Use the slope formula to calculate the average velocity of te ball from \(1.5s\) to \(2s\text{.}\)
Section 3.5 Rates of Change
Subsection 3.5.1 Introduction
Did you know crows give gifts? Some people have found that if you routinely feed local crows or ravens, they will eventually return the favor by bringing trinkets. People have reported that their crow friends have brought bottle caps, sticks, feathers, and miscelaneous shiny objects.
Imagine your friend, Lucas, has a crow friend, and his crow brings him two small gifts a day. So the shoebox where Lucas keeps these gifts grows by two items a day. We can chart it like below.
We see here an example of linear data. That is, if we chart this information, we can see a straight line with equation \(y = mx + b\text{,}\) where \(b\) is our initial amount (or \(y\)-intercept) and \(m\) is our slope.
What does slope mean exactly? The graph in Figure 3.5.1 is \(y = 2x\text{,}\) and so the slope is \(2\text{.}\) But what does that mean? The slope refers to how quickly our line is changing... a rate of change. In this case, \(m = 2\) gifts per day.
This is exactly why the slope formula is
\begin{equation*}
m = \dfrac{y_2-y_1}{x_2-x_1} \text{ with points } (x_1,y_1) \text{ and } (x_2,y_2).
\end{equation*}
The rate of change is measured by "change in \(y\) over change in \(x\)".
Subsection 3.5.2 Throwing a Ball
A lot of situations aren’t linear, though, right? For example, if our \(6ft\) tall friend throws a ball straight up at \(30\)mph (or \(44\) feet per second), thanks to gravity which accelerates over time, after \(t\) seconds, the ball is at height \(s = -16t^2 + 44t + 6\) feet.
This obviously has a rate of change, but our slope formula doesn’t seem to make sense anymore. So how do we find how quickly a ball is moving in the air at a given time? Unlike the crow-friend graph earlier, we cannot talk about slope... or maybe we should consider the slope to be variable. So when we talk about velocity, we have to talk about velocity at a given time, since velocity is not constant.
Suppose we want to know how fast the ball is travelling after \(2s\text{.}\) This is a difficult question at first, but maybe we can take an average velocity over a short time interval and use this average velocity as an estimate for the velocity at exactly \(2s\text{.}\) For example, let’s find the average velocity from \(1.5s\) to \(2s\text{.}\) In other words, we look for the slope of the line passing through the points \((1.5, s(1.5))\) and \((2, s(2))\text{.}\) We call such a line a secant line.
Question 3.5.3. Our First Average.
f(x) = -16*x^2+44*x+6
sl(x) = (f(2)-f(x))/(2-x)
g2 = plot(f, (0, 3),
color='blue', linestyle='-',
aspect_ratio=.06);
g2 += plot(sl(1)*(x-1)+f(1), (0, 3),
color='red', linestyle='-',
aspect_ratio=.06); g2
We can see the graph of the function modelling the ball and the secant line in Figure 3.5.4. The idea is that the slope of the secant line gives us an estimate for the rate the ball is travelling at \(2s\text{.}\) Is this a good estimate?
Probably not.
Question 3.5.5.
How could we get a better estimate?
Solution.
Maybe we should find secant line slopes that pass through points closer to \(2s\text{,}\) like the secant line corresponding to times \(1.9s\) and \(2s\text{.}\)
If we take an average over a shorter distance, we can get a more accurate measurement for the instantaneous rate of change at \(2s\text{.}\) The Geogebra applet below shows the trend of slopes of the secant lines as our time interval gets smaller.
Question 3.5.7. .
Use the Geogebra applet to estimate how quickly the ball is traveling at \(2s\text{.}\)
Solution.
It appears to be trending towards \(20 ft/s\text{.}\)
Notice that the secant line disappears when the time interval is \(0s\text{.}\) This is part of the problem, right? We can get a trend of slopes when the time interval is close to \(0\text{,}\) but the slope formula is undefined once we have a time interval length \(0\text{.}\)
There is, however, a special line we call the tangent line that intersects our graph right at \(2\) and whose slope \(m\) is the number that the secant slopes are trending towards. The tangent line’s slope gives us the instantaneous rate of change that we are looking for. Wouldn’t it be nice to have a good way of finding the tangent line’s slope?
PIC OF GRAPH WITH TANGENT LINE.
Subsection 3.5.3 Derivatives and Instantaneous Rate of Change
Remember from Section 3.1 that when we talk about trends, we’re actually talking about limits. In this case, we’re looking at slopes of secant lines as the time interval trends towards 0. This limit gives us the instantaneous rate of change of a function at a given time.
Mathematically, with help of the slope formula, we could say the instantaneous rate of change is
\begin{equation*}
\lim_{x \rightarrow a} \dfrac{s(x)-s(a)}{x-a},
\end{equation*}
and amazingly the simple slope formula we learned in algebra becomes the tool to finding instantaneous rate of change!
This limit is called the derivative, and it is the gateway to understanding movement and change in mathematical terms. Your first semester of Calculus will focus heavily on the derivative, including nice formulas for calculating derivatives and a variety of derivative applications.
Typically, for function \(f(x)\text{,}\) we denote the derivative as \(f'(x)\text{,}\) and with this new definition, we see that the tangent line at \(x = a\) has equation
\begin{equation*}
y = f'(a)(x-a) + f(a)
\end{equation*}
for any function. With the derivative in our arsenal, we have a more sophisticated method to finding a rate of change than using secant line.
Subsection 3.5.4 Applications of Derivatives
Subsubsection 3.5.4.1 Introduction
You encountere derivatives everywhere you go.... everytime you hit your car breaks... everytime you take medicine... and everytime you launch a rocket into space. The derivative is an extremely powerful tool. Let’s look at a few applications of derivatives.
Question 3.5.8. Applications of Derivatives.
With a group of classmates, find four applications of derivatives either through a textbook or the internet.
Subsubsection 3.5.4.2 More Rates of Change
We were not kidding around when we said that derivates are all around us. Any rate of change can be classified as a derivative. We can find examples from biology, chemistry, population dynamics, ecology, economy, and astronomy (among many others).
Let’s look at an example.
Question 3.5.9.
Suppose you have a tank holds \(1000\) gallons of water, which drains from the bottom in \(60\)min. By Torricelli’s Law , the volume \(V\) of water remaining in the tank after \(t\) minutes is
\begin{equation*}
V = 1000\left( 1 - \dfrac{1}{60}t\right)^2, \; 0 \leq t \leq 60.
\end{equation*}
What does the derivative \(V'\) model?
Solution.
Since \(V\) gives us the volume of water at time \(t\text{,}\) \(V'\) gives us the rate of change in the water volume at time \(t\text{.}\) That is, \(V'\) tells us how quickly the water is draining from the tank at time \(t\text{.}\)
Question 3.5.10.
Suppose that
\begin{equation*}
V' = -\dfrac{100}{3}\left(1-\dfrac{1}{60}t\right)\text{,}
\end{equation*}
where \(V\) is the volume of water remaining in Question 3.5.9.
How quickly is the water draining after \(5s\text{.}\)
Subsubsection 3.5.4.3 Highest and Lowest
If the derivative of \(f(x)\) is \(0\) at the value \(x = a\text{,}\) the instaneous rate of change at \(x = a\) is \(0\) as well. That means the tangent line at \(x = a\) has slope \(0\) and thus is horizontal. What does this mean? When does a function have no change? Or a horizontal tangent line? Many times, these are the places where \(f(x)\) has either reached a high point or a low point. For this reason, derivatives often help us find the highs and lows.
Suppose we find \(1600ft\) of fencing at a local flea market (for a price we can’t turn down), and we are wanting to build an enclosed field with it. We have the perfect place along a local creek that will serve as one border of our field. The question calculus can help with is "How do we build a field with the most area possible?" Should this field be long and then? Should it be square shaped? Somewhere in between? Let’s figure that out.
ADD GEOGEBRA ANIMATION OF A FIELD?
We can label the rectangular sides as \(x\) and \(y\) (with \(being the long side\)), but we need to also setup equations for area and perimeter.
Question 3.5.11.
What’s the area \(A\) of the field in terms of \(x\) and \(y\text{?}\)
Solution.
\(A = xy\)
Question 3.5.12.
What’s the perimeter \(p\) in terms of \(x\) and \(y\text{?}\) Don’t forget that the total amount of fence is \(1600ft\text{.}\)
Solution.
Since one side of the rectangle is the creek, we have perimeter \(p = 2x + y\text{,}\) and so in total we have \(1600 = 2x + y\) as a representation of perimeter.
We can use our perimeter equation to express \(A\) as a one variable function. From there, the equation \(A' = 0\) will help us pinpoint exactly what dimension field has the maximum area.
Question 3.5.13.
Express \(A\) as a function of \(x\text{.}\)
Solution.
We solve \(1600 = 2x + y\) for \(y\) to get \(y = 1600 - 2x,\) and so \(A = x(1600-2x) = 1600x-2x^2.\)
So now we know the area of a our field is \(A(x) = 1600x - 2x^2\text{.}\) After some simple derivative rules, we obtain \(A'(x) = 1600 - 4x\text{.}\)
Question 3.5.14.
Solve the equation \(A' = 0\text{.}\)
Solution.
We solve \(1600 - 4x = 0\) to get \(x = 400\text{.}\)
So \(x = 400\) is a potential maximum or minimum of \(A(x)\text{,}\) and since \(A(x)\) is a downward opening parabola, it makes sense that \(x = 400\) is in fact the maximum. Putting it all together, the largest field we could possibly build has two sides length \(x =400ft\) and one more side length \(y = 800ft\text{.}\) The area of this field is \(A = 400(800) = 320,000ft^2\text{.}\)
You can imagine there a lot of valuable applications similar to this.
