.5) \approx 2.1\text{.}\) One estimate for the distance traveled is the area of the pictured rectangle,

Note that because \(v\) is decreasing on \([1,1.5]\text{,}\) \(A_2 = 1.25\) is an over-estimate of the actual distance traveled.

To estimate the area under this non-constant velocity function on a wider interval, say \([0,3]\text{,}\) one rectangle will not give a good approximation. Instead, we could use the six rectangles pictured in Figure 4.1.3, find the area of each rectangle, and add up the total. Obviously there are choices to make and issues to understand: How many rectangles should we use? Where should we evaluate the function to decide the rectangle's height? What happens if the velocity is sometimes negative? Can we find the exact area under any non-constant curve?

We will study these questions and more in what follows; for now it suffices to observe that the simple idea of the area of a rectangle gives us a powerful tool for estimating distance traveled from a velocity function, as well as for estimating the area under an arbitrary curve. To explore the use of multiple rectangles to approximate area under a non-constant velocity function, see the applet found at http://gvsu.edu/s/9U.²

Activity 4.1.2.

Suppose that a person is walking in such a way that her velocity varies slightly according to the information given in Table 4.1.4 and graph given in Figure 4.1.5.

\(t\)	\(v(t)\)
\(0.00\)	\(1.500\)
\(0.25\)	\(1.789\)
\(0.50\)	\(1.938\)
\(0.75\)	\(1.992\)
\(1.00\)	\(2.000\)
\(1.25\)	\(2.008\)
\(1.50\)	\(2.063\)
\(1.75\)	\(2.211\)
\(2.00\)	\(2.500\)

1. Using the grid, graph, and given data appropriately, estimate the distance traveled by the walker during the two hour interval from \(t = 0\) to \(t = 2\text{.}\) You should use time intervals of width \(\Delta t = 0.5\text{,}\) choosing a way to use the function consistently to determine the height of each rectangle in order to approximate distance traveled.

2. How could you get a better approximation of the distance traveled on \([0,2]\text{?}\) Explain, and then find this new estimate.

3. Now suppose that you know that \(v\) is given by \(v(t) = 0.5t^3-1.5t^2+1.5t+1.5\text{.}\) Remember that \(v\) is the derivative of the walker's position function, \(s\text{.}\) Find a formula for \(s\) so that \(s' = v\text{.}\)

4. Based on your work in (c), what is the value of \(s(2) - s(0)\text{?}\) What is the meaning of this quantity?

Hint

1. For instance, the approximate distance traveled on \([0,0.5]\) can be computed by \(v(0) \cdot 0.5 = 1.5 \cdot 0.5 = 0.75\) miles.

2. Think about the possibility of using a larger number of rectangles.

3. If \(v(t) = t^3\) and we seek a function \(s\) such that \(s' = v\text{,}\) note that \(s\) has to involve \(t^4\text{.}\)

4. Observe that this quantity is measuring a change in position.

Answer

1. 
   \begin{align*} A =\mathstrut \amp v(0.0) \cdot 0.5 + v(0.5) \cdot 0.5 + v(1.0) \cdot 0.5 + v(1.5) \cdot 0.5\\ =\mathstrut \amp 1.500 \cdot 0.5 + 1.9375 \cdot 0.5 + 2.000 \cdot 0.5 + 2.0625 \cdot 0.5\\ =\mathstrut \amp 3.75 \end{align*}

   Thus, \(D \approx 3.75\) miles.

2. Using 8 rectangles of width \(0.25\text{,}\) \(D \approx 3.875\text{.}\)

3. \(s(t) = \frac{1}{8}t^4 - \frac{1}{2} t^3 + \frac{3}{4} t^2 + \frac{3}{2}t\text{.}\)

4. \(s(2) - s(0) = \frac{1}{8}2^4 - \frac{1}{2}2^3 + \frac{3}{4}2^2 + \frac{3}{2} 2 = 4\text{.}\)

Solution

1. Using rectangles of width \(\Delta t = 0.5\) and choosing to set the heights of the rectangles from the function value at the left end of the interval, we see the following graph and find the sum of the areas of the rectangles to be

   \begin{align*} A =\mathstrut \amp v(0.0) \cdot 0.5 + v(0.5) \cdot 0.5 + v(1.0) \cdot 0.5 + v(1.5) \cdot 0.5\\ =\mathstrut \amp 1.500 \cdot 0.5 + 1.9375 \cdot 0.5 + 2.000 \cdot 0.5 + 2.0625 \cdot 0.5\\ =\mathstrut \amp 3.75 \end{align*}

   Thus, the distance traveled is approximately \(D \approx 3.75\) miles.

   

2. It appears that a better approximation could be found using narrower rectangles. If we move to 8 rectangles of width \(0.25\text{,}\) similar computations show that \(D \approx 3.875\text{.}\)

3. By thinking about how the power rule for differentiation works, we can undo this rule and find a position function \(s\) whose derivative is \(v\text{.}\) For instance, since \(\frac{d}{dt}[t^4] = 4t^3\text{,}\) we see that \(\frac{d}{dt}[\frac{1}{8}t^4] = \frac{1}{2}t^3\text{.}\) Thus, if we let

   \begin{equation*} s(t) = \frac{1}{8}t^4 - \frac{1}{2} t^3 + \frac{3}{4} t^2 + \frac{3}{2}t\text{,} \end{equation*}

   then it is straightforward to check that \(s'(t) = \frac{1}{2}t^3 - \frac{3}{2}t^2 + \frac{3}{2}t + \frac{3}{2}\text{,}\) which is precisely the formula for \(v(t)\) that we were given.

4. By the rule found in (c) for \(s\text{,}\) we have that \(s(2) - s(0) = \frac{1}{8}2^4 - \frac{1}{2}2^3 + \frac{3}{4}2^2 + \frac{3}{2} 2 = 2 - 4 + 3 + 3 = 4\text{.}\) This is the change in the walker's position over the time interval \([0,2]\text{,}\) and since the velocity is always positive, this is actually the exact distance traveled. We see how both earlier estimates (\(3.75\) and \(3.785\)) are good approximations to this value.

Subsection 4.1.2 Two approaches: area and antidifferentiation

When the velocity of a moving object is positive, the object's position is always increasing. (We will soon consider situations where velocity is negative; for now, we focus on the situation where velocity is always positive.) We have established that whenever \(v\) is constant on an interval, the exact distance traveled is the area under the velocity curve. When \(v\) is not constant, we can estimate the total distance traveled by finding the areas of rectangles that approximate the area under the velocity curve.

Thus, we see that finding the area between a curve and the horizontal axis is an important exercise: besides being an interesting geometric question, if the curve gives the velocity of a moving object, the area under the curve tells us the exact distance traveled on an interval. We can estimate this area if we have a graph or a table of values for the velocity function.

In Activity 4.1.2, we encountered an alternate approach to finding the distance traveled. If \(y = v(t)\) is a formula for the instantaneous velocity of a moving object, then \(v\) must be the derivative of the object's position function, \(s\text{.}\) If we can find a formula for \(s(t)\) from the formula for \(v(t)\text{,}\) we will know the position of the object at time \(t\text{,}\) and the change in position over a particular time interval tells us the distance traveled on that interval.

For a simple example, consider the situation from Preview Activity 4.1.1, where a person is walking along a straight line with velocity function \(v(t) = 3\) mph.

On the left-hand graph of the velocity function in Figure 4.1.6, we see the relationship between area and distance traveled,

In addition, we observe³ that if \(s(t) = 3t\text{,}\) then \(s'(t) = 3\text{,}\) so \(s(t) = 3t\) is the position function whose derivative is the given velocity function, \(v(t) = 3\text{.}\) The respective locations of the person at times \(t = 0.25\) and \(t = 1.5\) are \(s(1.5) = 4.5\) and \(s(0.25) = 0.75\text{,}\) and therefore

This is the person's change in position on \([0.25,1.5]\text{,}\) which is precisely the distance traveled. In this example there are profound ideas and connections that we will study throughout Chapter 4.

For now, observe that if we know a formula for a velocity function \(v\text{,}\) it can be very helpful to find a function \(s\) that satisfies \(s' = v\text{.}\) We say that \(s\) is an antiderivative of \(v\text{.}\) More generally, we have the following formal definition.

For example, if \(g(x) = 3x^2 + 2x\text{,}\) \(G(x) = x^3 + x^2\) is an antiderivative of \(g\text{,}\) because \(G'(x) = g(x)\text{.}\) Note that we say “an” antiderivative of \(g\) rather than “the” antiderivative of \(g\text{,}\) because \(H(x) = x^3 + x^2 + 5\) is also a function whose derivative is \(g\text{,}\) and thus \(H\) is another antiderivative of \(g\text{.}\)

Subsection 4.1.3 When velocity is negative

The assumption that its velocity is positive on a given interval guarantees that the movement of an object is always in a single direction, and hence ensures that its change in position is the same as the distance it travels. As we saw in Activity 4.1.3, there are natural settings in which an object's velocity is negative, and we would like to understand this scenario as well.

Consider a simple example where a woman goes for a walk on the beach along a stretch of very straight shoreline that runs east-west. We assume that her initial position is \(s(0) = 0\text{,}\) and that her position function increases as she moves east from her starting location. For instance, \(s = 1\) mile represents one mile east of the start location, while \(s = -1\) tells us she is one mile west of where she began walking on the beach.

Now suppose she walks in the following manner. From the outset at \(t = 0\text{,}\) she walks due east at a constant rate of \(3\) mph for 1.5 hours. After 1.5 hours, she stops abruptly and begins walking due west at a constant rate of \(4\) mph and does so for 0.5 hours. Then, after another abrupt stop and start, she resumes walking at a constant rate of \(3\) mph to the east for one more hour. What is the total distance she traveled on the time interval from \(t = 0\) to \(t = 3\text{?}\) What the total change in her position over that time?

These questions are possible to answer without calculus because the velocity is constant on each interval. From \(t = 0\) to \(t = 1.5\text{,}\) she traveled

On \(t = 1.5\) to \(t = 2\text{,}\) the distance traveled is

Finally, in the last hour she walked

so the total distance she traveled is

Since the velocity for \(1.5 \lt t \lt 2\) is \(v = -4\text{,}\) indicating motion in the westward direction, the woman first walked 4.5 miles east, then 2 miles west, followed by 3 more miles east. Thus, the total change in her position is

We have been able to answer these questions fairly easily, and if we think about the problem graphically, we can generalize our solution to the more complicated setting when velocity is not constant, and possibly negative.

In Figure 4.1.9, we see how the distances we computed can be viewed as areas: \(A_1 = 4.5\) comes from multiplyimg rate times time (\(3 \cdot 1.5\)), as do \(A_2\) and \(A_3\text{.}\) But while \(A_2\) is an area (and is therefore positive), because the velocity function is negative for \(1.5 \lt t \lt 2\text{,}\) this area has a negative sign associated with it. The negative area distinguishes between distance traveled and change in position.

The distance traveled is the sum of the areas,

But the change in position has to account for travel in the negative direction. An area above the \(t\)-axis is considered positive because it represents distance traveled in the positive direction, while one below the \(t\)-axis is viewed as negative because it represents travel in thenegative direction. Thus, the change in the woman's position is

In other words, the woman walks 4.5 miles in the positive direction, followed by two miles in the negative direction, and then 3 more miles in the positive direction.

Negative velocity is also seen in the graph of the position function \(y=s(t)\text{.}\) Its slope is negative (specifically, \(-4\)) on the interval \(1.5\lt t\lt 2\) because the velocity is \(-4\) on that interval. The negative slope shows the position function is decreasing because the woman is walking east, rather than west.

To summarize, we see that if velocity is sometimes negative, a moving object's change in position different from its distance traveled. If we compute separately the distance traveled on each interval where velocity is positive or negative, we can calculate either the total distance traveled or the total change in position. We assign a negative value to distances traveled in the negative direction when we calculate change in position, but a positive value when we calculate the total distance traveled.

Activity 4.1.4.

Suppose that an object moving along a straight line path has its velocity \(v\) (in meters per second) at time \(t\) (in seconds) given by the piecewise linear function whose graph is pictured at left in Figure 4.1.10. We view movement to the right as being in the positive direction (with positive velocity), while movement to the left is in the negative direction.

Suppose further that the object's initial position at time \(t = 0\) is \(s(0) = 1\text{.}\)

1. Determine the total distance traveled and the total change in position on the time interval \(0 \le t \le 2\text{.}\) What is the object's position at \(t = 2\text{?}\)

2. On what time intervals is the moving object's position function increasing? Why? On what intervals is the object's position decreasing? Why?

3. What is the object's position at \(t = 8\text{?}\) How many total meters has it traveled to get to this point (including distance in both directions)? Is this different from the object's total change in position on \(t = 0\) to \(t = 8\text{?}\)

4. Find the exact position of the object at \(t = 1, 2, 3, \ldots, 8\) and use this data to sketch an accurate graph of \(y = s(t)\) on the axes provided at right in Figure 4.1.10. How can you use the provided information about \(y = v(t)\) to determine the concavity of \(s\) on each relevant interval?

Hint

1. Find the area of each triangular region formed between \(y = v(t)\) and the \(t\)-axis.

2. Recall that \(v = s'\text{,}\) and here we are given complete information about \(v\text{.}\)

3. Be careful to address whether \(v\) is positive or negative when calculating areas and adding the results.

4. Consider finding the area bounded by \(y = v(t)\) and the \(t\)-axis on each interval \([0,1]\text{,}\) \([1,2]\text{,}\) \(\ldots\text{.}\)

Answer

1. Total distance traveled is \(2\text{;}\) change in position is \(0\text{.}\)

2. \(0 \lt t \lt 1\) and \(4 \lt t \lt 8 \text{.}\)

3. \(s(8) - s(0) = 1 - 4 + 8 = 5 \ \mbox{m} \text{.}\)

4. See the figure below.

Solution

1. By finding the area of the triangular regions formed between \(y = v(t)\) and the \(t\)-axis on \([0,1]\) and \([1,2]\) (each of which is \(1\)), it follows that the object's total distance traveled is \(2\text{,}\) while its change in position is \(0\text{.}\) The latter is true since the net signed area bounded by \(v\) on \([0,2]\) is \(1 - 1 = 0\text{.}\)

2. The object's position is increasing wherever its velocity is positive, hence for \(0 \le t \lt 1\) and \(4 \lt t \lt 8 \text{.}\)

3. By calculating the area bounded by the curve, we find 1 unit of area on \([0,1]\text{,}\) 4 units of area on \([1,4]\text{,}\) and 8 units of area on \([4,8]\text{,}\) thus the total distance traveled on \(0 \le t \le 8\) is \(D = 1 + 4 + 8\) meters. As the change in position is given by the net signed area on this interval, we find that the change in position is

   \begin{equation*} s(8) - s(0) = 1 - 4 + 8 = 5 \ \mbox{m}\text{.} \end{equation*}

4. In the figure below, at left we list all of the areas bounded by \(v\) on each one-unit subinterval. Along with the given starting point that \(s(0) = 1\text{,}\) we use the resulting changes in position to plot points for the function \(s\text{.}\) For instance, we know \(s(1) - s(0) = 1\text{,}\) hence \(s(1) = 2\text{.}\) Similarly, \(s(2) - s(1) = -1\text{,}\) thus \(s(2) = 1\text{.}\) Continuing across the interval, we generate the function \(s\) that is pictured at right. Note that the portion of \(s\) from \(t = 2\) to \(t = 3\) is linear because \(v\) is constant there, while the other parts of \(s\) appear to be quadratic, as they correspond to intervals where \(v\) is linear.

Subsection 4.1.4 Summary

• If we know the velocity of a moving body at every point in a given interval and the velocity is positive throughout, we can estimate the object's distance traveled and in some circumstances determine this value exactly.

• In particular, when velocity is positive on an interval, we can find the total distance traveled by finding the area under the velocity curve and above the \(t\)-axis on the given time interval. We may only be able to estimate this area, depending on the shape of the velocity curve.

• An antiderivative of a function \(f\) is a new function \(F\) whose derivative is \(f\text{.}\) That is, \(F\) is an antiderivative of \(f\) provided that \(F' = f\text{.}\) In the context of velocity and position, if we know a velocity function \(v\text{,}\) an antiderivative of \(v\) is a position function \(s\) that satisfies \(s' = v\text{.}\) If \(v\) is positive on a given interval, say \([a,b]\text{,}\) then the change in position, \(s(b) - s(a)\text{,}\) measures the distance the moving object traveled on \([a,b]\text{.}\)

• If its velocity is sometimes negative, a moving object is sometimes traveling in the opposite direction or backtracking. To determine distance traveled, we have to think compute the distance separately on intervals where velocity is positive or negative, and account for the change in position on each such interval.