Differential-Equations · Unit 4 · Video 6 · Interactive Practice

Solved but Not Morally: The Boat's Path Is the Spiral $r = Ce^{\theta}$

IKey Formulas

Formula	Step	Remark
$x\,\dfrac{dz}{dx} = \dfrac{1 + z}{1 - z} - z = \dfrac{1 + z^2}{1 - z}$	Substitute $y = zx$ , so $y' = z'x + z$	Combine into one fraction before flipping
$\dfrac{1 - z}{1 + z^2}\,dz = \dfrac{dx}{x} \;\Rightarrow\; \arctan z - \tfrac{1}{2}\ln(1 + z^2) = \ln x + c$	Separate, integrate ( $x > 0$ )	$1 + z^2 \neq 0$ , so dividing loses no solution
$\arctan\dfrac{y}{x} = \ln\sqrt{x^2 + y^2} + c$	Put back $z = \dfrac{y}{x}$	Correct, but it hides the shape
$\theta = \ln r + c \;\Rightarrow\; r = Ce^{\theta}$	Polar coordinates	$C = e^{-c} > 0$ ; $\dfrac{dr}{d\theta} = r$ is the 45° condition

Key Insight: In polar coordinates the 45° heading says the outward step $dr$ equals the sideways step $r\,d\theta$ , so $\dfrac{dr}{d\theta} = r$ and $r = Ce^{\theta}$ in one line. Each half-turn of the beam multiplies the boat's distance by $e^{\pi} \approx 23.1$ .

IIThe Solution, Line by Line, in the Picture

Every line of the solution says something about the beam from the lighthouse $O$ to the boat $P$ .

Step 1

Step 1 — Substitute

y' = z'x + z, \qquad \frac{x + zx}{x - zx} = \frac{1 + z}{1 - z}

x\,\frac{dz}{dx} = \frac{1 + z}{1 - z} - z

Step 2 — One fraction, then flip

x\,\frac{dz}{dx} = \frac{z + 1 - z(1 - z)}{1 - z} = \frac{1 + z^2}{1 - z}

\frac{1 - z}{1 + z^2}\,dz = \frac{dx}{x}

Step 3 — Integrate

\int \frac{dz}{1 + z^2} - \int \frac{z\,dz}{1 + z^2} = \int \frac{dx}{x}

\arctan z - \tfrac{1}{2}\ln(1 + z^2) = \ln x + c

Step 4 — Undo the substitution

\arctan z = \ln\big(x\sqrt{1 + z^2}\,\big) + c

x\sqrt{1 + \tfrac{y^2}{x^2}} = \sqrt{x^2 + y^2} \quad (x > 0)

\arctan\frac{y}{x} = \ln\sqrt{x^2 + y^2} + c

Step 5 — Read it in polar

\theta = \ln r + c \;\Rightarrow\; r = e^{\theta - c} = Ce^{\theta}

P = (1, \sqrt{3})

r = 2

and

\theta = \tfrac{\pi}{3}

, so

C = e^{-c} = 2e^{-\pi/3} \approx 0.70

IIIThe Spirals Are the Integral Curves

The spiral $r = Ce^{\theta}$ through any boat position follows the direction field of $y' = \frac{x + y}{x - y}$ , for $x < 0$ too.

Boat x 2 Boat y 1

IVThe 45° Check: $dr = r\,d\theta$

Turning the beam by $\Delta\theta$ moves the boat out by $\Delta r$ and across by $r\,\Delta\theta$ , legs that become equal as $\Delta\theta \to 0$ .

Beam angle θ 2.20 Turn Δθ 0.30

VQuiz Questions

Problem 1 · Substitute and Simplify

Given: $y' = \dfrac{x + y}{x - y}$ and the substitution $y = zx$ , so $y' = z'x + z$ . Find the right side of $x\,\dfrac{dz}{dx} = \;?$

✅ Correct! Moving

z

across and writing everything over

1 - z

leaves the numerator

z + 1 - z + z^2 = 1 + z^2

❌ The $+z$ is still missing. That is the right side before the

z

from the product rule moves across:

x\,\dfrac{dz}{dx} = \dfrac{1 + z}{1 - z} - z

, and the

-z

still has to be combined in.

❌ Check the sign of $z^2$ .

-z(1 - z) = -z + z^2

, not

-z - z^2

. With the correct sign the

z

terms cancel and

+z^2

survives.

❌ The product rule gives $+z$ .

(zx)' = z'x + z

. Writing

z'x - z

makes you add

z

instead of subtracting it, which produces

1 + 2z - z^2

❌ Not quite. Substitute, cancel the

x

's, move

z

to the right, and combine over

1 - z

Show solution

With $y = zx$ the left side is $y' = z'x + z$ , and the right side is

\frac{x + zx}{x - zx} = \frac{x(1 + z)}{x(1 - z)} = \frac{1 + z}{1 - z}.

So $z'x + z = \dfrac{1 + z}{1 - z}$ . Move $z$ to the right and use the common denominator $1 - z$ :

x\,\frac{dz}{dx} = \frac{z + 1 - z(1 - z)}{1 - z} = \frac{z + 1 - z + z^2}{1 - z} = \frac{1 + z^2}{1 - z}.

Problem 2 · Flip It the Right Way

Given: $x\,\dfrac{dz}{dx} = \dfrac{z + 1}{1 - z} - z$ . Separate the variables, and decide which constant solutions $z = z_0$ (rays $y = z_0 x$ from the lighthouse) the division could lose.

Which separated form is correct?

Which constant solutions does the division lose?

✅ Correct! One fraction first,

\dfrac{1 + z^2}{1 - z}

, then flip it:

\dfrac{1 - z}{1 + z^2}\,dz = \dfrac{dx}{x}

, and since

1 + z^2 > 0

no ray is lost.

❌ The reciprocal of a difference is not the difference of the reciprocals.

\dfrac{1}{a - b} \neq \dfrac{1}{a} - \dfrac{1}{b}

: combine

\dfrac{z + 1}{1 - z} - z

into one fraction first, then flip that single fraction.

❌ The fraction has to be flipped. From

x\,\dfrac{dz}{dx} = \dfrac{1 + z^2}{1 - z}

, dividing by the right side puts its reciprocal

\dfrac{1 - z}{1 + z^2}

next to

dz

❌ The $x$ goes into the denominator.

x\,\dfrac{dz}{dx} = g(z)

separates as

\dfrac{dz}{g(z)} = \dfrac{dx}{x}

: you divide by

x

, you do not multiply by it.

❌ Where $1 - z = 0$ the equation itself is undefined.

z = 1

is the line

y = x

, where

\dfrac{x + y}{x - y}

divides by zero; a lost solution would be a zero of the numerator

1 + z^2

❌ That numerator comes from a sign slip. The numerator is

1 + z^2

, not

1 - z^2

, and

1 + z^2 = 0

has no real roots.

❌ The $x$ -axis is not a path. On

y = 0

the equation asks for

y' = \dfrac{x}{x} = 1

, but the line

y = 0

has slope

0

❌ Not quite. Combine into one fraction, then divide both sides by it and by

x

❌ Not quite. Dividing by

g(z)

can only lose constants

z_0

with

g(z_0) = 0

Show solution

One fraction first. Over the common denominator $1 - z$ :

\frac{z + 1}{1 - z} - z = \frac{z + 1 - z(1 - z)}{1 - z} = \frac{1 + z^2}{1 - z}.

Separate. Multiply both sides of $x\,\dfrac{dz}{dx} = \dfrac{1 + z^2}{1 - z}$ by $\dfrac{1 - z}{1 + z^2}$ and by $\dfrac{dx}{x}$ :

\frac{1 - z}{1 + z^2}\,dz = \frac{dx}{x}.

Lost solutions. Dividing by $\dfrac{1 + z^2}{1 - z}$ could only lose constant solutions $z = z_0$ at which it is zero, that is $1 + z_0^2 = 0$ . There is no real $z_0$ , so nothing is lost. Geometrically, no ray $y = z_0 x$ is a path: the direction field crosses every ray from the lighthouse at 45°.

Problem 3 · Where Is the Boat Next?

Given: the boat passes through $(\sqrt{3}, 1)$ holding its 45° heading, so its path is $r = Ce^{\theta}$ . Find its distance from the lighthouse when the beam next points straight up, $\theta = \tfrac{\pi}{2}$ .

✅ Correct!

C = 2e^{-\pi/6}

, so

r\big(\tfrac{\pi}{2}\big) = 2e^{-\pi/6}e^{\pi/2} = 2e^{\pi/3}

❌ $C$ is not the starting distance.

C

is the distance at

\theta = 0

. The boat is at

r = 2

when

\theta = \tfrac{\pi}{6}

, so

C = 2e^{-\pi/6}

❌ The angle adds to $\ln r$ , not to $r$ .

\theta = \ln r + c

means turning the beam by

\tfrac{\pi}{3}

raises

\ln r

\tfrac{\pi}{3}

, which multiplies

r

e^{\pi/3}

❌ Check the sign when you solve for $C$ . From

2 = Ce^{\pi/6}

C = 2e^{-\pi/6}

, not

2e^{+\pi/6}

❌ Not quite. Find

r

and

\theta

at the given point, solve for

C

, then evaluate at

\theta = \tfrac{\pi}{2}

Show solution

Polar coordinates of the point. $r = \sqrt{3 + 1} = 2$ , and since $x > 0$ , $\theta = \arctan\dfrac{1}{\sqrt{3}} = \dfrac{\pi}{6}$ .

The constant. $2 = Ce^{\pi/6}$ gives $C = 2e^{-\pi/6}$ .

Evaluate. The beam turns counterclockwise from $\tfrac{\pi}{6}$ and first reaches $\tfrac{\pi}{2}$ after a turn of $\tfrac{\pi}{3}$ :

r = 2e^{-\pi/6}\,e^{\pi/2} = 2e^{\pi/3} \approx 5.70.

Problem 4 · A Different Heading

Given: the 45° check generalizes. If the heading makes angle $\beta$ with the outward beam, the sideways and outward legs satisfy $\tan\beta = \dfrac{r\,d\theta}{dr}$ . A second boat holds $\beta = 60\degree$ . Find its path.

✅ Correct!

\dfrac{dr}{d\theta} = \dfrac{r}{\tan 60\degree} = \dfrac{r}{\sqrt{3}}

, so

r = Ce^{\theta/\sqrt{3}}

, a spiral that opens more slowly than the 45° one.

❌ The legs are swapped.

\tan\beta

is sideways over outward,

\dfrac{r\,d\theta}{dr}

, so

\dfrac{dr}{d\theta} = r\cot\beta = \dfrac{r}{\sqrt{3}}

, not

\sqrt{3}\,r

❌ The ratio of the legs is $\cot\beta$ , not $\cos\beta$ .

\dfrac{dr}{r\,d\theta} = \cot 60\degree = \dfrac{1}{\sqrt{3}}

;

\cos 60\degree = \tfrac{1}{2}

would compare a leg with the hypotenuse.

❌ The angle changes the spiral.

r = Ce^{\theta}

has

\dfrac{dr}{d\theta} = r

, equal legs, which is

\beta = 45\degree

only.

❌ Not quite. Solve

\tan 60\degree = \dfrac{r\,d\theta}{dr}

for

\dfrac{dr}{d\theta}

and separate.

Show solution

The polar equation. With $\tan 60\degree = \sqrt{3}$ :

\sqrt{3} = \frac{r\,d\theta}{dr} \quad\Rightarrow\quad \frac{dr}{d\theta} = \frac{r}{\sqrt{3}}.

Separate and integrate. $\dfrac{dr}{r} = \dfrac{d\theta}{\sqrt{3}}$ gives $\ln r = \dfrac{\theta}{\sqrt{3}} + c$ , so

r = Ce^{\theta/\sqrt{3}}, \qquad C = e^{c} > 0.

Check. At $\beta = 45\degree$ the same steps give $\dfrac{dr}{d\theta} = r$ and $r = Ce^{\theta}$ . At $60\degree$ each half-turn multiplies the distance by $e^{\pi/\sqrt{3}} \approx 6.13$ instead of $e^{\pi} \approx 23.1$ .

Solved: 0 / 4

Solved but Not Morally: The Boat's Path Is the Spiral r=Ceθr = Ce^{\theta}r=Ceθ

IKey Formulas

IIThe Solution, Line by Line, in the Picture

IIIThe Spirals Are the Integral Curves

IVThe 45° Check: dr=r dθdr = r\,d\thetadr=rdθ

VQuiz Questions

Solved but Not Morally: The Boat's Path Is the Spiral $r = Ce^{\theta}$

IVThe 45° Check: $dr = r\,d\theta$