Applying the Net Change Theorem

The net change theorem can be applied to the flow and consumption of fluids, as shown in Example 5.26.

Example 5.26

How Many Gallons of Gasoline Are Consumed?

If the motor on a motorboat is started at \(t = 0\) and the boat consumes gasoline at the rate of \(5 - 0.1 t^{3}\) gal/hr, how much gasoline is used in the first 2 hours?

Solution

Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval \(\left[\right. 0 , 2 \left]\right. .\) We have

\[\begin{aligned} \int_{0}^{2} \left(\right. 5 - 0 . 1 t^{3} \left.\right) d t = \left(\right. 5 t – 0 . 1 \frac{t^{4}}{4} \left.\right) \left(\left|\right.^{2}\right)_{0} \\ = \left[\right. 5 \left(\right. 2 \left.\right) – 0 . 1 \frac{\left(\left(\right. 2 \left.\right)\right)^{4}}{4} \left]\right. – 0 \\ = 10 – 0 . 4 \\ = 9 . 6 \end{aligned}\]

Thus, the motorboat uses 9.6 gal of gas in 2 hours.

Example 5.27

Chapter Opener: Iceboats

Figure 5.34 (credit: modification of work by Carter Brown, Flickr)

As we saw at the beginning of the chapter, top iceboat racers (Figure 5.1) 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 \(v \left(\right. t \left.\right) = 20 t + 5 .\) 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

\[\begin{aligned} v \left(\right. t \left.\right) = \left{\right. 20 t + 5 & \text{for} & 0 \leq t \leq \frac{1}{2} \\ 15 & \text{for} & \frac{1}{2} \leq t \leq 1 . \end{aligned}\]

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

Distance \(= \int_{0}^{1} 2 v \left(\right. t \left.\right) d t .\)

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

\[\begin{aligned} \int_{0}^{1} 2 v \left(\right. t \left.\right) d t & = \int_{0}^{1 / 2} 2 v \left(\right. t \left.\right) d t + \int_{1 / 2}^{1} 2 v \left(\right. t \left.\right) d t \\ & = \int_{0}^{1 / 2} 2 \left(\right. 20 t + 5 \left.\right) d t + \int_{1 / 2}^{1} 2 \left(\right. 15 \left.\right) d t \\ & = \int_{0}^{1 / 2} \left(\right. 40 t + 10 \left.\right) d t + \int_{1 / 2}^{1} 30 d t \\ & = \left[\right. 20 t^{2} + 10 t \left]\right. \left|\right._{0}^{1 / 2} + \left[\right. 30 t \left]\right. \left|\right._{1 / 2}^{1} \\ & = \left(\right. \frac{20}{4} + 5 \left.\right) - 0 + \left(\right. 30 - 15 \left.\right) \\ & = 25. \end{aligned}\]

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

Checkpoint 5.23

Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the wind starts to die down according to the function \(v \left(\right. t \left.\right) = −10 t + 15 .\) In other words, the wind speed is given by

\[\begin{aligned} v \left(\right. t \left.\right) = \left{\right. 20 t + 5 & \text{for} & 0 \leq t \leq \frac{1}{2} \\ - 10 t + 15 & \text{for} & \frac{1}{2} \leq t \leq 1 . \end{aligned}\]

Under these conditions, how far from his starting point is Andrew after 1 hour?

This lesson is part of:

Integration

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