CCM Product rule and integrating factors

$\begin{aligned} \frac{d}{d t} (t \ln | t |) & = t \cdot \frac{d}{d t} (\ln | t |) + \frac{d t}{d t} \cdot \ln | t | \\ = t \cdot \frac{1}{t} + 1 \cdot \ln | t | \\ = 1 + \ln | t | \end{aligned}$
$\begin{aligned} \frac{d}{d t} (e^{t} \tan (t)) & = e^{t} \cdot \frac{d}{d t} (\tan (t)) + \frac{d e^{t}}{d t} \cdot \tan (t) \\ = e^{t} \sec^{2} (t) + e^{t} \cdot \tan (t) \\ = e^{t} (\sec^{2} (t) + \tan (t)) \end{aligned}$
For

$\frac{d}{d t} (t \sin (t^{2})),$

set $u = t^{2} .$ Then

$\begin{aligned} \frac{d}{d t} (t \sin (t^{2})) & = t \cdot \frac{d}{d t} (\sin (t^{2})) + \frac{d t}{d t} \cdot \sin (t^{2}) \\ = t \cdot \frac{d (\sin (u))}{d u} \cdot \frac{d u}{d t} + 1 \cdot \sin (t^{2}) \\ = t \cos (u) \cdot (2 t) + 1 \cdot \sin (t^{2}) \\ = 2 t^{2} \cos (t^{2}) + \sin (t^{2}) . \end{aligned}$

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The first calculation in Example 17.1.3 actually tells us something about the antiderivatives of the logarithm.

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Pattern 17.1.4. Integral of the natural logarithm.

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We have

\int_{}^{} \ln | t | d_{} t = t \ln | t | - t + C .

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Justification.

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Turning the first calculation in Example 17.1.3 around, we see

\begin{matrix} 1 + \ln | t | = \frac{d}{d t} (t \ln | t |) \\ \ln | t | = \frac{d}{d t} (t \ln | t |) - 1 \\ \ln | t | = \frac{d}{d t} (t \ln | t |) - \frac{d t}{d t} \\ \ln | t | = \frac{d}{d t} (t \ln | t | - t) . \end{matrix}

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Since

\ln | t |

is the derivative of

t \ln | t | - t,

then also

t \ln | t | - t

is an antiderivative of

\ln | t |,

and every other antiderivative must be a vertical shift of that particular one:

\int_{}^{} \ln | t | d_{} t = t \ln | t | - t + C,

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as required.

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We can also combine the product rule with logarithmic differentiation to tackle derivatives that would have been more difficult to compute before.

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Example 17.1.5. An exponential functin with a varying base.

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Consider

\begin{aligned} q (t) & = [\cos (t)]^{t} & - \frac{π}{2} & < t < \frac{π}{2} . \end{aligned}

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(We restrict the domain to ensure the base is always positive.)

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We can’t apply Pattern 12.4.3, because the exponent is varying. We can’t apply Pattern 12.5.7 directly, because the base is varying. So instead we’ll take the logarithm of each side so that we can turn that varying exponent into a factor:

\begin{matrix} q = [\cos (t)]^{t} \\ \ln (q) = t \ln [\cos (t)] \end{matrix}

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(We haven’t bothered with the absolute value brackets this time, since we have already restricted the domain so that

q

is always positive.)

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Now we differentiate both sides:

\begin{matrix} \frac{d}{d t} [\ln (q)] = \frac{d}{d t} [t \ln (\cos (t))] \\ \frac{1}{q} \cdot \frac{d q}{d t} = t \cdot \frac{d}{d t} [\ln (\cos (t))] + \frac{d t}{d t} \cdot \ln (\cos (t)) \\ \frac{1}{q} \cdot \frac{d q}{d t} = t \cdot \frac{1}{\cos (t)} \cdot \frac{d}{d t} (\cos (t)) + 1 \cdot \ln (\cos (t)) \\ \frac{1}{q} \cdot \frac{d q}{d t} = \frac{t}{\cos (t)} \cdot (- \sin (t)) + \ln (\cos (t)) \\ \frac{1}{q} \cdot \frac{d q}{d t} = \ln (\cos (t)) - t \tan (t) . \end{matrix}

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Isolating

d q / d t

and subsituting back in for

q,

we have

\frac{d q}{d t} = [\cos (t)]^{t} \cdot [\ln (\cos (t)) - t \tan (t)] .

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Remark 17.1.6.

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Another approach to Example 17.1.5 would be to express

q (t)

q (t) = \exp (t \ln (\cos (t)))

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and then combining the Chain Rule with the Product Rule. This alternative approach is essentially equivalent to what we have done in Example 17.1.5, but doesn’t involve implicit differentiation.

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Checkpoint 17.1.7. A Quotient Rule.

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Suppose

q (t)

is a rational function

q (t) = \frac{f (t)}{g (t)} .

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Rewriting this as

q (t) = f (t) {[g (t)]}^{- 1},

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combine the Product Rule with one of our two versions of the Chain Rule (Pattern 14.2.1 or Pattern 16.1.7) to produce a “Quotient Rule.” You should be able to simplify your final result into a rational expression involving

f,

g,

and their derivatives.

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Section 17.2 Integrating factors

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Example 17.2.1. Mixing problem with variable source.

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Let’s revisit our Mixing problem from Chapter 4, where fresh brine was entering a vat, and the volume of the vat was held constant by simultaneously draining liquid from the vat at the same rate that fresh brine was entering.

Diagram of brine entering and being drained from a tank so that volume is held constant. — Figure 17.2.2. Brine entering and being drained from a tank so that volume is held constant.

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Now consider a variation of this problem. Perhaps the incoming brine is being produced by a machine that introduces salt into a fresh water supply. When the system is first started up, we might assume that opening the fresh water valve brings the flow rate up to speed nearly instantaneously. But salt moves more slowly, so perhaps the rate at which the salt is supplied causes it to take some time until the concentration of the incoming brine is near the design specifications.

Diagram of brine entering and being drained from a tank so that volume is held constant, where the incoming brine is assumed to be of varying concentration. — Figure 17.2.3. Brine entering and being drained from a tank so that volume is held constant, where the incoming brine is assumed to be of varying concentration.

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We might model the incoming concentration

k (t)

with a function that begins at

k (0) = 0,

but then quickly ramps up to approach a horizontal asymptote at the desired concentration of 15 ^g⁄_L. Maybe something like

k (t) = 15 (1 - e^{- t}) .

Graph of a model function for the ramping up of salt concentration in the incoming brine supply. — Figure 17.2.4. A model function for the ramping up of salt concentration in the incoming brine supply.

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To obtain a rate equation for this system, we use the same

r (t) = (rate in) - (rate out)

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analysis that we used in Example 4.2.4, but now our rate in is varying instead of constant. Remember that our analysis in that example was based on amount of salt in grams, not concentration. So letting

q (t)

represent the amount of salt in the tank at time

t,

we have

\begin{aligned} rate in & = (k (t) g / L) (2 L / s) \\ = 2 k (t) g / s \\ = 30 (1 - e^{- t}) g / s \\ rate out & = (\frac{q (t)}{1000} g / L) (2 L / s) \\ = \frac{q (t)}{500} g / s \\ r (t) & = 30 (1 - e^{- t}) - \frac{q (t)}{500} . \end{aligned}

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Writing

r (t) = \frac{d q}{d t}

as usual, let’s rearrange this equation to

\frac{d q}{d t} + \frac{q}{500} = 30 (1 - e^{- t}) .

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Just as when we solved separable rate equations (Section 15.3), we would like to interpret each side as somehow the result of a differentiation. With the addition on the left-hand side, it could be the result of a Product Rule computation. It would look even more like that if the first term was

u \cdot \frac{d q}{d t}

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instead of just

d q / d t,

for some function

u (t) .

To introduce this new factor, let’s multiply the whole rate equation by this (currently unknown) function

u (t) :

u \cdot \frac{d q}{d t} + \frac{u}{500} \cdot q = 30 u (1 - e^{- t}) .

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But the Product Rule pattern is

\frac{d}{d t} (u q) = u \cdot \frac{d q}{d t} + \frac{d u}{d t} \cdot q .

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So for this to work, we would need

\frac{d u}{d t} = \frac{u}{500} .

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Luckily, we already know from Pattern 15.3.5 that the function

u (t) = e^{t / 500}

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satisfies this new rate equation. (Note that we have chosen

u_{0} = u (0) = 1,

since we only need one appropriate function

u

to use, not every possible such function.)

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So with

u

as above, our rate equation becomes

\begin{matrix} e^{t / 500} \cdot \frac{d q}{d t} + \frac{e^{t / 500}}{500} \cdot q = 30 e^{t / 500} (1 - e^{- t}) \\ \frac{d}{d t} (e^{t / 500} \cdot q) = 30 e^{t / 500} - 30 e^{- 499 t / 500} . \end{matrix}

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So now we explicitly have the left-hand side as a derivative calculation. Since the two sides are equal, let’s make them appear more the same by also interpreting the right-hand side as the result of a derivative calculation. Since

\begin{aligned} \frac{d}{d t} (15000 e^{t / 500} + \frac{15000}{499} e^{- 499 t / 500}) & = \frac{15000}{500} e^{t / 500} + \frac{15000}{499} \cdot (\frac{- 499}{500}) e^{- 499 t / 500} \\ = 30 e^{t / 500} - 30 e^{- 499 t / 500}, \end{aligned}

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we can rewrite our modified rate equation as

\frac{d}{d t} (e^{t / 500} \cdot q) = \frac{d}{d t} (15000 e^{t / 500} + \frac{15000}{499} e^{- 499 t / 500}) .

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This now says that the functions

\begin{aligned} e^{t / 500} & \cdot q & \frac{15000}{499} & (499 e^{t / 500} + e^{- 499 t / 500}) \end{aligned}

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have the same derivative. From Fact 13.4.6, we know that this means that they must be vertical shifts of each other:

e^{t / 500} \cdot q = \frac{15000}{499} (499 e^{t / 500} + e^{- 499 t / 500}) + C

\begin{aligned} q & = \frac{1}{e^{t / 500}} \cdot \frac{15000}{499} (499 e^{t / 500} + e^{- 499 t / 500}) + C \\ = \frac{15000}{499} (499 + e^{- t}) + C e^{- t / 500}, \end{aligned}

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where

C

is an arbitrary constant.

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We can use the same method to solve rate equations of a similar form.

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Definition 17.2.5. First-order, linear rate equation.

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A rate equation that can be expressed in the form

\frac{d q}{d t} + p (t) \cdot q = f (t)

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for some functions

p

and

f

is called a first-order, linear rate equation. The function

p (t)

is called the coefficient function, and the function

f (t)

is called the forcing function for the system.

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We can use the same method as in Example 17.2.1 to solve such a rate equation. We would like to determine a function

u (t)

so that

u \cdot \frac{d q}{d t} + u p q

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is the same as the result of a Product Rule computation

\frac{d}{d t} (u q) = u \cdot \frac{d q}{d t} + \frac{d u}{d t} \cdot q .

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For this to be true, we need

u

to satisfy

\frac{d u}{d t} = u \cdot p (t),

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which is a Separable rate equation in

u

and

t .

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Procedure 17.2.6. Solving a first order, linear rate equation.

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To solve

\frac{d q}{d t} + p (t) \cdot q = f (t) :

First use Procedure 15.3.3 to solve the separable rate equation
$\frac{1}{u} \cdot \frac{d u}{d t} = p (t) .$
Choose one particular solution function $u (t)$ and multiply the original rate equation through by it:
$u (t) \frac{d q}{d t} + u (t) p (t) q (t) = u (t) f (t) .$
Integrate both sides:
$\begin{matrix} \frac{d}{d t} (u (t) q (t)) = u (t) f (t) \\ \int_{}^{} \frac{d}{d t} (u (t) q (t)) d_{} t = \int_{}^{} u (t) f (t) d_{} t \\ u (t) q (t) = \int_{}^{} u (t) f (t) d_{} t, \end{matrix}$
introducing an arbitrary constant of integration on the right side only.
Finally, isolate $q (t) .$

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Remark 17.2.7.

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The chosen function

u (t)

is called an integrating factor for the rate equation, as introducing it as an extra factor in each term allows us to undo the differentiation on the left-hand side — that is, it allows us to solve the equation by integrating.

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Example 17.2.8. Using the procedure.

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Suppose we would like to solve

r (t) = \frac{1 - 2 t q (t)}{1 + t^{2}} .

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It certainly doesn’t like a Separable rate equation, and at first glance it also doesn’t look it’s of the form in Definition 17.2.5. But we can rearrange: