Ulearngo

Velocities and Rates of Change

Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if \(s \left(\right. t \left.\right)\) is the position of an object moving along a coordinate axis, the average velocity of the object over a time interval \(\left[\right. a , t \left]\right.\) if \(t > a\) or \(\left[\right. t , a \left]\right.\) if \(t < a\) is given by the difference quotient

\[v_{\text{ave}} = \frac{s \left(\right. t \left.\right) - s \left(\right. a \left.\right)}{t - a} .\]

As the values of \(t\) approach \(a ,\) the values of \(v_{\text{ave}}\) approach the value we call the instantaneous velocity at \(a .\) That is, instantaneous velocity at \(a ,\) denoted \(v \left(\right. a \left.\right) ,\) is given by

\[v \left(\right. a \left.\right) = s^{'} \left(\right. a \left.\right) = \underset{t \rightarrow a}{\text{lim}} \frac{s \left(\right. t \left.\right) - s \left(\right. a \left.\right)}{t - a} .\]

To better understand the relationship between average velocity and instantaneous velocity, see Figure 3.8. In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time \(t = a\) whose position at time \(t\) is given by the function \(s \left(\right. t \left.\right) .\) The slope of the secant line (shown in green) is the average velocity of the object over the time interval \(\left[\right. a , t \left]\right. .\)

This figure consists of the Cartesian coordinate plane with 0, a, and t1 marked on the t-axis. The function y = s(t) is graphed in the first quadrant along with two lines marked tangent and secant. The tangent line touches y = s(t) at only one point, (a, s(a)). The secant line touches y = s(t) at two points: (a, s(a)) and (t1, s(t1)).
Figure 3.8 The slope of the secant line is the average velocity over the interval \(\left[\right. a , t \left]\right. .\) The slope of the tangent line is the instantaneous velocity.

We can use Equation 3.5 to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using Equation 3.7.

Example 3.7

Estimating Velocity

A lead weight on a spring is oscillating up and down. Its position at time \(t\) with respect to a fixed horizontal line is given by \(s \left(\right. t \left.\right) = \text{sin} t\) (Figure 3.9). Use a table of values to estimate \(v \left(\right. 0 \left.\right) .\) Check the estimate by using Equation 3.5.

A picture of a spring hanging down with a weight at the end. There is a horizontal dashed line marked 0 a little bit above the weight.
Figure 3.9 A lead weight suspended from a spring in vertical oscillatory motion.

Solution

We can estimate the instantaneous velocity at \(t = 0\) by computing a table of average velocities using values of \(t\) approaching \(0 ,\) as shown in Table 3.1.

\(t\) \(\frac{\text{sin} t - \text{sin} 0}{t - 0} = \frac{\text{sin} t}{t}\)
\(−0.1\) \(0.998334166\)
\(−0.01\) \(0.9999833333\)
\(−0.001\) \(0.999999833\)
\(0.001\) \(0.999999833\)
\(0.01\) \(0.9999833333\)
\(0.1\) \(0.998334166\)
Table 3.1 Average velocities using values of t approaching 0

From the table we see that the average velocity over the time interval \(\left[\right. −0.1 , 0 \left]\right.\) is \(0.998334166 ,\) the average velocity over the time interval \(\left[\right. −0.01 , 0 \left]\right.\) is \(0.9999833333 ,\) and so forth. Using this table of values, it appears that a good estimate is \(v \left(\right. 0 \left.\right) = 1 .\)

By using Equation 3.5, we can see that

\[v \left(\right. 0 \left.\right) = s^{'} \left(\right. 0 \left.\right) = \underset{t \rightarrow 0}{\text{lim}} \frac{\text{sin} t - \text{sin} 0}{t - 0} = \underset{t \rightarrow 0}{\text{lim}} \frac{\text{sin} t}{t} = 1 .\]

Thus, in fact, \(v \left(\right. 0 \left.\right) = 1 .\)

Checkpoint 3.4

A rock is dropped from a height of \(64\) feet. Its height above ground at time \(t\) seconds later is given by \(s \left(\right. t \left.\right) = −16 t^{2} + 64 , 0 \leq t \leq 2 .\) Find its instantaneous velocity \(1\) second after it is dropped, using Equation 3.5.

As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function.

Definition

The instantaneous rate of change of a function \(f \left(\right. x \left.\right)\) at a value \(a\) is its derivative \(f^{'} \left(\right. a \left.\right) .\)

Example 3.8

Chapter Opener: Estimating Rate of Change of Velocity

The same sports car speeding along a winding road from the beginning of the chapter.
Figure 3.10 (credit: modification of work by Codex41, Flickr)

Reaching a top speed of \(270.49\) mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from \(0\) to \(60\) mph in \(3.05\) seconds, from \(0 \text{to} 100\) mph in \(5.88\) seconds, from \(0 \text{to} 200\) mph in \(14.51\) seconds, and from \(0 \text{to} 229.9\) mph in \(19.96\) seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration) as it approaches \(229.9\) mph. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?

Solution

First observe that \(60\) mph = \(88\) ft/s, \(100\) mph \(\approx 146.67\) ft/s, \(200\) mph \(\approx 293.33\) ft/s, and \(229.9\) mph \(\approx 337.19\) ft/s. We can summarize the information in a table.

\(t\) \(v \left(\right. t \left.\right)\)
\(0\) \(0\)
\(3.05\) \(88\)
\(5.88\) \(146.67\)
\(14.51\) \(293.33\)
\(19.96\) \(337.19\)
Table 3.2 \(v \left(\right. t \left.\right)\) at different values of t

Now compute the average acceleration of the car in feet per second per second on intervals of the form \(\left[\right. t , 19.96 \left]\right.\) as \(t\) approaches \(19.96 ,\) as shown in the following table.

\(t\) \(\frac{v \left(\right. t \left.\right) - v \left(\right. 19.96 \left.\right)}{t - 19.96} = \frac{v \left(\right. t \left.\right) - 337.19}{t - 19.96}\)
\(0.0\) \(16.89\)
\(3.05\) \(14.74\)
\(5.88\) \(13.53\)
\(14.51\) \(8.05\)
Table 3.3 Average acceleration

The rate at which the car is accelerating is decreasing as its velocity approaches \(229.9\) mph \(( 337.19\) ft/s).

Example 3.9

Rate of Change of Temperature

A homeowner sets the thermostat so that the temperature in the house begins to drop from \(70 ° \text{F}\) at \(9\) p.m., reaches a low of \(60 °\) during the night, and rises back to \(70 °\) by \(7\) a.m. the next morning. Suppose that the temperature in the house is given by \(T \left(\right. t \left.\right) = 0.4 t^{2} - 4 t + 70\) for \(0 \leq t \leq 10 ,\) where \(t\) is the number of hours past \(9\) p.m. Find the instantaneous rate of change of the temperature at midnight.

Solution

Since midnight is \(3\) hours past \(9\) p.m., we want to compute \(T^{'} \left(\right. 3 \left.\right) .\) Refer to Equation 3.5.

\[T^{'} \left(\right. 3 \left.\right) & = \underset{t \rightarrow 3}{\text{lim}} \frac{T \left(\right. t \left.\right) - T \left(\right. 3 \left.\right)}{t - 3} & & & \text{Apply the definition}. \\ & = \underset{t \rightarrow 3}{\text{lim}} \frac{0.4 t^{2} - 4 t + 70 - 61.6}{t - 3} & & & \begin{matrix}\text{Substitute} T \left(\right. t \left.\right) = 0.4 t^{2} - 4 t + 70 \text{and} \\ T \left(\right. 3 \left.\right) = 61.6 .\end{matrix} \\ & = \underset{t \rightarrow 3}{\text{lim}} \frac{0.4 t^{2} - 4 t + 8.4}{t - 3} & & & \text{Simplify}. \\ & = \underset{t \rightarrow 3}{\text{lim}} \frac{0.4 \left(\right. t - 3 \left.\right) \left(\right. t - 7 \left.\right)}{t - 3} & & & = \underset{t \rightarrow 3}{\text{lim}} \frac{0.4 \left(\right. t - 3 \left.\right) \left(\right. t - 7 \left.\right)}{t - 3} \\ & = \underset{t \rightarrow 3}{\text{lim}} 0.4 \left(\right. t - 7 \left.\right) & & & \text{Cancel}. \\ & = −1.6 & & & \text{Evaluate the limit}.\]

The instantaneous rate of change of the temperature at midnight is \(−1.6 ° \text{F}\) per hour.

Example 3.10

Rate of Change of Profit

A toy company can sell \(x\) electronic gaming systems at a price of \(p = −0.01 x + 400\) dollars per gaming system. The cost of manufacturing \(x\) systems is given by \(C \left(\right. x \left.\right) = 100 x + 10,000\) dollars. Find the rate of change of profit when \(10,000\) games are produced. Should the toy company increase or decrease production?

Solution

The profit \(P \left(\right. x \left.\right)\) earned by producing \(x\) gaming systems is \(R \left(\right. x \left.\right) - C \left(\right. x \left.\right) ,\) where \(R \left(\right. x \left.\right)\) is the revenue obtained from the sale of \(x\) games. Since the company can sell \(x\) games at \(p = −0.01 x + 400\) per game,

\[R \left(\right. x \left.\right) = x p = x \left(\right. −0.01 x + 400 \left.\right) = −0.01 x^{2} + 400 x .\]

Consequently,

\[P \left(\right. x \left.\right) = −0.01 x^{2} + 300 x - 10,000 .\]

Therefore, evaluating the rate of change of profit gives

\[\begin{aligned} P^{'} \left(\right. 10000 \left.\right) & = \underset{x \rightarrow 10000}{\text{lim}} \frac{P \left(\right. x \left.\right) - P \left(\right. 10000 \left.\right)}{x - 10000} \\ & = \underset{x \rightarrow 10000}{\text{lim}} \frac{−0.01 x^{2} + 300 x - 10000 - 1990000}{x - 10000} \\ & = \underset{x \rightarrow 10000}{\text{lim}} \frac{−0.01 x^{2} + 300 x - 2000000}{x - 10000} \\ & = 100 . \end{aligned}\]

Since the rate of change of profit \(P^{'} \left(\right. 10,000 \left.\right) > 0\) and \(P \left(\right. 10,000 \left.\right) > 0 ,\) the company should increase production.

Checkpoint 3.5

A coffee shop determines that the daily profit on scones obtained by charging \(s\) dollars per scone is \(P \left(\right. s \left.\right) = −20 s^{2} + 150 s - 10 .\) The coffee shop currently charges \(\$ 3.25\) per scone. Find \(P^{'} \left(\right. 3.25 \left.\right) ,\) the rate of change of profit when the price is \(\$ 3.25\) and decide whether or not the coffee shop should consider raising or lowering its prices on scones.

This lesson is part of:

Derivatives

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