CCM Reversing the Chain Rule

is now being considered as the independent variable instead of the dependent one. Finally, we apply Property 1 of Pattern 6.3.11 to this new integral to obtain

\begin{aligned} \int_{a}^{b} \frac{d q}{d t} d_{} t & = \int_{q (a)}^{q (b)} d_{} q \\ = \int_{q (a)}^{q (b)} 1 d_{} q \\ = 1 \cdot (q (b) - q (a)) \\ = q (b) - q (a), \end{aligned}

🔗

just as in (✶).

🔗

Example 15.1.1. Integrating a derivative.

🔗

Consider

q (t) = \sin (t),

so that

q^{'} (t) = \cos (t) .

Then

d q = \frac{d q}{d t} d t = \cos (t) d t,

🔗

so we can compute

\begin{aligned} \int_{- π / 2}^{π / 2} \cos (t) d_{} t & = \int_{\sin (- π / 2)}^{\sin (π / 2)} d_{} q \\ = \int_{\sin (- π / 2)}^{\sin (π / 2)} d_{} q \\ = \int_{- 1}^{1} d_{} q \\ = 1 - (- 1) \\ = 2, \end{aligned}

🔗

which agrees with the many previous computations of this accumulation that we have made before.

🔗

What if we apply all of the above to a derivative that has been calculated using the chain rule? Suppose

q

is a function of

u

and

u

is a function of

t .

Then ultimately

q

is a function of

t,

and the Chain Rule says that

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

🔗

If we are computing an integral

\int_{a}^{b} \frac{d q}{d t} d_{} t,

🔗

we would need to know

q (t)

in order to use an analysis like that in (✶✶). So instead, we will use

u

as an intermediate variable:

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

🔗

giving us

\int_{a}^{b} \frac{d q}{d t} d_{} t = \int_{u (a)}^{u (b)} \frac{d q}{d u} d_{} u .

🔗

The hope is that this second integral is either something for which we already have a formula, or for which it is simpler to guess an antiderivative.

🔗

Example 15.1.2. Integrating the Chain Rule.

🔗

Consider

q (t) = \sqrt{t^{2} + 1} .

In Example 14.2.5 we used the substitution

u = t^{2} + 1

to compute

q^{'} (t) = \frac{t}{\sqrt{t^{2} + 1}} = \frac{2 t}{2 \sqrt{t^{2} + 1}},

🔗

where

d u / d t = 2 t .

Let’s pretend we don’t know the formula for

q (t)

and work in reverse. We can use

u

to compute

\begin{aligned} \int_{2}^{3} \frac{t}{\sqrt{t^{2} + 1}} d_{} t & = \int_{2}^{3} \frac{2 t}{2 \sqrt{t^{2} + 1}} d_{} t \\ = \int_{2}^{3} \frac{1}{2 \sqrt{t^{2} + 1}} \cdot (2 t) d_{} t \\ = \int_{2}^{3} \frac{1}{2 \sqrt{t^{2} + 1}} \cdot \frac{d u}{d t} d_{} t \\ = \int_{5}^{10} \frac{1}{2 \sqrt{u}} d_{} u, \end{aligned}

🔗

where we have adjusted the bounds of integration to

\begin{aligned} u (2) & = 2^{2} + 1 & u (3) & = 3^{2} + 1 \\ = 5 & = 9 \end{aligned}

🔗

when we converted to an integral in terms of

u .

This new integral is one we know how to deal with:

\begin{aligned} \int_{2}^{3} \frac{t}{\sqrt{t^{2} + 1}} d_{} t & = \int_{5}^{10} \frac{1}{2 \sqrt{u}} d_{} u \\ = \frac{1}{2} \int_{5}^{10} u^{- 1 / 2} d_{} u \end{aligned}

🔗

Using antiderivative

U (t) = \frac{u^{1 / 2}}{1 / 2},

🔗

we have

\begin{aligned} \int_{2}^{3} \frac{t}{\sqrt{t^{2} + 1}} d_{} t & = \frac{1}{2} \int_{5}^{10} u^{- 1 / 2} d_{} u \\ = \frac{1}{2} (\frac{10^{1 / 2}}{1 / 2} - \frac{5^{1 / 2}}{1 / 2}) \\ = \sqrt{10} - \sqrt{5} . \end{aligned}

🔗

Remembering that we actually do know

q (t),

notice that this result is the same as computing

\int_{2}^{3} \frac{d q}{d t} d_{} t = q (3) - q (2) .

🔗

Section 15.2 Method of substitution

🔗

In general, it is difficult to recognize an integrand as a derivative function that is the result of a chain rule. Even in Example 15.1.2, we had to go backwards one step from the final result of Example 14.2.5 by reintroducing the

2

in both numerator and denominator in order to make it work. So we create a slightly more general procedure that can be tried when we merely suspect that we have such an integrand that might be the result of a chain rule.

🔗

Procedure 15.2.1. Method of substitution.

🔗

f (t)

is a function that can be expressed as

f (t) = g (t) \frac{d u}{d t},

🔗

where

g (t)

can be converted into a function of

u

alone by substituting

u = u (t),

then

\int_{a}^{b} f (t) d_{} t = \int_{u (a)}^{u (b)} g (u) d_{} u .

🔗

Remark 15.2.2. Method of substitution is structured “trial-and-error”.

🔗

When we try this method, usually we guess at

u

first and tweak

g

to match. But it is often not obvious what

u

to try — sometimes we try some version of

u

and it leads to a dead end. In that case, we either try a different substitution by choosing a different

u,

or we try a different method all together. No matter what we try, our strategy when using substitution is to recognize the integrand as a product of two functions, where the second is related to the derivative of some “part” of the first.

🔗

Example 15.2.3. Integral of a horizontal scaling.

🔗

For

\int_{0}^{π / 10} \cos (5 t) d_{} t,

🔗

consider

u = 5 t

with

d u / d t = 5 .

Since our integrand doesn’t involve a multiple of

5

outside of the cosine function, we need to introduce one:

\cos (t) = \frac{\cos (5 t)}{5} \cdot 5 = \frac{\cos (5 t)}{5} \cdot \frac{d u}{d t} .

🔗

So our

g (t)

g (t) = \frac{\cos (5 t)}{5} ⟹ g (u) = \frac{\cos (u)}{5},

🔗

and our new bounds of integration are

\begin{aligned} u (0) & = 5 \cdot 0 & u (π / 10) & = 5 \cdot π / 10 \\ = 0 & = π / 2 \end{aligned}

🔗

and we can calculate

\begin{aligned} \int_{0}^{π / 10} \cos (5 t) d_{} t & = \int_{0}^{π / 2} \frac{\cos (u)}{5} d_{} u \\ = \frac{1}{5} \int_{0}^{π / 2} \cos (u) d_{} u \\ = \frac{1}{5} (\sin (π / 2) - \sin (0)) \\ = \frac{1}{5} (1 - 0) \\ = \frac{1}{5}, \end{aligned}

🔗

where we have used the antiderivative

\sin (u)

for

\cos (u) .

🔗

Example 15.2.4. Integral of a horizontal shift.

🔗

For

\int_{0}^{e^{3} - 1} \frac{1}{t + 1} d_{} t,

🔗

we could probably guess an antiderivative involving a logarithm. But we can also be more systematic using substitution. Using

u = t + 1 ⟹ \frac{d u}{d t} = 1

🔗

and

\begin{aligned} u (0) & = 0 + 1 & u (e^{3} - 1) & = (e^{3} - 1) + 1 \\ = 1 & = e^{3} \end{aligned}

🔗

we have

\begin{aligned} \int_{0}^{e^{3} - 1} \frac{1}{t + 1} d_{} t & = \int_{0}^{e^{3} - 1} \frac{d u / d t}{t + 1} d_{} t \\ = \int_{1}^{e^{3}} \frac{1}{u} d_{} u . \end{aligned}

🔗

Using the antiderivative

\ln | u |

for

1 / u,

we have

\begin{aligned} \int_{0}^{e^{3} - 1} \frac{1}{t + 1} d_{} t & = \ln | e^{3} | - \ln | 1 | \\ = 3 - 0 \\ = 3 . \end{aligned}

🔗

Example 15.2.5. Another trig example.

🔗

For

\int_{0}^{π / 2} \cos (t) \sin^{2} (t) d_{} t,

🔗

we recognize that

\cos (t)

is the derivative of

\sin (t),

so if we substitute

\begin{aligned} u & = \sin (t) & \frac{d u}{d t} & = \cos (t) \end{aligned}

🔗

and

\begin{aligned} u (0) & = 0 & u (π / 2) = 1 \end{aligned}

🔗

we have

\begin{aligned} \int_{0}^{π / 2} \cos (t) \sin^{2} (t) d_{} t & = \int_{0}^{π / 2} {(\sin (t))}^{2} \frac{d u}{d t} d_{} t \\ = \int_{0}^{1} u^{2} d_{} u \\ = \frac{1^{3}}{3} - \frac{0^{3}}{3} \\ = \frac{1}{3} . \end{aligned}

🔗

We can also use substitution to compute indefinite integrals.

🔗

Example 15.2.6. Substitution in an indefinite integral.

🔗

For

\int_{}^{} t e^{t^{2}} d_{} t

🔗

we recognize

d (t^{2}) / d t = 2 t,

so let’s try substituting

u = t^{2} .

As with a previous example, the factor of

2

d u / d t

is missing in our integrand, so we must artificially introduce it.

\begin{aligned} \int_{}^{} t e^{t^{2}} d_{} t & = \int_{}^{} \frac{e^{t^{2}}}{2} \cdot (2 t) d_{} t \\ = \int_{}^{} \frac{e^{t^{2}}}{2} \cdot \frac{d u}{d t} d_{} t \\ = \int_{}^{} \frac{e^{u}}{2} d_{} u \\ = \frac{e^{u}}{2} + C . \end{aligned}

🔗

However, we weren’t interested in the antidervivatives of

g (u) = e^{u} / 2,

we wanted the antiderivatives of

f (t) = t e^{t^{2}} .

We need to substitute back in for

u :

\begin{aligned} \int_{}^{} t e^{t^{2}} d_{} t & = \frac{e^{u}}{2} + C \\ = \frac{e^{t^{2}}}{2} + C . \end{aligned}

🔗

Sometimes there are multiple choices for the substitution.

🔗

Example 15.2.7. Choosing the substitution.

🔗

For

\int_{}^{} \sin (t) \cos (t) d_{} t,

🔗

each of the two factors in the integrand is related to the derivative of the other. Let’s choose

u = \sin (t)

so that

d u / d t = \cos (t) .

Then,

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

🔗

Checkpoint 15.2.8. Choosing a different substitution.

🔗

Repeat Example 15.2.7, but this time choose

u = \cos (t) .

You will arrive at a seemingly different answer, but Fact 13.4.6 says that your result should be a vertical shift of the result computed in Example 15.2.7. Determine how.

🔗

Somewhat similar to Example 15.2.4, we can sometimes use substitution to “shift” a complicated expression to another place in the integrand.

🔗

Example 15.2.9. Shifting a sum from the denominator to the numerator.

🔗

For

\int_{}^{} \frac{t^{5}}{t^{2} + 1} d_{} t,

🔗

the integrand would be easier to deal with if there wasn’t a plus sign in the denominator. So let’s try substituting

u = t^{2} + 1 .

Since

d u / d t = 2 t,

we first reorganize the integrand, artificially introducing that factor of

2 .

\begin{aligned} \int_{}^{} \frac{t^{5}}{t^{2} + 1} d_{} t & = \int_{}^{} \frac{2 t \cdot t^{4}}{2 (t^{2} + 1)} d_{} t \\ = \int_{}^{} \frac{t^{4}}{2 (t^{2} + 1)} \cdot (2 t) d_{} t \\ = \int_{}^{} \frac{t^{4}}{2 (t^{2} + 1)} \cdot \frac{d u}{d t} d_{} t . \end{aligned}

🔗

What about that numerator of

t^{4} ?

Luckily, it can be expressed as a function of

u :

\begin{matrix} t^{2} + 1 = u \\ t^{2} = u - 1 \\ t^{4} = (u - 1)^{2} . \end{matrix}

🔗

\begin{aligned} \int_{}^{} \frac{t^{5}}{t^{2} + 1} d_{} t & = \int_{}^{} \frac{t^{4}}{2 (t^{2} + 1)} \cdot \frac{d u}{d t} d_{} t \\ = \int_{}^{} \frac{(u - 1)^{2}}{2 u} d_{} u . \end{aligned}

🔗

Now that we have shifted the sum in the original denominator to be a difference in the numerator, we can manipulate the integrand algebraically before integrating:

\begin{aligned} \int_{}^{} \frac{t^{5}}{t^{2} + 1} d_{} t & = \int_{}^{} \frac{(u - 1)^{2}}{2 u} d_{} u \\ = \int_{}^{} \frac{u^{2} - 2 u + 1}{2 u} d_{} u \\ = \int_{}^{} \frac{u}{2} - 1 + \frac{1}{2 u} d_{} u \\ = \frac{u^{2}}{4} - u + \frac{1}{2} \ln | u | + C \\ = \frac{(t^{2} + 1)^{2}}{4} - (t^{2} + 1) + \frac{1}{2} \ln | t^{2} + 1 | + C . \end{aligned}

🔗

Section 15.3 Separable rate equations

🔗

Recall that a rate equation relates a quantity to its rate of variation. Our goal in solving a rate equation is to relate the quantity directly to the independent variable. Since a rate of variation is a derivative function, ultimately solving a rate equation involves recognizing and reversing the differentiation. When we can “separate” the variables involved in a rate function to either side of a rate equation, on one side we can reverse the differentiation by reversing the chain rule.

🔗

Example 15.3.1. A first example.

🔗

Consider the rate equation

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

🔗

Let’s simply write

q

instead of

q (t) .

But let’s also replace

r (t)

with

d q / d t,

since

d q / d t

is the rate of variation of

q .

Finally, we’ll also do a little algebraic rearrangement.

\begin{matrix} r (t) = \frac{t^{2} + 1}{2 q (t) + 1} \\ \frac{d q}{d t} = \frac{t^{2} + 1}{2 q + 1} \\ (†) & (2 q + 1) \frac{d q}{d t} = t^{2} + 1 \\ 2 q \cdot \frac{d q}{d t} + \frac{d q}{d t} = t^{2} + 1 . \end{matrix}

🔗

The first term on the left now looks the type of result we would obtain when applying the chain rule during an implicit differentiation computation:

\begin{aligned} formula in q & = formula in t \\ ↓ d / d t ? \\ (2 q + 1) \frac{d q}{d t} & = t^{2} + 1 . \end{aligned}

🔗

Our goal is to relate quantity

q

to the independent variable

t

directly, and if we could reverse the process above and return to an equality involving

q

and

t

then we could attempt to isolate

q

to achieve our goal.

🔗

To reverse a suspected differentiation process, let us antidifferentiate both sides of our re-arranged rate equation. First, we’ll tackle the right-hand side of (†), since it is straight-forward:

\int_{}^{} t^{2} + 1 d_{} t = \frac{t^{3}}{3} + t + C .

🔗

For now, let us choose the specific antiderivative with

C = 0 .

By definition of antiderivative, our calculation above says

t^{2} + 1 = \frac{d}{d t} (\frac{t^{3}}{3} + t) .

🔗

Antidifferentiating the left-hand side of (†), involves computing

\int_{}^{} (2 q + 1) \frac{d q}{d t} d_{} t .

🔗

But this looks like an integral involving the method of substitution, just with

q

in place of

u :

\int_{}^{} (2 q + 1) \frac{d q}{d t} d_{} t = \int_{}^{} 2 q + 1 d_{} q = q^{2} + q + C .

🔗

Again, let us choose the specific antiderivative with

C = 0 .

Then our calculation above says that

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

🔗

Combining these separate expressions for the two sides of (†) back into an equality, we have

\begin{matrix} (2 q + 1) \frac{d q}{d t} = t^{2} + 1 \\ \frac{d}{d t} (q^{2} + q) = \frac{d}{d t} (\frac{t^{3}}{3} + t) . \end{matrix}

🔗

However, Corollary 13.4.7 states that two derivatives can only be equal when the two functions being differentiated are vertical shifts of one another. In other words, we must have

q^{2} + q = \frac{t^{3}}{3} + t + K

🔗

for some unknown constant

K .

🔗

We now at least have obtained an implicit expression relating

q

t .

It’s not always possible to isolate

q,

but in this case we can use the Quadratic formula to obtain a “lower” and an “upper” solution:

\begin{matrix} q^{2} + q = \frac{t^{3}}{3} + t + K \\ 3 q^{2} + 3 q = t^{3} + 3 t + K \\ 3 q^{2} + 3 q - (t^{3} + 3 t + K) = 0 \end{matrix}

\begin{aligned} q (t) & = \frac{- 3 - \sqrt{9 + 12 (t^{3} + 3 t + K)}}{6} & q (t) & = \frac{- 3 + \sqrt{9 + 12 (t^{3} + 3 t + K)}}{6} \\ = \frac{- 3 - \sqrt{K + 12 (t^{3} + 3 t)}}{6}, & = \frac{- 3 + \sqrt{K + 12 (t^{3} + 3 t)}}{6} . \end{aligned}

🔗

Notice that we have replaced

9 + 12 K

with just

K,

since if

K

is an arbitrary value, then so is

9 - 12 K .

🔗

Our two classes of solutions both involve the parameter

K,

which would have to be determined using a known or desired initial value. For example, if we want a solution of the “upper” type (with the plus sign before the square root) that satisfies

q (0) = 2,

we could solve for

K :

\begin{matrix} 2 = \frac{- 3 + \sqrt{K + 12 (0^{3} + 3 \cdot 0)}}{6} \\ 15 = \sqrt{K} \\ 225 = K . \end{matrix}

🔗

So the solution that satisfies

q (0) = 2

q (t) = \frac{- 3 + \sqrt{225 + 12 (t^{3} + 3 t)}}{6} .

🔗

Note that there is no solution of the “lower” type (with the minus sign before the square root) that satisfies

q (0) = 2,

since that would require

\sqrt{K}

to be a negative value.

🔗

Definition 15.3.2. Separable rate equation.

🔗

A rate equation that can be expressed in the form

m (q) \cdot \frac{d q}{d t} = n (t)

🔗

for some functions

m

and

n

is called separable.

🔗

Ultimately, the problem in Example 15.3.1 was solved by reversing differentiation, also known as antidifferentiating.

🔗

Procedure 15.3.3. Solving a separable rate equation.

🔗

To solve

m (q) \cdot \frac{d q}{d t} = n (t),

🔗

integrate both sides:

\begin{matrix} m (q) \cdot \frac{d q}{d t} = n (t) \\ \int_{}^{} m (q) \cdot \frac{d q}{d t} d_{} t = \int_{}^{} n (t) d_{} t \\ \int_{}^{} m (q) d_{} q = \int_{}^{} n (t) d_{} t \\ M (q) = N (t) + C, \end{matrix}

🔗

where

M (q)

is an antiderivative for

m (q)

(with respect to

q

) and

N (t)

is an antiderivative for

n (t)

(with respect to

t

). Notice that we have introduced an arbitrary constant of integration on one side only.

🔗

Example 15.3.4. Solving exponential decay.

🔗

Let’s once again revisit Example 4.2.3, which involved the rate equation

\frac{r (t)}{q (t)} = - k .

🔗

In Example 12.5.8 we verified that an horizontally-transformed exponential function can be a solution to this type of rate equation. Now we’ll solve this equation to verify that exponential functions are the only possible solutions.

\begin{matrix} \frac{1}{q} \cdot \frac{d q}{d t} = - k \\ \int_{}^{} \frac{1}{q} \cdot \frac{d q}{d t} d_{} t = \int_{}^{} - k d_{} t \\ \int_{}^{} \frac{1}{q} d_{} q = \int_{}^{} - k d_{} t \\ \ln | q | = - k t + C . \end{matrix}

🔗

To isolate

q,

we undo the logarithm by taking the exponential of each side.

\begin{matrix} \exp (\ln | q |) = e^{- k t + C} \\ | q | = e^{C} \cdot e^{- k t} . \end{matrix}

🔗

Now, the exponentials on the left are always positive, but the

q

inside the absolute value brackets could be positive or negative, so again we really have a “lower” and “upper” solution

\begin{aligned} q & = e^{C} \cdot e^{- k t} & q & = - e^{C} \cdot e^{- k t} \end{aligned}

🔗

However, as

C

is an arbitrary constant, together

\pm e^{C}

can take on any non-zero value, so we unify these two solutions by letting

A

represent either of

\pm e^{C} :

q (t) = A e^{- k t} .

🔗

Notice that

A

represents the initial value

q (0),

since

e^{0} = 1 .

🔗

One last thing to consider is that if

A = \pm e^{C},

then

A = 0

is not possible. And that’s good, because

q = 0

would not work in

\frac{r (t)}{q (t)} = - k .

🔗

However, if we rearrange this to

r (t) = - k q (t),

🔗

then

q (t) = 0

would in fact be a solution, since then

q^{'} (t) = 0

also, so

\begin{aligned} LHS & = r (t) & RHS & = - k q (t) \\ = q^{'} (t) & RHS & = - k \cdot 0 \\ = 0 & RHS & = 0 . \end{aligned}

🔗

Let’s record the pattern of the solution to the previous example.

🔗

Pattern 15.3.5. Solution to proportional rate equation.

🔗

q (t)

is a varying quantity so that its rate of variation

r (t)

is always proportional to the amount

q (t)

according to the rate equation

r (t) = k q (t),

🔗

then

q (t)

is the exponential function

q (t) = q_{0} e^{k t},

🔗

where

q_{0} = q (0)

is the initial quantity.

🔗

As one last example, let’s return to the rate equation from Example 4.2.11.

🔗

Example 15.3.6.

🔗

In Example 4.2.11 we considered the rate equation

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

🔗

This is somewhat similar to Example 15.3.1, except this time the solution will involve logarithms, because after separating variables we will have reciprocal functions on both sides.

\frac{1}{q (1 - q)} \cdot \frac{d q}{d t} = \frac{1}{t + 1} .

🔗

Because of the product in the denominator, the left-hand side looks like something you get after obtaining a common denominator. In fact, combining

\frac{1}{q} + \frac{1}{1 - q}

🔗

into a single fraction leads to our left-hand side above (without the

d q / d t

\begin{matrix} (\frac{1}{q} + \frac{1}{1 - q}) \cdot \frac{d q}{d t} = \frac{1}{t + 1} \\ \int_{}^{} \frac{1}{q} + \frac{1}{1 - q} d_{} q = \int_{}^{} \frac{1}{t + 1} d_{} t \end{matrix}

🔗

For each of the right-hand side and the second term on the left-hand side, we could perform a substitution similarly to Example 15.2.4, but we can probably just guess-and-check these antiderivatives using the logarithm:

\ln | q | - \ln | 1 - q | = \ln | t + 1 | + C,

🔗

where the original addition on the left has turned into subtraction to correct the extra negative we would get from the chain rule if we differentiated

\ln | 1 - q |

to check our answer.

🔗

Finally, we would like to solve for

q

if possible. We use the properties of the logarithm to combine the two terms on the left, and then apply the exponential function to “undo” the logarithm.

\begin{matrix} \ln | \frac{q}{1 - q} | = \ln | t + 1 | + C \\ \exp (\ln | \frac{q}{1 - q} |) = \exp (\ln | t + 1 | + C) \\ | \frac{q}{1 - q} | = \exp (\ln | t + 1 |) \cdot e^{C} \\ | \frac{q}{1 - q} | = e^{C} | t + 1 | \\ \frac{q}{1 - q} = \pm e^{C} (t + 1) \end{matrix}

🔗

Replace

\pm e^{C}

with arbitrary

A :

\begin{matrix} q = A (t + 1) (1 - q) \\ q = A (t + 1) - A (t + 1) q \\ q + A (t + 1) q = A (t + 1) \\ (1 + A (t + 1)) q = A (t + 1) \\ q (t) = \frac{A (t + 1)}{1 + A (t + 1)}, \end{matrix}

🔗

where

A

is an arbitrary constant. This general solution appears different from the one previously provided in Example 4.2.11, but this solution can be converted into that solution by substituting