Implicit differentiation and related rates

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1. Implicit differentiation

Warm-up

Three quick questions from earlier lessons.

Discover

The circle $x^2 + y^2 = 25$ is not the graph of one function, but it has a tangent line at each point. If $y$ depends on $x$ , what is $\dfrac{d}{dx}\big(y^2\big)$ ?

$2y \cdot \dfrac{dy}{dx}$
$2y$

Key idea

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ .

Implicit differentiation: every term with

y

picks up a factor

\dfrac{dy}{dx}

(chain rule); products like

xy

need the product rule. Then solve the equation for

\dfrac{dy}{dx}

Worked example

$x^2 + y^2 = 25$ . Find $\dfrac{dy}{dx}$ .

2x + 2y\,y' = 0

Differentiate each term; (y²)' = 2y y'.

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

Isolate y'.

y'(3, 4) = -\dfrac{3}{4}

At (3, 4): the tangent's slope.

Your turn, guided

Your turn: $y^3 + x = 7$ .

3y^2\,y' + 1 = 0

(y³)' = 3y² y'; (x)' = 1.

y' = -\dfrac{1}{3y^2}

Isolate y'.

Practice ★

$x^2 + 6y^2 = 63$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
$x^2 + 7y^2 = 19$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

$y^2 = x^3 + 5x$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
$y^2 = x^3 + 6x$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

Watch out!

Practice ★★

$xy + 7x = 19$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
$xy + 3x = 15$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

Exit ticket

Three questions to check you have got it.

$x^2 + 6y^2 = 63$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
$x^2 + 7y^2 = 19$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
$x^2 + 7y^2 = 69$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

Summary card

Implicit differentiation

Differentiate both sides with respect to $x$ .
Each $y$ term gets a factor $y'$ ; use the product rule for $xy$ .
Solve for $y'$ .

Watch out: $(y^2)' = 2y\,y'$ , not $2y$ .

2. Tangent lines to curves defined implicitly

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is the circle $x^2 + y^2 = 25$ ; the blue line is its tangent at the point at angle $a$ . Move $a$ .

The circle x² + y² = 25 and its tangent line at the point (5 cos a, 5 sin a). The tangent is perpendicular to the radius, with slope −x/y.

The radius to $(3, 4)$ has slope $\tfrac{4}{3}$ . What is the slope of the tangent there?

$-\tfrac{3}{4}$
$\tfrac{4}{3}$

Key idea

After implicit differentiation, $y'$ depends on both coordinates.

To find the tangent at

(x_0, y_0)

: compute

y'

implicitly, substitute both

x_0

and

y_0

to get the slope

m

, then

y = y_0 + m(x - x_0)

Worked example

Tangent to $x^2 + xy + y^2 = 7$ at $(1, 2)$ .

2x + y + x\,y' + 2y\,y' = 0

Product rule on xy.

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

Solve for y'.

y = 2 - \tfrac{4}{5}(x - 1)

At (1, 2): −4/5.

Your turn, guided

Your turn: $x^2 + y^2 = 13$ at $(2, 3)$ .

slope

= -\tfrac{2}{3}

y' = −x/y at (2, 3).

y = 3 - \tfrac{2}{3}(x - 2)

Point-slope form.

Practice ★

Find the slope of the tangent line to the circle $x^2 + y^2 = 25$ at the point $(4, 3)$ .
Find the slope of the tangent line to the circle $x^2 + y^2 = 40$ at the point $(2, 6)$ .
Find the slope of the tangent line to the circle $x^2 + y^2 = 26$ at the point $(1, 5)$ .

Watch out!

Practice ★★

Write the tangent line to $x^2 + y^2 = 20$ at $(4, 2)$ : enter the right-hand side of $y = \ldots$
Write the tangent line to $x^2 + y^2 = 45$ at $(6, 3)$ : enter the right-hand side of $y = \ldots$

The point $(1, 2)$ is on the curve $x^2 + xy + y^2 = 7$ . Find the slope of the tangent line there.
The point $(1, 5)$ is on the curve $x^2 + xy + y^2 = 31$ . Find the slope of the tangent line there.

Exit ticket

Three questions to check you have got it.

Find the slope of the tangent line to the circle $x^2 + y^2 = 25$ at the point $(4, 3)$ .
Find the slope of the tangent line to the circle $x^2 + y^2 = 40$ at the point $(2, 6)$ .
Find the slope of the tangent line to the circle $x^2 + y^2 = 26$ at the point $(1, 5)$ .

Summary card

Tangents to implicit curves

Slope: substitute the point into $y'$ (both $x$ and $y$ ).
Line: $y = y_0 + m(x - x_0)$ .

Watch out: the tangent to a circle is perpendicular to the radius.

3. The derivative of an inverse function

Warm-up

Three quick questions from earlier lessons.

Discover

$f(2) = 5$ and the tangent to $f$ at $(2, 5)$ has slope 3. The graph of $f^{-1}$ is the reflection in $y = x$ , so it passes through $(5, 2)$ . Reflecting swaps rise and run. What is its slope there?

$\tfrac{1}{3}$
$3$

Key idea

Slopes of inverse functions are reciprocals at matching points.

f(a) = b

and

f'(a) \ne 0

, then

(f^{-1})'(b) = \dfrac{1}{f'(a)}

. In general

(f^{-1})'(x) = \dfrac{1}{f'\big(f^{-1}(x)\big)}

Worked example

$f(x) = x^3 + 2x$ , $g = f^{-1}$ . Find $g'(3)$ .

g(3) = 1

Find a with f(a) = 3: 1 + 2 = 3.

f'(1) = 5

f'(x) = 3x² + 2.

g'(3) = \tfrac{1}{5}

Reciprocal.

Your turn, guided

Your turn: same $f$ , find $g'(12)$ . Note $f(2) = 12$ .

f'(2) = 14

f'(2) = 3·4 + 2.

g'(12) = \tfrac{1}{14}

Reciprocal.

Practice ★

$f$ is invertible with $f(3) = 9$ and $f'(3) = 2$ . Find $(f^{-1})'(9)$ .
$f$ is invertible with $f(1) = 7$ and $f'(1) = 3$ . Find $(f^{-1})'(7)$ .

$g$ is the inverse of $f$ . Which expression equals $g'(7)$ ?
1. $\dfrac{1}{f'(g(7))}$
2. $\dfrac{1}{f'(7)}$
3. $f'(g(7))$
4. $\dfrac{1}{f(7)}$

Watch out!

Practice ★★

$f(x) = x^3 + 8x$ has an inverse $g$ . Note $f(1) = 9$ . Find $g'(9)$ .
$f(x) = x^3 + 7x$ has an inverse $g$ . Note $f(1) = 8$ . Find $g'(8)$ .

Exit ticket

Three questions to check you have got it.

$f$ is invertible with $f(3) = 9$ and $f'(3) = 2$ . Find $(f^{-1})'(9)$ .
$f$ is invertible with $f(1) = 7$ and $f'(1) = 3$ . Find $(f^{-1})'(7)$ .
$f$ is invertible with $f(4) = 7$ and $f'(4) = 6$ . Find $(f^{-1})'(7)$ .

Summary card

Derivative of an inverse

Find $a = f^{-1}(b)$ .
Compute $f'(a)$ .
$(f^{-1})'(b) = 1 / f'(a)$ .

Watch out: evaluate $f'$ at $a$ , the input of $f$ , not at $b$ .

4. Related rates: area and volume

Warm-up

Three quick questions from earlier lessons.

Discover

A stone drops in a pond; the ripple's radius grows at 2 cm/s. Is the area growing at a constant rate?

No, faster and faster
Yes, 2 cm²/s

Key idea

When quantities are linked by an equation, their rates are linked by its derivative with respect to time.

Related rates: write an equation linking the quantities, differentiate it with respect to

t

(chain rule: each quantity gets its own rate), then substitute the values at the moment of interest.

Worked example

A ripple's radius grows at 2 cm/s. How fast is the area growing when $r = 10$ cm?

A = \pi r^2

The link between the quantities.

\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}

Differentiate with respect to t.

\dfrac{dA}{dt} = 40\pi \approx 126

cm²/s

Now substitute r = 10, dr/dt = 2.

Your turn, guided

Your turn: a square's side grows at 3 cm/s. $A = s^2$ . Find $\dfrac{dA}{dt}$ when $s = 5$ .

\dfrac{dA}{dt} = 2s\dfrac{ds}{dt}

Differentiate.

\dfrac{dA}{dt} = 30

cm²/s

2 · 5 · 3.

Practice ★

A circular ripple's radius grows at 5 cm/s. How fast is its area growing when the radius is 11 cm? (cm²/s, exact)
A circular ripple's radius grows at 2 cm/s. How fast is its area growing when the radius is 6 cm? (cm²/s, exact)

A balloon's radius increases at 5 cm/s. With $V = \tfrac{4}{3}\pi r^3$ , find $\dfrac{dV}{dt}$ when $r = 3$ cm. (cm³/s, exact)
A balloon's radius increases at 2 cm/s. With $V = \tfrac{4}{3}\pi r^3$ , find $\dfrac{dV}{dt}$ when $r = 7$ cm. (cm³/s, exact)

Watch out!

Practice ★★

Air is pumped into a spherical balloon at 29 cm³/s. How fast is the radius increasing when $r = 6$ cm? (cm/s, exact)
Air is pumped into a spherical balloon at 26 cm³/s. How fast is the radius increasing when $r = 3$ cm? (cm/s, exact)

Exit ticket

Three questions to check you have got it.

A circular ripple's radius grows at 5 cm/s. How fast is its area growing when the radius is 11 cm? (cm²/s, exact)
A circular ripple's radius grows at 2 cm/s. How fast is its area growing when the radius is 6 cm? (cm²/s, exact)
A circular ripple's radius grows at 3 cm/s. How fast is its area growing when the radius is 11 cm? (cm²/s, exact)

Summary card

Related rates

Equation linking the quantities.
Differentiate with respect to $t$ .
Substitute the values at that moment; solve for the unknown rate.

Watch out: substitute changing values only after differentiating.

5. Related rates with right triangles

Warm-up

Three quick questions from earlier lessons.

Discover

A 5-m ladder leans against a wall; its foot is $a$ m from the wall. Slide the foot outwards at a steady pace.

A 5-m ladder from (a, 0) on the ground to the wall at x = 0. As a grows by equal steps, the top drops slowly at first, then faster and faster.

The foot moves at a steady speed. Does the top slide down at a steady speed?

No: it speeds up near the ground
Yes, the same speed as the foot

Key idea

Right triangles link distances by Pythagoras.

x^2 + y^2 = D^2

, then

x\,x' + y\,y' = D\,D'

(all with respect to

t

). A constant length (a ladder) has rate 0; a decreasing length has a negative rate.

Worked example

A 10-ft ladder: foot 6 ft from the wall, sliding out at 2 ft/s. How fast is the top moving?

y = 8

At that moment y = √(100 − 36).

2x\,x' + 2y\,y' = 0

Differentiate x² + y² = 100.

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

ft/s: the top slides down

6 · 2 + 8 · y' = 0.

Your turn, guided

Your turn: same ladder, when the foot is 8 ft out (so $y = 6$ ).

y' = -\tfrac{8}{3}

ft/s

8 · 2 + 6 · y' = 0.

Practice ★

A 15-ft ladder leans against a wall. Its foot slides away from the wall at 4 ft/s. How fast is the top moving when the foot is 9 ft from the wall? (ft/s; negative means down)
A 10-ft ladder leans against a wall. Its foot slides away from the wall at 3 ft/s. How fast is the top moving when the foot is 6 ft from the wall? (ft/s; negative means down)
A 15-ft ladder leans against a wall. Its foot slides away from the wall at 2 ft/s. How fast is the top moving when the foot is 9 ft from the wall? (ft/s; negative means down)

Watch out!

Practice ★★

Two cars leave the same crossing at noon: one drives north at 50 mph, the other east at 51 mph. How fast is the distance between them growing? (mph, exact)

A plane flies at an altitude of 6 km, horizontally away from a radar station, at 11 km/min. How fast is its distance to the station increasing when the horizontal distance is 8 km? (km/min)
A plane flies at an altitude of 9 km, horizontally away from a radar station, at 10 km/min. How fast is its distance to the station increasing when the horizontal distance is 12 km? (km/min)

Boss problem. A 15-ft ladder slides: its foot moves away from the wall at 1 ft/s. Look at the moment the foot is 9 ft from the wall.

Exit ticket

Three questions to check you have got it.

A 15-ft ladder leans against a wall. Its foot slides away from the wall at 4 ft/s. How fast is the top moving when the foot is 9 ft from the wall? (ft/s; negative means down)
A 10-ft ladder leans against a wall. Its foot slides away from the wall at 3 ft/s. How fast is the top moving when the foot is 6 ft from the wall? (ft/s; negative means down)
A 15-ft ladder leans against a wall. Its foot slides away from the wall at 2 ft/s. How fast is the top moving when the foot is 9 ft from the wall? (ft/s; negative means down)

Summary card

Right-triangle rates

$x^2 + y^2 = D^2 \Rightarrow x\,x' + y\,y' = D\,D'$ .
Find the missing side at that moment first.

Watch out: rates at right angles do not add; decreasing lengths have negative rates.

6. Related rates with similar triangles

Warm-up

Three quick questions from earlier lessons.

Discover

A 6-ft person walks away from a 15-ft lamp. With $x$ the distance to the lamp and $s$ the shadow's length, two right triangles share the angle at the shadow's tip. Which equation links $s$ and $x$ ?

$\dfrac{6}{s} = \dfrac{15}{x + s}$
$\dfrac{6}{s} = \dfrac{15}{x}$

Key idea

Similar triangles remove an extra variable.

When a problem has more changing quantities than rates you know, use a fixed ratio (similar triangles, a cone's shape) to write one quantity in terms of another before differentiating.

Worked example

6-ft person, 15-ft lamp, walking at 3 ft/s. How fast does the shadow grow?

6x + 6s = 15s

, so

s = \tfrac{2}{3}x

Cross-multiply 6/s = 15/(x + s).

s' = \tfrac{2}{3}x'

Differentiate.

s' = 2

ft/s

x' = 3.

Your turn, guided

Your turn: a cone tank whose radius is half the depth: $V = \tfrac{\pi}{12}h^3$ . Differentiate.

\dfrac{dV}{dt} = \dfrac{\pi}{4}h^2 \cdot \dfrac{dh}{dt}

Chain rule: 3h²/12 = h²/4.

Practice ★

Put the steps for finding $\dfrac{dA}{dt}$ when $r = 3$ in order.
- Draw a picture and name the changing quantities
- Write an equation linking them: $A = \pi r^2$
- Differentiate with respect to $t$ : $A' = 2\pi r \, r'$
- Substitute $r = 3$ and the known rate

A person 6 ft tall walks away from a 13-ft street lamp at 4 ft/s. How fast is the length of their shadow growing? (ft/s, exact)
A person 5 ft tall walks away from a 18-ft street lamp at 4 ft/s. How fast is the length of their shadow growing? (ft/s, exact)

Watch out!

Practice ★★

Water flows at 9 m³/min into a cone-shaped tank, vertex down, whose radius is always half the water's depth: $V = \tfrac{\pi}{12}h^3$ . How fast is the depth rising when $h = 4$ m? (m/min, exact)
Water flows at 3 m³/min into a cone-shaped tank, vertex down, whose radius is always half the water's depth: $V = \tfrac{\pi}{12}h^3$ . How fast is the depth rising when $h = 3$ m? (m/min, exact)

Exit ticket

Three questions to check you have got it.

Water flows at 9 m³/min into a cone-shaped tank, vertex down, whose radius is always half the water's depth: $V = \tfrac{\pi}{12}h^3$ . How fast is the depth rising when $h = 4$ m? (m/min, exact)
Water flows at 3 m³/min into a cone-shaped tank, vertex down, whose radius is always half the water's depth: $V = \tfrac{\pi}{12}h^3$ . How fast is the depth rising when $h = 3$ m? (m/min, exact)
Water flows at 7 m³/min into a cone-shaped tank, vertex down, whose radius is always half the water's depth: $V = \tfrac{\pi}{12}h^3$ . How fast is the depth rising when $h = 4$ m? (m/min, exact)

Summary card

Similar triangles

Use the fixed ratio to write one variable in terms of another.
Differentiate with respect to $t$ .
Substitute the moment's values.

Watch out: the big triangle's base runs from the lamp to the shadow's tip.

Answers: Implicit differentiation and related rates

1. Implicit differentiation

$- \frac{x}{6 y}$
$- \frac{x}{7 y}$

$\frac{3 x^{2} + 5}{2 y}$
$\frac{3 x^{2} + 6}{2 y}$

$\frac{- y - 7}{x}$
$\frac{- y - 3}{x}$

$- \frac{x}{6 y}$
$- \frac{x}{7 y}$
$- \frac{x}{7 y}$

2. Tangent lines to curves defined implicitly

$- \frac{4}{3}$
$- \frac{1}{3}$
$- \frac{1}{5}$

$10 - 2 x$
$15 - 2 x$

$- \frac{4}{5}$
$- \frac{7}{11}$

$- \frac{4}{3}$
$- \frac{1}{3}$
$- \frac{1}{5}$

3. The derivative of an inverse function

$\frac{1}{2}$
$\frac{1}{3}$

$\dfrac{1}{f'(g(7))}$

$\frac{1}{11}$
$\frac{1}{10}$

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{6}$

4. Related rates: area and volume

$110 \pi$
$24 \pi$

$180 \pi$
$392 \pi$

$\frac{29}{144 \pi}$
$\frac{13}{18 \pi}$

$110 \pi$
$24 \pi$
$66 \pi$

5. Related rates with right triangles

$-3$
$- \frac{9}{4}$
$- \frac{3}{2}$

$\sqrt{5101}$

$\frac{44}{5}$
$8$

$- \frac{1}{12}$

$-3$
$- \frac{9}{4}$
$- \frac{3}{2}$

6. Related rates with similar triangles

1. Draw a picture and name the changing quantities 2. Write an equation linking them: $A = \pi r^2$ 3. Differentiate with respect to $t$ : $A' = 2\pi r \, r'$ 4. Substitute $r = 3$ and the known rate

$\frac{24}{7}$
$\frac{20}{13}$

$\frac{9}{4 \pi}$
$\frac{4}{3 \pi}$

$\frac{9}{4 \pi}$
$\frac{4}{3 \pi}$
$\frac{7}{4 \pi}$