Calculus Concepts and Modelling: An introduction to the concepts of calculus

Consider functions

Then

Notice that if we form the composition in the other order, we don’t get the same result:

Consider functions

Then composing in the order $g\circ f$ has the effect of horizontally transforming $g(t)\text{:}$

On the other hand, composing in the order $f\circ g$ has the effect of vertically transforming $g(t)\text{:}$

The function $h(t)=e^{t^2+1}$ is a composition of the natural exponential function with a polynomial. Writing $\exp (t)=e^t$ as usual, and also $f(t)=t^2+1\text{,}$ then

so we can say

Consider the function

Working from the outside inward, we can write

To verify that this is correct, we can compute:

which is exactly the formula for $h(t)\text{.}$

Suppose a reservoir is filled with water to a depth of 3 m. A period of steady, heavy rain begins, and it is observed that the depth of water is increasing by 50 ^cm⁄_h. Knowing that a constant variation must mean a linear quantity, we can create quantity model

for the depth of water in metres at time $t$ hours after it began raining.

With our model expressed as a function of time, we can directly answer questions like “What will the depth be 1 h after the rain began?” But often we want to answer questions like “If it keeps raining at this rate, how long until the reservoir overflows?” In other words, “How long must it rain at this rate for the water to reach a specific depth $D_0\text{?}$” To answer this type of question directly, we need a formula where $D$ is the input and $t$ is the output:

Consider functions

Let’s compute both compositions $f\circ g$ and $g\circ f\text{:}$

Consider functions

Let’s compute both compositions $f\circ g$ and $g\circ f\text{:}$

So the square and square root functions are not inverses of each other.

Consider the function

Let’s write $y$ as the output variable, so that $(t,y)$ is an input-output pair whenever

If $f(t)$ is invertible, then, exactly as in (✶), the inverse function should maintain the same relationship between input $t$ and output $y\text{,}$ just with their roles reversed:

In this form, the relationship between $t$ and $y$ is preserved, but with $t$ expressed as a formula in $y$ we can view $y$ as the input and $t$ as the output.

The reversed formula above tells us the formula for the inverse function:

If we consider an input-output formula as a sequence of calculation instructions (paying proper mind to BEDMAS), it makes sense that these two formulas are inverse: to compute an output for $f(t)\text{,}$ one must

whereas to compute an output for $f^{-1}(t)\text{,}$ one must

Notice that the two sequences of operations involve the opposite operations in the reverse order, which confirms that if we bothered to compute both compositions, we should find

as required by Definition 16.2.3.

If we try to apply the same procedure as in Example 16.2.10 to

we run into a problem: there are two different ways we could isolate $t\text{.}$ If we set

then certainly $y$ can’t be negative and we can take its square root. But there are two different numbers whose square is the right-hand side above:

Neither function

will universally satisfy

for all $t$ in the domain of $f\text{,}$ so we cannot say that either one is the inverse function of $f\text{.}$

Consider function $f(t)=t^4+2t^3-7t^2+t-5\text{.}$ The Fundamental Theorem of Algebra says that the equation

could have up to $4$ solutions for $t\text{,}$ depending on the value of $y\text{.}$

The range of the function $\sin (t)$ is $-1\le y\le 1\text{,}$ but there are output values in that range which have many corresponding input values. For example, $\sin (t)=1$ occurs at positive inputs

and also at negative inputs

This tells us that

We already knew that

and this now also confirms that

As we see in Figure 16.4.5.(a), the graph of $g(t)=\sqrt{t}$ is in fact the reflection of part of the graph of $f(t)=t^2\text{,}$ and the graph of $g(t)$ passes the Vertical Line Test because the “bottom half” of this sideways parabola is not actually part of the graph. This indicates how we should “fix” the square function so that the square root function is actually its inverse: just remove the “left” half of the upright parabola. In other words, suppose we create a new square function with a restricted domain that cuts off the portion of the graph that is causing the Horizontal Line Test to fail:

Then the new graph of this new function becomes one-to-one and hence the function is invertible. And, in fact, $h^{-1}(t)=\sqrt{t}\text{.}$

We have fixed also fixed the problem where

(as in Example 16.2.8). When we now compute $g(h(t))\text{,}$ we assume that $t$ is in the domain of $h\text{,}$ which has been restricted to $t\ge 0\text{.}$ In that case,

To compute a Derivative value like $q^{\prime }(2)$ from the definition, we should first approximate that instantaneous rate with an average rate calculation over a short time interval containing $2\text{.}$ So suppose we choose $t_1\le 2$ and $t_2\ge 2$ so that $\mathrm{\Delta }t=t_2-t_1$ is small but not zero. The corresponding average rate value

is the slope of the secant line between the points $(t_1,q_1)$ and $(t_2,q_2)\text{.}$

Notice that in Figure 16.5.2.(b), which is the graph of the cube root function, we have labelled the horizontal axis as $q$ and the vertical axis as $t\text{,}$ to match our idea that the inverse function interchanges the roles of input and output. So while $q_1\le 8\le q_2$ creates a small range around $q=8$ on the vertical axis for the cube graph, it also creates a small domain around $q=8$ on the horizontal axis for the cube root graph. This allows us to relate the derivative value $q^{\prime }(2)$ not to $(q^{-1}{)}^{\prime }(2)$ but to $(q^{-1}{)}^{\prime }(8)\text{.}$ On the second graph, the slope of the reflected secant line must be calculated as

The result is that an average rate for the inverse function is the reciprocal of a corresponding average rate for the original function. Since this is true about every average rate function calculation, we conclude that there should also be a reciprocal relationship for the derivatives of the two functions.

Let’s test out this conclusion. On the one hand, we know $q^{\prime }(t)=3t^2\text{,}$ so $q^{\prime }(2)=12\text{.}$ On the other hand, we can also use Pattern 12.4.3 to directly compute that

so

which is indeed the reciprocal of our computed value for $q^{\prime }(2)\text{.}$

With the knowledge of the Fundamental Theorem of Calculus, we might interpret our definition of the natural logarithm (Definition 8.1.1) as specifically designed so that

Viewing the natural exponential function as the inverse of the natural logarithm, we take $f=\ln$ in Pattern 16.5.3, so that $f^{-1}=\exp \text{.}$ Using ${\ln }^{\prime }(t)=1/t$ in the formula from that pattern, we have

as expected.