Single-Variable-Calculus · Unit 5 · Video 2 · Interactive Practice

The Slope of a Circle, Two Ways

IKey Formulas

Formula	Name	What it says
$x^2 + y^2 = 1$	The unit circle, implicitly	A condition on $x$ and $y$ — not a formula for $y$
$y = \pm\sqrt{1 - x^2}$	Explicit form, two branches	$+$ is the top half, $-$ is the bottom half
$y' = \dfrac{-x}{\sqrt{1 - x^2}}$	Explicit derivative (top half)	Chain rule, with the power rule at exponent $\tfrac{1}{2}$
$2x + 2y\,y' = 0 \;\Longrightarrow\; y' = -\dfrac{x}{y}$	Implicit derivative	$\dfrac{d}{dx}\big(y^2\big) = 2y\,y'$ — the factor $y'$ is never dropped

Key Insight: The implicit route never takes a square root, so it never picks a sign: $y' = -x/y$ is the slope on the top half and on the bottom half at once. It returns no value exactly where the circle has no slope — at $(\pm 1, 0)$ , where $y = 0$ and the tangent is vertical.

IIVisualization 1 — One Formula, Both Halves

At every point of the circle the tangent slope is $-x/y$ — top half and bottom half alike.

Angle θ 53.1°

💡 The tangent is perpendicular to the radius: the radius through $(x, y)$ has slope $y/x$ , and $-x/y$ is exactly its negative reciprocal.

IIIVisualization 2 — Two Routes to the Same Slope

Solving for $y$ first and leaving the equation alone land on the same number at $x = 0.6$ .

Route Step 1

Step 1 — Solve for $y$ , choose a branch

y^2 = 1 - x^2 \qquad y = \pm\sqrt{1 - x^2}

Top half:

y = \sqrt{1 - x^2}

. The bottom half waits its turn.

Step 2 — Rewrite the root as a power

y = \big(1 - x^2\big)^{1/2}

Step 3 — Chain rule: outside, then inside

y' = \tfrac{1}{2}\big(1 - x^2\big)^{-1/2} \cdot (-2x)

Outside: the power rule at exponent

\tfrac{1}{2}

, inside left alone. Inside:

\dfrac{d}{dx}\big(1 - x^2\big) = -2x

Step 4 — Simplify, then evaluate

y' = \frac{-x}{\sqrt{1 - x^2}}

x = 0.6

\ y' = \dfrac{-0.6}{\sqrt{1 - 0.36}} = \dfrac{-0.6}{0.8} = -0.75

Step 1 — Differentiate the equation as it stands

\frac{d}{dx}\big(x^2 + y^2\big) = \frac{d}{dx}(1) = 0

Step 2 — Term by term, chain rule on $y^2$

2x + 2y\,y' = 0

y

is a function of

x

, so

\dfrac{d}{dx}\big(y^2\big) = 2y\,y'

Step 3 — Solve for $y'$

2y\,y' = -2x \qquad y' = \frac{-2x}{2y} = -\frac{x}{y}

Step 4 — No branch was ever chosen

(0.6,\ 0.8)

\ y' = \dfrac{-0.6}{0.8} = -0.75

(0.6,\ -0.8)

\ y' = \dfrac{-0.6}{-0.8} = 0.75

IVVisualization 3 — Predict the Slope

At this point, does the circle tilt up or down — and how steeply?

Point Your slope 0.00

VQuiz Questions

Problem 1 · Differentiate the Equation

Given: $x^2 + y^2 = 1$ , with $y$ a function of $x$ — apply $\dfrac{d}{dx}$ to both sides. Which equation results?

✅ Correct! The chain rule turns

y^2

into

2y\,y'

, and solving gives

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

❌ The factor $y'$ is missing.

\dfrac{d}{dx}\big(y^2\big)

2y

times

\dfrac{dy}{dx}

, because

y

is itself a function of

x

❌ Check the right-hand side. The derivative of the constant

1

0

, not

1

❌ Not quite. Differentiate term by term:

\dfrac{d}{dx}\big(x^2\big) = 2x

and

\dfrac{d}{dx}\big(y^2\big) = 2y\,y'

Show solution

Apply $\dfrac{d}{dx}$ to each side of $x^2 + y^2 = 1$ :

\frac{d}{dx}\big(x^2\big) + \frac{d}{dx}\big(y^2\big) = \frac{d}{dx}(1)

The first term is $2x$ . The second needs the chain rule, since $y$ depends on $x$ : the derivative of $y^2$ with respect to $y$ is $2y$ , times $\dfrac{dy}{dx} = y'$ . The right-hand side is the derivative of a constant, which is $0$ :

2x + 2y\,y' = 0

Solving: $2y\,y' = -2x$ , so $y' = \dfrac{-2x}{2y} = -\dfrac{x}{y}$ .

Problem 2 · A Point on the Bottom Half

Given: the point $\left(\tfrac{3}{5},\ -\tfrac{4}{5}\right)$ on $x^2 + y^2 = 1$ — find the slope $y'$ there.

✅ Correct! Two minus signs cancel, and the bottom-right of the circle really does tilt upward.

❌ Watch the sign of $y$ . Here

y = -\tfrac{4}{5}

is negative, so

-x/y

is a negative divided by a negative — the slope comes out positive.

❌ The fraction is upside down. The formula is

-\dfrac{x}{y}

, not

-\dfrac{y}{x}

: the

x

-coordinate goes on top.

❌ Not quite. Substitute carefully into

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

with

x = \tfrac{3}{5}

and

y = -\tfrac{4}{5}

Show solution

The implicit formula holds at every point of the circle where $y \neq 0$ :

y' = -\frac{x}{y} = -\frac{3/5}{-4/5} = \frac{3/5}{4/5} = \frac{3}{4}

Check by the explicit route. This point is on the bottom half, $y = -\sqrt{1 - x^2}$ , whose derivative is $y' = \dfrac{x}{\sqrt{1 - x^2}}$ . At $x = \tfrac{3}{5}$ the root is $\sqrt{1 - \tfrac{9}{25}} = \tfrac{4}{5}$ , so

y' = \frac{3/5}{4/5} = \frac{3}{4}

Same answer. Geometrically, moving right along the lower-right arc takes you upward toward $(1, 0)$ , so a positive slope is what the picture demands.

Problem 3 · The Explicit Route on the Bottom Half

Given: the bottom half of the unit circle, $y = -\sqrt{1 - x^2}$ .

Differentiate it — what is $y'$ ?

Now evaluate it at $x = \tfrac{1}{2}$ .

✅ Correct! And

-x/y

agrees: at

x = \tfrac{1}{2}

the bottom half has

y = -\tfrac{\sqrt{3}}{2}

, so

-x/y = \dfrac{1/2}{\sqrt{3}/2} = \dfrac{1}{\sqrt{3}}

❌ That is the top half's answer. The leading minus sign of

y = -\sqrt{1 - x^2}

is a constant factor, so it flips the sign of the whole derivative.

❌ Check the chain rule. Outside:

\tfrac{1}{2}(1 - x^2)^{-1/2}

. Inside:

-2x

. The

\tfrac{1}{2}

and the

2

cancel.

❌ Check the arithmetic.

\sqrt{1 - \tfrac{1}{4}} = \sqrt{\tfrac{3}{4}} = \tfrac{\sqrt{3}}{2}

, and

\tfrac{1}{2}

divided by

\tfrac{\sqrt{3}}{2}

\tfrac{1}{\sqrt{3}}

Show solution

Step 1 — Rewrite as a power.

y = -\big(1 - x^2\big)^{1/2}

Step 2 — Chain rule, carrying the constant factor $-1$ along:

y' = -\tfrac{1}{2}\big(1 - x^2\big)^{-1/2} \cdot (-2x) = x\big(1 - x^2\big)^{-1/2} = \frac{x}{\sqrt{1 - x^2}}

Step 3 — Evaluate at $x = \tfrac{1}{2}$ :

y' = \frac{1/2}{\sqrt{1 - 1/4}} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}

Step 4 — Compare with the implicit answer. On this branch $y = -\sqrt{1 - x^2}$ , so

-\frac{x}{y} = -\frac{x}{-\sqrt{1 - x^2}} = \frac{x}{\sqrt{1 - x^2}}

Both derivatives picked up the extra minus sign, so the two routes still agree — and $-x/y$ needed no branch at all.

Problem 4 · A Circle of Radius 5

Given: $x^2 + y^2 = 25$ .

Differentiate implicitly and solve for $y'$ .

At which points of this circle does that formula return no value?

✅ Correct! The radius never enters the derivative, and the formula goes silent exactly where the tangent is vertical: at

(\pm 5, 0)

the circle has a tangent line but no slope.

❌ The $25$ is a constant. Its derivative is

0

, so it never reaches the formula for

y'

❌ Both twos cancel. From

2x + 2y\,y' = 0

you get

y' = \dfrac{-2x}{2y}

, and the

2

s divide out completely.

❌ Differentiate each side first.

2x + 2y\,y' = 0

— the right-hand side contributes nothing.

❌ Look at the denominator. A fraction fails only when its denominator is

0

, so ask where

y = 0

on this circle.

Show solution

Part 1. Apply $\dfrac{d}{dx}$ to both sides of $x^2 + y^2 = 25$ . The constant $25$ differentiates to $0$ , exactly as the $1$ did:

2x + 2y\,y' = 0 \qquad \Longrightarrow \qquad y' = \frac{-2x}{2y} = -\frac{x}{y}

The radius has vanished: every circle centred at the origin has tangent slope $-x/y$ .

Part 2. The formula fails only where $y = 0$ . On this circle that means $x^2 = 25$ , so $x = \pm 5$ : the two points $(5, 0)$ and $(-5, 0)$ .

There the circle does have a tangent line — the vertical lines $x = 5$ and $x = -5$ — but a vertical line has no slope, so a formula for $y'$ must return nothing. The picture and the algebra fail in the same place.

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