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In Section FunctionArithmetic, we saw how the arithmetic of real numbers carried over into an arithmetic of functions. In this section, we discuss
another way to combine functions which is unique to functions and isn’t shared with real numbers - function
composition.
Let \(f\) and \(g\) be functions where the real number \(x\) is in the domain of \(f\) and the real number \(f(x)\) is in the domain of \(g\). The composite
of \(g\) with \(f\), denoted \(g \circ f\), and read ‘\(g\) composed with \(f\)’ is defined by the formula: \((g \circ f) (x) = g(f(x))\).
To compute \((g \circ f)(x) \), we use the formula given in Defintion functioncompositiondefn: \((g \circ f) (x) = g(f(x))\). However, from a procedural viewpoint, Defintion functioncompositiondefn tells us the output from
\(g \circ f\) is found by taking the output from \(f\), \(f(x)\), and then making that the input to \(g\). From this perspective, we see \(g \circ f\) as a two
step process taking an input \(x\) and first applying the procedure \(f\) then applying the procedure \(g\). Abstractly, we
have:
In the expression \(g(f(x))\), the function \(f\) is often called the ‘inside’ function while \(g\) is often called the ‘outside’ function. When
evaluating composite function values we present two methods in the example below: the ‘inside out’ and ‘outside in’
methods.
Let \(f(x) = x^2-4x\), \(g(t) = 2-\sqrt {t+3}\), and \(h(s) = \frac {2s}{s+1}\).
In numbers fcexformfirst - fcexformlast, find and simplify the indicated composite functions. State the domain of each.
\((g \circ f)(x)\)
\((f \circ g)(t)\)
\((g \circ h)(s)\)
\((h \circ g)(t)\)
\((h \circ h)(x)\)
\((h \circ (g \circ f))(x)\)
\(((h \circ g) \circ f)(x)\)
Using Definition functioncompositiondefn, \((g \circ f)(1) = g(f(1))\). Since \(f(1) = (1)^2 - 4(1) = -3\) and \(g(-3) = 2 - \sqrt {(-3)+3} = 2\), we have \((g \circ f)(1) = g(f(1)) = g(-3) = 2\).
By definition, \((f \circ g)(1) = f(g(1))\). We find \(g(1) = 2 - \sqrt {1+3} = 0\), and \(f(0) = (0)^2-4(0) = 0\), so \((f \circ g)(1) = f(g(1)) = f(0) = 0\). Comparing this with our answer to the last problem, we see that \((g \circ f)(1) \neq (f \circ g)(1)\) which
tells us function composition is not commutative, (That is, in general, \(g \circ f \neq f \circ g\). This shouldn’t be too surprising,
since, in general, the order of processes matters: adding eggs to a cake batter then baking the cake batter has
a much different outcome than baking the cake batter then adding eggs.)
Since \((g \circ g)(6) = g(g(6))\), we ‘iterate’ the process \(g\): that is, we apply the process \(g\) to \(6\), then apply the process \(g\) again. We find \(g(6) = 2 - \sqrt {6+3} = -1\), and
\(g(-1) = 2 - \sqrt {(-1)+3} = 2 - \sqrt {2}\), so \((g \circ g)(6) = g(g(6)) = g(-1) = 2-\sqrt {2}\).
By definition, \((g \circ f)(x) = g(f(x))\). We now illustrate two ways to approach this problem.
inside out: We substitute \(f(x) = x^2-4x\) in for \(t\) in the expression \(g(t)\) and get
We get the same answer as before, \((g \circ f)(x) = 2 - \sqrt {x^2-4x+3}\).
To find the domain of \(g \circ f\), we need to find the elements in the domain of \(f\) whose outputs \(f(x)\) are in the domain of \(g\). Since the
domain of \(f\) is all real numbers, we focus on finding the range elements compatible with \(g\). Owing to the
presence of the square root in the formula \(g(t) = 2 - \sqrt {t+3}\) we require \(t \geq -3\). Hence, we need \(f(x) \geq -3\) or \(x^2-4x \geq -3\). To solve this inequality we
rewrite as \(x^2-4x + 3 \geq 0\) and use a sign diagram. Letting \(r(x) = x^2-4x+3\), we find the zeros of \(r\) to be \(x = 1\) and \(x = 3\) and obtain the sign diagram
below.
Our solution to \(x^2-4x+3 \geq 0\), and hence the domain of \(g \circ f\), is \((-\infty , 1] \cup [3,\infty )\).
To find \((f \circ g)(t)\), we find \(f(g(t))\).
inside out: We substitute the expression \(g(t) = 2 - \sqrt {t+3}\) in for \(x\) in the formula \(f(x)\) and get
Thus we get \((f \circ g)(t) = t-1\). To find the domain of \(f \circ g\), we look for the elements \(t\) in the domain of \(g\) whose outputs, \(g(t)\) are in the domain of \(f\).
As mentioned previously, the domain of \(g\) is limited by the presence of the square root to \(\{ t \in \mathbb {R} \, | \, t \geq -3\}\) while the domain of \(f\) is all real
numbers. Hence, the domain of \(f \circ g\) is restricted only by the domain of \(g\) and is \(\{ t \in \mathbb {R} \, | \, t \geq -3\}\) or, using interval notation, \([-3, \infty )\). Note that as with
Example funcarithex in Section FunctionArithmetic, had we used the simplified formula for \((f \circ g)(t) = t-1\) to determine domain, we would have arrived at the
incorrect answer.
To find \((g \circ h)(s)\), we compute \(g(h(s))\).
inside out: We substitute \(h(s)\) in for \(t\) in the expression \(g(t)\) to get
outside in: We use the formula for \(g(t)\) first to get
\((g \circ h)(s)\)
=
\(g(h(s))\) = \(2 - \sqrt {h(s)+3}\)
=
\(2 - \sqrt {\left (\frac {2s}{s+1}\right )+3}\)
=
\(2 - \sqrt {\frac {5s+3}{s+1}}\)
get common denominators as before
To find the domain of \(g \circ h\), we need the elements in the domain of \(h\) so that \(h(s)\) is in the domain of \(g\). Owing to the \(s+1\) in the
denominator of the expression \(h(s)\), we require \(s \neq -1\). Once again, because of the square root in \(g(t) = 2 - \sqrt {t+3}\), we need \(t \geq -3\) or, in this case \(h(s) \geq -3\). To
use a sign diagram to solve, we rearrange this inequality:
Defining \(r(s) = \frac {5s+3}{s+1}\), we see \(r\) is undefined at \(s=-1\) (a carry over from the domain restriction of \(h\)) and \(r(s) = 0\) at \(s = -\frac {3}{5}\). Our sign diagram
is:
To find the domain of \(h \circ g\), we need the elements of the domain of \(g\) so that \(g(t)\) is in the domain of \(h\). As we’ve seen already, for \(t\)
to be in the domain of \(g\), \(t \geq -3\). For \(s\) to be in the domain of \(h\), \(s \neq -1\), so we require \(g(t) \neq -1\). Hence, we solve \(g(t) = 2-\sqrt {t+3} = -1\) with the intent of excluding the
solutions. Isolating the radical expression gives \(\sqrt {t+3} = 3\) or \(t = 6\). Sure enough, we check \(g(6)=-1\) so we exclude \(t=6\) from the domain of \(h \circ g\). Our
final answer is \([-3, 6) \cup (6, \infty )\).
To find \((h \circ h)(s)\) we find \(h(h(s))\):
inside out: We substitute the expression \(h(s)\) for \(s\) in the expression \(h(s)\) into \(h\) to get
To find the domain of \(h \circ h\), we need to find the elements in the domain of \(h\) so that the outputs, \(h(s)\) are also in the domain of \(h\).
The only domain restriction for \(h\) comes from the denominator: \(s \neq -1\), so in addition to this, we also need \(h(s) \neq -1\). To this end, we
solve \(h(s) = -1\) and exclude the answers. Solving \(\frac {2s}{s+1} = -1\) gives \(s = -\frac {1}{3}\). The domain of \(h \circ h\) is \((-\infty , -1) \cup \left (-1, -\frac {1}{3}\right ) \cup \left (-\frac {1}{3}, \infty \right )\).
The expression \((h \circ (g \circ f))(x)\) indicates that we first find the composite, \(g \circ f\) and compose the function \(h\) with the result. We know from
number fcexformfirst that \((g \circ f)(x) = 2 - \sqrt {x^2-4x+3}\) with domain \((-\infty , 1] \cup [3,\infty )\). We now proceed as usual.
inside out: We substitute the expression \((g \circ f)(x)\) for \(s\) in the expression \(h(s)\) first to get
To find the domain of \(h \circ (g \circ f)\), we need the domain elements of \(g \circ f\), \((-\infty , 1] \cup [3,\infty )\), so that \((g \circ f)(x)\) is in the domain of \(h\). As we’ve seen several times
already, the only domain restriction for \(h\) is \(s \neq -1\), so we set \((g \circ f)(x) = 2 - \sqrt {x^2-4x+3} = -1\) and exclude the solutions. We get \(\sqrt {x^2-4x+3} = 3\), and, after squaring both
sides, we have \(x^2-4x+3 = 9\). We solve \(x^2-4x-6 = 0\) using the quadratic formula and obtain \(x = 2 \pm \sqrt {10}\). The reader is encouraged to check
that both of these numbers satisfy the original equation, \(2 - \sqrt {x^2-4x+3} = -1\) and also belong to the domain of \(g \circ f\), \((-\infty , 1] \cup [3,\infty )\), and so
must be excluded from our final answer. (We can approximate \(\sqrt {10} \approx 3\) so \(2-\sqrt {10} \approx -1\) and \(2+\sqrt {10} \approx 5\).) Our final domain for \(h \circ (f \circ g)\) is
\((-\infty , 2 -\sqrt {10}) \cup (2 - \sqrt {10}, 1] \cup \left [3, 2 + \sqrt {10}\right ) \cup \left (2+\sqrt {10}, \infty \right )\).
The expression \(((h \circ g) \circ f)(x)\) indicates that we first find the composite \(h \circ g\) and then compose that with \(f\). From number hogex, we have
Since the domain of \(f\) is all real numbers, the challenge here to find the domain of \((h \circ g) \circ f\) is to determine the
values \(f(x)\) which are in the domain of \(h \circ g\), \([-3, 6) \cup (6, \infty )\). At first glance, it appears as if we have two (or three!) inequalities to
solve: \(-3 \leq f(x) < 6\) and \(f(x) > 6\). Alternatively, we could solve \(f(x) = x^2-4x \geq -3\) and exclude the solutions to \(f(x) = x^2-4x = 6\) which is not only easier from a
procedural point of view, but also easier since we’ve already done both calculations. In number fcexformfirst, we
solved \(x^2-4x \geq -3\) and obtained the solution \((-\infty , 1] \cup [3, \infty )\) and in number hogoffirst, we solved \(x^2-4x-6 = 0\) and obtained \(x = 2 \pm \sqrt {10}\). Hence, the domain of \((h \circ g) \circ f\) is
\((-\infty , 2 -\sqrt {10}) \cup (2 - \sqrt {10}, 1] \cup \left [3, 2 + \sqrt {10}\right ) \cup \left (2+\sqrt {10}, \infty \right )\). □
As previously mentioned, it should be clear from Example functioncompex1 that, in general, \(g \circ f \neq f \circ g\), in other words, function composition is not
commutative. However, numbers hogoffirst and fcexformlast demonstrate the associative property of function composition. That is,
when composing three (or more) functions, as long as we keep the order the same, it doesn’t matter which two
functions we compose first. We summarize the important properties of function composition in the theorem
below.
Properties of Function Composition: Suppose \(f\), \(g\), and \(h\) are functions.
Associative Law of Composition:\(h \circ (g \circ f) = (h \circ g) \circ f\), provided the composite functions are defined.
Composition Identity: The function \(I(x) = x\) satisfies: \( I \circ f = f \circ I =f\) for all functions, \(f\).
By repeated applications of Definition functioncompositiondefn, we find \((h \circ (g \circ f))(x) = h((g \circ f)(x)) = h(g(f(x)))\). Similarly, \(((h \circ g) \circ f)(x) = (h \circ g)(f(x)) = h(g(f(x)))\). This establishes that the formulas for the two functions are the
same. We leave it to the reader to think about why the domains of these two functions are identical, too. These two facts
establish the equality \(h \circ (g \circ f) = (h \circ g) \circ f\). A consequence of the associativity of function composition is that there is no need for parentheses
when we write \(h \circ g \circ f\). The second property can also be verified using Definition functioncompositiondefn. Recall that the function \(I(x) = x\) is called the identity
function and was introduced in Exercise identityexercise in Section ConstantandLinearFunctions. If we compose the function \(I\) with a function \(f\), then we have \((I \circ f)(x) = I(f(x)) = f(x)\), and a similar
computation shows \((f\circ I)(x) = f(I(x)) = f(x)\). This establishes that we have an identity for function composition much in the same way the function \(I(x) = 1\) is
an identity for function multiplication.
As we know, not all functions are described by formulas, and, moreover, not all functions are described by just one formula.
The next example applies the concept of function composition to functions represented in various and sundry
ways.
Find and simplify a formula for \((g \circ f)(x)\).
Write \(s \circ h\) as a set of ordered pairs.
To find \((g \circ f)(2) = g(f(2))\) we first find \(f(2) = 6(2)-(2)^2 = 8\). Since \(8 \geq 3\), we use the rule \(g(t) = t^2\) so \(g(8) = (8)^2 = 64\). Hence, \((g \circ f)(3) = g(f(3)) = g(8) = 64\).
Since \((h\circ g)(-1) = h(g(-1))\) we first need \(g(-1)\). Since \(-1 \leq -1 < 3\), we use the rule \(g(t) = 2t-1\) and find \(g(-1) = 2(-1)-1 = -3\). Next, we need \(h(-3)\). Since \((-3, 1) \in h\), we have that \(h(-3) =1\). Putting it
all together, we find \((h\circ g)(-1) = h(g(-1)) = h(-3) = 1\).
To find \((h \circ s)(-2) = h(s(-2))\) , we first need \(s(-2)\). We see the point \((-2,3)\) is on the graph of \(s\), so \(s(-2) = 3\). Next, we see \((3,-1) \in h\), so \(h(3) = -1\). Hence, \((h \circ s)(-2) = h(s(-2)) = h(3) = -1\).
To find \((f \circ s)(0) = f(s(0))\) we infer from the graph of \(s\) that it contains the point \((0,3)\), so \(s(0) = 3\). Sine \(f(3) = 6(3) - (3)^2 = 9\), we have \((f \circ s)(0) = f(s(0)) = f(3) = 9\).
To find a formula for \((g \circ f)(x) = g(f(x))\), we substitute \(f(x) = 6x-x^2\) in for \(t\) in the formula for \(g(t)\):
Simplifying each expression, we get \( 2(6x-x^2) -1 = -2x^2+12x-1\) for the first piece and \((6x-x^2)^2 = x^4 - 12x^3 +36x^2\) for the second piece. The real challenge comes in solving
the inequalities \(-1 \leq 6x-x^2 < 3\) and \(6x-x^2 \geq 3\). While we could solve each individually using a sign diagram, a graphical approach works best
here. We graph the parabola \(y = 6x-x^2\), finding the vertex is \((3, 9)\) with intercepts \((0,0)\) and \((6,0)\) along with the horizontal lines \(y = -1\) and \(y=3\) below. We
determine the intersection points by solving \(6x-x^2=-1\) and \(6x-x^2=3\). Using the quadratic formula, we find the solutions to each equation
are \(x = 3 \pm \sqrt {10}\) and \(x = 3 \pm \sqrt {6}\), respectively.
From the graph, we see the parabola \(y = 6x - x^2\) is between the lines \(y = -1\) and \(y=3\) from \(x = 3- \sqrt {10}\) to \(x= 3 - \sqrt {6}\) and again from \(x = 3+\sqrt {6}\) to \(x=3+\sqrt {10}\). Hence
the solution to \(-1 \leq 6x-x^2 < 3\) is \([3- \sqrt {10}, 3 - \sqrt {6}) \cup (3+\sqrt {6}, 3+\sqrt {10}]\). We also note \(y = 6x-x^2\) is above the line \(y=3\) for all \(x\) between \(x=3-\sqrt {6}\) and \(3+\sqrt {6}\). Hence, the solution to \(6x-x^2 \geq 3\) is \([3-\sqrt {6}, 3+\sqrt {6}]\).
Hence,
Last but not least, we are tasked with representing \(s \circ h\) as a set of ordered pairs. Since \(h\) is described by the discrete set of
points \(h = \{ (-3,1), (-2,6), (0,-2), (1,5), (3,-1) \}\), we find \(s \circ h\) point by point. We keep the graph of \(s\) handy and construct the table below to help us organize our
work.
Since neither \(6\) nor \(5\) are in the domain of \(s\), \(-2\) and \(1\) are not in the domain of \(s \circ h\). Hence, we get \(s \circ h = \{ (-3,3), (0,3), (3,3) \}\).
A useful skill in Calculus is to be able to take a complicated function and break it down into a composition of
easier functions which our last example illustrates. As with Example funcarithdecompex, we want to avoid trivial decompositions,
which, when it comes to function composition, are those involving the identity function \(I(x) = x\) as described in Theorem
functioncompprops.
\(~\)
Write each of the following functions as a composition of two or more (non-identity) functions. Check your
answer by performing the function composition.
\(F(x) = |3x-1|\)
\(G(t) = \frac {2}{t^2+1}\)
\(H(s) = \frac {\sqrt {s}+1}{\sqrt {s}-1}\)
For \(F(x) = \sqrt {\frac {2x-1}{x^2+4}}\), find functions \(f\), \(g\), and \(h\) to decompose \(F\) nontrivially as \(F = f \circ \left (\frac {g}{h} \right )\).
There are many approaches to this kind of problem, and we showcase a different methodology in each of the solutions
below.
Our goal is to express the function \(F\) as \(F = g \circ f\) for functions \(g\) and \(f\). From Definition functioncompositiondefn, we know \(F(x) = g(f(x))\), and we can think
of \(f(x)\) as being the ‘inside’ function and \(g\) as being the ‘outside’ function. Looking at \(F(x) = |3x-1|\) from an ‘inside versus
outside’ perspective, we can think of \(3x-1\) being inside the absolute value symbols. Taking this cue, we define
\(f(x) = 3x-1\). At this point, we have \(F(x) = |f(x)|\). What is the outside function? The function which takes the absolute value of its
input, \(g(x) = |x|\). Sure enough, this checks: \((g \circ f)(x) = g(f(x)) = |f(x)| = |3x-1| = F(x)\).
We attack deconstructing \(G\) from an operational approach. Given an input \(t\), the first step is to square \(t\),
then add \(1\), then divide the result into \(2\). We will assign each of these steps a function so as to write \(G\) as a
composite of three functions: \(f\), \(g\) and \(h\). Our first function, \(f\), is the function that squares its input, \(f(t) = t^2\). The next
function is the function that adds \(1\) to its input, \(g(t) = t+1\). Our last function takes its input and divides it into \(2\), \(h(t) = \frac {2}{t}\). The
claim is that \(G = h \circ g \circ f\) which checks:
If we look \(H(s) = \frac {\sqrt {s}+1}{\sqrt {s}-1}\) with an eye towards building a complicated function from simpler functions, we see the
expression \(\sqrt {s}\) is a simple piece of the larger function. If we define \(f(s) = \sqrt {s}\), we have \(H(s) = \frac {f(s)+1}{f(s)-1}\). If we want to decompose \(H = g \circ f\),
then we can glean the formula for \(g(s)\) by looking at what is being done to \(f(s)\). We take \(g(s) = \frac {s+1}{s-1}\), and check below:
Working from the inside out, we have a rational expression with numerator \(g(x)\) and denominator \(h(x)\). Looking at the formula for
\(F(x)\), one choice is \(g(x) = 2x-1\) and \(h(x) = x^2+4\). Making these identifications, we have
Since \(F\) takes the square root of \(\frac {g(x)}{h(x)}\), the our last function \(f\) is the function that takes the square root of its input, i.e., \(f(x) = \sqrt {x}\). We
leave it to the reader to check that, indeed, \(F = f \circ \left (\frac {g}{h} \right )\).
We close this section of a real-world application of function composition.
The surface area of a sphere is a function of its radius \(r\) and is given by the formula \(S(r) = 4 \pi r^2\). Suppose a spherical balloon is inflated
so that the radius of the sphere is increasing according to the formula \(r(t) = 2t\), where \(t\) is measured in minutes (min), \(t \geq 0\), and \(r\) is
measured in centimeters (cm). Find and interpret \((S \circ r)(t)\).
The function \(S(r)\) gives the surface area of the sphere and \(r(t)\) gives the radius of the sphere at a given time. Given a specific time, \(t\),
we find the radius at that time, \(r(t)\) and feed that into \(S(r)\) to find the surface area. Hence, the surface area \(S\) is ultimately a function of
time \(t\) and we find \((S \circ r)(t) = S(r(t)) = 4 \pi (r(t))^2 = 4 \pi \left (2t\right )^2 = 16 \pi t^{2}\). This formula allows us to compute the surface area directly given the time without going through the
‘intermediary variable’ \(r\).
1 Related Rates
In Section ??, we studied the difference operator, \(\Delta \) and showed how average rates of change operate with the basic function
arithmetic. In this section, we explore how rates of change of composite functions are related to the rates of change of their
constituent functions. As in that section, we’ll use the formulation:
adjusting the names of functions and independent variables as needed.
As a motivational example, we revisit the scenario in Example 4.
The surface area of a sphere is a function of its radius \(r\) and is given by the formula \(S(r) = 4 \pi r^2\). Suppose a spherical balloon is inflated
so that the radius of the sphere is increasing according to the formula \(r(t) = 2t\), where \(t\) is measured in minutes (min), \(t \geq 0\), and \(r\) is
measured in centimeters (cm).
Find and simplify an expression for the average rate of change of \(S\) with respect to \(r\). Find and interpret the
average rate of change of \(S\) with respect to \(r\) over the interval \([1, 3]\).
Find, simplify, and interpret an expression for the average rate of change of \(r\) with respect to \(t\).
Find and simplify an expression for the average rate of change of \(S\) with respect to \(t\). Find and interpret the
average rate of change of \(S\) with respect to \(t\) over the interval \(\left [\frac {1}{2}, \frac {3}{2}\right ]\).
Multiply your answers to 1 and 2 and compare those to your answer in 3.
It is important to note that as we work through the expressions below, the variables \(r\) and \(\Delta r\) as well as \(t\) and \(\Delta t\) are distinct. That is,
they do not combine as ‘like terms.’
We start by simplifying \(\frac {\Delta [S(r)]}{\Delta r} = \frac {S(r+ \Delta r) - S(r)}{\Delta r}\):
This means as the radius of the balloon increases from \(1\) centimeter to \(3\) centimeters, the surface area is increasing at an
average rate of \(16 \pi \, \frac {\text {cm}^2}{\text {cm}} \).
Note that the units here, cm, do cancel and we could write the average rate of change as \(16 \pi \) cm. This somewhat hides the
fact this number represents a ratio. Any time area and length are measured in compatible units, the ratio of units \(\frac {\text {area}}{\text {length}}\) will
simplify to units of length. (As always, context is key!)
The fact that the average rate of change here is constant shouldn’t be too surprising. Note \(r(t) = 2t\) is a linear function with
slope \(2\). Hence, ‘\(2\)’ is the (constant) rate of change of \(r\). (We could probably have lead with that and avoided some
tedious computations …) This means that the radius of the balloon is increasing at a constant rate of \(2\)
\(\frac {\text {cm}}{\text {min}}\).
To find \(\frac {\Delta [S(t)]}{\Delta t} = \frac {S(t+ \Delta t) - S(t)}{\Delta t}\), we start with our answer from Example 4: \(S(t) = 16 \pi t^{2}\) :
To find the average rate of change of \(S\) over the interval \(\left [\frac {1}{2}, \frac {3}{2}\right ]\), we take \(t = \frac {1}{2}\) and \(\Delta t = \frac {3}{2} - \frac {1}{2} = 1\):
This means the surface area of the balloon is increasing at an average rate of \(32 \pi \) \(\frac {\text {cm}^2}{\text {min}}\) over the time span of \(\frac {1}{2}\) minute (\(30\) seconds)
after the start of inflation to \(\frac {3}{2}\) (\(90\) seconds) after the start of inflation.
Moreover, we note that the time interval \(\frac {1}{2} \leq t \leq \frac {3}{2}\) corresponds to the interval \(1 \leq r \leq 3\) so it makes sense to multiply our numerical
answers as well:
Example 5 verifies a property of rates we formalize below.
Related Rates: Let \(f\) and \(u\) be functions where \(u\) is defined over an interval containing \(x\) and \(x+\Delta x\) and \(f\) is defined over an interval
containing \(u(x)\) and \(u(x + \Delta x)\). Then:
If we think of \(u\) as being an ‘intermediary’ variable, Theorem 2 allows us to determine the rate of change of \(f\) with respect to \(x\) by
multiplying the rate of change of \(f\) with respect to this ‘intermediary’ \(u\) by the rate of change of the ‘intermediary’ \(u\) with respect to \(x\).
That is, we are looking for rate information on how \(f\) depends on \(x\) by decomposing the rate into two rates as visualized
below.
We close the section with one last example.
The drag force \(F\), in Newtons (N), of a perfectly fine OER Precalculus Textbook which has been discarded off of a cliff because
it didn’t have enough Calculus in it is given by: \(F(v) = 0.6 \, v^2\), where \(v\) is the speed of the book as it hurtles towards the Earth. If the speed is
increasing at a constant rate of \(9.8\) meters per second per second, \(\frac {\text {m/s}}{\text {s}}\), determine the rate of change of \(F\) with respect to time as the
speed changes from \(5\) \(\frac {\text {m}}{\text {s}}\) to \(6\) \(\frac {\text {m}}{\text {s}}\)
In this scenario, the drag force, \(F\) depends directly on the speed, \(v\) and the speed, \(v\) depends directly on time. (The longer the
book falls, the faster it falls. (Well until it reaches terminal velocity …) ) By Theorem 2, we know
We are told that the
speed is increasing at a constant rate of \(9.8\) meters per second per second, so we know \(\frac {\Delta v}{\Delta t} = 9.8 \, \frac {\text {m/s}}{\text {s}}\) for all time (and hence,
speeds.).
To find \(\frac {\Delta [F(v)]}{\Delta v}\) as the speed changes from \(5\) \(\frac {\text {m}}{\text {s}}\) to \(6\) \(\frac {\text {m}}{\text {s}}\), we calculate:
The units on \(\frac {\Delta [F(v)]}{\Delta v}\) would be the units of \(F\), N, divided by the units of \(v\), \(\frac {\text {m}}{\text {s}}\) which works out (Note: we could simplify this a bit farther.
A Newton, N, has units \(\frac {\text {kg m}}{\text {s}^2}\), so some units could cancel to give \(\frac {\text {kg}}{\text {s}}\). We leave things as they are for now for a more simple
calculation later.) to \(\frac {\text {N s}}{\text {m}}\).
The force is increasing at an average rate of \(64.68\) Newtons per second.
Note that we never needed to know explicitly how the speed, \(v\), directly depended on time in order to answer the question
posed in Example 6. All we needed was the rate.
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