5.6: The Net Change Theorem

The Net Change Theorem

The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.

Subtracting \(F(a)\) from both sides of the Equation \ref{Net1} yields Equation \ref{Net2}. Since they are equivalent formulas, which one we use depends on the application.

The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.

We looked at a simple example of this in The Definite Integral section. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in Figure \(\PageIndex{1}\).

Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by

\[ ∫^5_2v(t)\,dt=∫^4_240\,dt+∫^5_4−30\,dt=80−30=50. \nonumber \]

Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by

\[ ∫^5_2|v(t)|\,dt=∫^4_240\,dt+∫^5_430\,dt=80+30=110. \nonumber \]

Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.

To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.

Example \(\PageIndex{2}\): Finding Net Displacement

Given a velocity function \(v(t)=3t−5\) (in meters per second) for a particle in motion from time \(t=0\) to time \(t=3,\) find the net displacement of the particle.

Solution

Applying the net change theorem, we have

\[ ∫^3_0(3t−5)\,dt=\left(\frac{3t^2}{2}−5t\right)\bigg|^3_0=\left[\frac{3(3)^2}{2}−5(3)\right]−0=\frac{27}{2}−15=\frac{27}{2}−\frac{30}{2}=−\frac{3}{2}. \nonumber \]

The net displacement is \( −\frac{3}{2}\) m (Figure \(\PageIndex{2}\)).

Applying the Net Change Theorem

The net change theorem can be applied to the flow and consumption of fluids, as shown in Example \(\PageIndex{4}\).

Example \(\PageIndex{5}\): Chapter Opener: Iceboats

As we saw at the beginning of the chapter, top iceboat racers can attain speeds of up to five times the wind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed.

Suppose Andrew takes his iceboat out one morning when a light \(5\)-mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind speed increases according to the function \(v(t)=20t+5.\) For the second half hour of Andrew’s outing, the wind remains steady at \(15\) mph. In other words, the wind speed is given by

\[ v(t)=\begin{cases}20t+5, & \text{for } 0≤t≤\frac{1}{2}\\15, & \text{for } \frac{1}{2}≤t≤1\end{cases} \nonumber \]

Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after \(1\) hour?

Solution

To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then

\[\text{Distance} = ∫^1_02v(t)\,dt. \nonumber \]

Substituting the expressions we were given for \(v(t)\), we get

\[ \begin{align*} ∫^1_02v(t)\,dt &=∫^{1/2}_02v(t)\,dt+∫^1_{1/2}2v(t)\,dt \\[4pt]
&=∫^{1/2}_02(20t+5)\,dt+∫^1_{1/3}2(15)\,dt \\[4pt]
&=∫^{1/2}_0(40t+10)\,dt+∫^1_{1/2}30\,dt \\[4pt]
&=\big[20t^2+10t\big]\bigg|^{1/2}_0+\big[30t\big]\bigg|^1_{1/2} \\[4pt]
&=\left(\frac{20}{4}+5\right)−0+(30−15) \\[4pt]
&=25. \end{align*}\]

Andrew is 25 mi from his starting point after 1 hour.

Key Concepts

• The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.
• The area under an even function over a symmetric interval can be calculated by doubling the area over the positive \(x\)-axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.

Key Equations

• Net Change Theorem\[F(b)=F(a)+∫^b_aF'(x)\,dx\nonumber \] or \[∫^b_aF'(x)\,dx=F(b)−F(a) \nonumber \]

Glossary

net change theorem

if we know the rate of change of a quantity, the net change theorem says the future quantity is equal to the initial quantity plus the integral of the rate of change of the quantity