Differential-Equations · Unit 4 · Video 5 · Interactive Practice

The Lighthouse Chase: Turning a 45-Degree Heading into a Homogeneous ODE

IKey Formulas

Formula	Name	What it says
$y' = \tan(\alpha + 45\degree), \quad \tan\alpha = \dfrac{y}{x}$	The rule as a slope	A slope is the tangent of the angle with the horizontal; the heading is the beam's angle $\alpha$ turned $45\degree$ counterclockwise
$\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$	Tangent of a sum	Not $\tan\alpha + \tan\beta$ ; at $\alpha = \beta = 45\degree$ the denominator is $1 - (1)(1) = 0$ , a vertical direction
$y' = \dfrac{y/x + 1}{1 - y/x} = F\!\left(\dfrac{y}{x}\right)$	The model, homogeneous form	Every $x$ and $y$ sits inside the ratio $y/x$
$y' = \dfrac{x + y}{x - y}$	The model, tidy form	Top and bottom times $x$ ; still homogeneous (divide by $x$ , or zoom $x \to ax,\ y \to ay$ ), and defined on the $y$ -axis

Key Insight: The rule fixes an angle, and angles add while slopes do not: the heading's slope is $\tan(\alpha + 45\degree)$ , not $\tan\alpha + 1$ . Pushed through the tangent-sum formula, every $x$ and $y$ lands inside $y/x$ , and that is what makes the equation homogeneous.

IIThe Heading as a Slope

The rule fixes an angle, $\alpha + 45\degree$ with the horizontal, so the slope at the boat is its tangent.

IIIAngles Add, Slopes Do Not

Adding $\tan 45\degree = 1$ to the beam's slope does not, in general, turn the beam through $45\degree$ ; the tangent-sum formula does.

Beam slope tan α 1/2

IVThe Zoom Test, Drawn

A zoom slides each point along its line through the lighthouse, where $y/x$ is fixed, so the slope survives.

💡 A right-hand side that depends on $y/x$ alone is exactly what the substitution $y = zx$ from the previous video needs; the next video carries it through to the boat's path.

VQuiz Questions

Problem 1 · The Slope at One Point

Given: the lighthouse at the origin and the boat at $(5, 2)$ , holding its heading $45\degree$ counterclockwise from the beam — find the slope $y'$ of its path there.

✅ Correct!

\tan\alpha = y/x = \tfrac{2}{5}

, so

y' = \dfrac{2/5 + 1}{1 - 2/5} = \dfrac{7/5}{3/5} = \tfrac{7}{3}

, the same as the tidy form

\dfrac{5 + 2}{5 - 2}

❌ That adds the slopes.

\tfrac{2}{5} + 1

\tan\alpha + \tan 45\degree

; the tangent of a sum also divides by

1 - \tan\alpha\tan 45\degree = \tfrac{3}{5}

❌ Numerator and denominator are swapped.

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

is not the model; the sum belongs on top:

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

❌ That is a clockwise turn.

\tan(\alpha - 45\degree) = \dfrac{y/x - 1}{1 + y/x} = -\tfrac{3}{7}

; the boat turns counterclockwise from the beam, so the angles add.

❌ Not quite. Find

\tan\alpha = y/x

first, then use

\tan(\alpha + 45\degree) = \dfrac{\tan\alpha + 1}{1 - \tan\alpha}

Show solution

The beam runs from the lighthouse to the boat, so its slope is

\tan\alpha = \frac{y}{x} = \frac{2}{5}

The heading is at $\alpha + 45\degree$ , and $\tan 45\degree = 1$ :

y' = \tan(\alpha + 45\degree) = \frac{\tan\alpha + 1}{1 - \tan\alpha} = \frac{\tfrac{2}{5} + 1}{1 - \tfrac{2}{5}} = \frac{7/5}{3/5} = \frac{7}{3}

The tidy form agrees: $y' = \dfrac{x + y}{x - y} = \dfrac{5 + 2}{5 - 2} = \dfrac{7}{3}$ .

Angle check: $\alpha = \arctan 0.4 \approx 21.8\degree$ , so the heading is at about $66.8\degree$ , and $\tan 66.8\degree \approx 2.33$ .

Problem 2 · The Slope-Sum Shortcut

Given: a student adds the slopes and writes $y' = \tan\alpha + \tan 45\degree = \dfrac{y}{x} + 1$ — find the points of the sea, away from the lighthouse, where the student's slope equals the correct one.

✅ Correct! With

t = y/x

, the equation

\dfrac{t + 1}{1 - t} = t + 1

reduces to

(t + 1)\,t = 0

, so

t = 0

(the

x

-axis) or

t = -1

(the line

y = -x

). Everywhere else the shortcut is wrong.

❌ The $x$ -axis is only one of the answers. On

y = -x

the heading is horizontal, and both formulas give

0

❌ That line is where they disagree most. On

y = x

the heading is vertical and has no finite slope, while the shortcut claims

1 + 1 = 2

❌ Test a point on the $x$ -axis. There

y/x = 0

, and both formulas give

1

❌ Not quite. Put

t = y/x

and solve

\dfrac{t + 1}{1 - t} = t + 1

Show solution

Write $t = y/x$ , so $x \neq 0$ . The correct slope is $\dfrac{t + 1}{1 - t}$ and the shortcut gives $t + 1$ . Set them equal, with $t \neq 1$ , and multiply by $1 - t$ :

t + 1 = (t + 1)(1 - t) \iff (t + 1)\big(1 - (1 - t)\big) = 0 \iff (t + 1)\,t = 0

So $t = 0$ or $t = -1$ :

$t = 0$ : $y = 0$ , the $x$ -axis. The beam is horizontal, the heading is at $45\degree$ or $225\degree$ , and both slopes are $1$ .
$t = -1$ : $y = -x$ . The beam is at $-45\degree$ or $135\degree$ , the heading is horizontal, and both slopes are $0$ .

Everywhere else the shortcut is wrong. On $y = x$ it says $2$ where the heading is vertical, and on the $y$ -axis it is not even defined, while the tidy form gives the true slope $-1$ .

Problem 3 · A 60-Degree Heading

Given: a bolder boat that holds its heading $60\degree$ counterclockwise from the beam, with $\tan 60\degree = \sqrt{3}$ — find the equation of its path and its slope on the positive $y$ -axis.

Which equation?

Its slope on the positive $y$ -axis?

✅ Correct!

\tan(\alpha + 60\degree) = \dfrac{y/x + \sqrt{3}}{1 - \sqrt{3}\,y/x}

, then times

x

top and bottom. At

x = 0

the tidy form gives

\dfrac{y}{-\sqrt{3}\,y} = -\tfrac{1}{\sqrt{3}}

, the slope of a

150\degree

heading.

❌ That adds the slopes: it is

y/x + \sqrt{3}

. The tangent of a sum also needs the denominator

1 - \sqrt{3}\,y/x

❌ That uses $\tan 60\degree = \tfrac{1}{\sqrt{3}}$ , which is

\tan 30\degree

. This equation describes a

30\degree

turn.

❌ That is $\tan(\alpha - 60\degree)$ , a clockwise turn. Counterclockwise, the angles add.

❌ $-\sqrt{3}$ is the slope of a $120\degree$ heading, a

30\degree

turn from the vertical beam. Use

\tan 60\degree = \sqrt{3}

❌ $\tfrac{1}{\sqrt{3}}$ is the slope of a $30\degree$ heading, the clockwise turn from a beam at

90\degree

❌ Only the $y/x$ form breaks down at $x = 0$ . The tidy form is defined there: put

x = 0

in it.

❌ Check the equation. Use

\tan(\alpha + 60\degree) = \dfrac{\tan\alpha + \tan 60\degree}{1 - \tan\alpha\tan 60\degree}

with

\tan\alpha = y/x

❌ Check the slope. On the positive

y

-axis the beam points straight up, at

90\degree

Show solution

Step 1: the rule as a slope. The heading is at $\alpha + 60\degree$ , so

y' = \tan(\alpha + 60\degree) = \frac{\tan\alpha + \tan 60\degree}{1 - \tan\alpha\tan 60\degree} = \frac{y/x + \sqrt{3}}{1 - \sqrt{3}\,y/x}

Step 2: tidy it. Multiply the top and the bottom by $x$ :

y' = \frac{y + \sqrt{3}\,x}{x - \sqrt{3}\,y}

It is still homogeneous: the first form is a function of $y/x$ alone.

Step 3: the positive $y$ -axis. There $y/x$ is undefined, so use the tidy form with $x = 0$ :

y' = \frac{y + 0}{0 - \sqrt{3}\,y} = -\frac{1}{\sqrt{3}}

Picture check: the beam points straight up, at $90\degree$ , the heading is at $150\degree$ , and $\tan 150\degree = -\tfrac{1}{\sqrt{3}}$ ✓.

Problem 4 · A Lighthouse Off the Origin

Given: the same boat, $45\degree$ counterclockwise from the beam, but the lighthouse stands at $(1, 0)$ instead of the origin — find the equation of the path and decide whether it passes the zoom test $x \to ax,\ y \to ay$ .

Which equation?

Does the zoom $x \to ax,\ y \to ay$ leave it unchanged?

✅ Correct! The beam now runs from

(1, 0)

, so

\tan\alpha = \dfrac{y}{x - 1}

. The

-1

does not scale with

a

, so a zoom about the origin changes the equation; a zoom about the lighthouse, in

u = x - 1

, does not.

❌ That is the lighthouse at the origin. The beam's slope is now rise

y

over run

x - 1

❌ Check the run. From

(1, 0)

(x, y)

it is

x - 1

, not

x + 1

❌ That is the clockwise turn,

\tan(\alpha - 45\degree)

❌ Try it. The zoomed right side is

\dfrac{ax - 1 + ay}{ax - 1 - ay}

, and the

-1

has no factor

a

to cancel. At the point

(3, 1)

the original right side is

3

, while the zoomed one with

a = 2

\tfrac{7}{3}

❌ Moving the origin to the lighthouse does it. With

u = x - 1

we have

du = dx

, and the equation is the original model in

u

and

y

❌ Check the equation. Measure the beam from the lighthouse at

(1, 0)

❌ Check the zoom test. Replace

x

ax

and

y

ay

and see whether every

a

cancels.

Show solution

Step 1: the beam. From $(1, 0)$ to $(x, y)$ the run is $x - 1$ and the rise is $y$ :

\tan\alpha = \frac{y}{x - 1}

Step 2: the heading. Multiply the top and the bottom by $x - 1$ :

y' = \frac{\tan\alpha + 1}{1 - \tan\alpha} = \frac{\frac{y}{x - 1} + 1}{1 - \frac{y}{x - 1}} = \frac{x - 1 + y}{x - 1 - y}

Step 3: the zoom test about the origin. Under $x \to ax$ , $y \to ay$ the left side $y'$ is unchanged, but the right side becomes

\frac{ax - 1 + ay}{ax - 1 - ay}

and the $-1$ carries no factor $a$ , so nothing cancels. At $(3, 1)$ with $a = 2$ it is $\tfrac{7}{3}$ instead of $3$ : not homogeneous in $x$ and $y$ .

Step 4: move the origin to the lighthouse. With $u = x - 1$ , $du = dx$ , so $\dfrac{dy}{du} = y'$ and

\frac{dy}{du} = \frac{u + y}{u - y}

This is the lighthouse model again, homogeneous in $u$ and $y$ . Homogeneity is invariance under a zoom about the lighthouse, which is why the video puts the lighthouse at the origin.

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