Multivariable-Calculus · Unit 13 · Video 2 · Interactive Practice

What Is Held Constant? Subscript Notation for Partial Derivatives and the Right-Triangle Area

IKey Formulas

Formula	Meaning	Held fixed
$\left(\dfrac{\partial f}{\partial x}\right)_y$	Rate of change of $f$ as $x$ varies	$y$ (the subscript)
$\left(\dfrac{\partial f}{\partial u}\right)_v = \left(\dfrac{\partial f}{\partial x}\right)_v = 2 \neq \left(\dfrac{\partial f}{\partial x}\right)_y = 1$	$f = x + y$ with $x = u$ , $y = u + v$	$v = y - x$ versus $y$
$A = \tfrac{1}{2}ab\sin\theta, \quad a = b\cos\theta$	Area of the triangle and the right-angle constraint	—
$A_\theta = \tfrac{1}{2}ab\cos\theta$ , $\ \left(\dfrac{\partial A}{\partial\theta}\right)_a$ , $\ \left(\dfrac{\partial A}{\partial\theta}\right)_b$	Three different rates of change of $A$ with respect to $\theta$	$a$ and $b$ / $a$ / $b$

Key Insight: When the variables are related, the subscript names the variable held fixed, and the letter missing from the symbol is the dependent variable.

IIVisualization 1 — Holding $y$ Fixed vs Holding $v = y - x$ Fixed

Moving $x$ by the same amount changes $f = x + y$ by different amounts, depending on what stays fixed.

Hold fixed Δx 1

IIIVisualization 2 — Three Motions of the Triangle, Three Rates

Increasing $\theta$ from the same right triangle changes its area at different rates, depending on what is held fixed.

Hold fixed Δθ 0°

💡 Challenge: with $\Delta\theta$ close to $0$ , rank the three difference quotients $\Delta A / \Delta\theta$ ; the next video computes the middle one, $\left(\partial A / \partial\theta\right)_a$ , exactly.

IVQuiz Questions

Problem 1 · Read the Symbol

Given: the right triangle with area $A = \tfrac{1}{2}ab\sin\theta$ and constraint $a = b\cos\theta$ . In the rate $\left(\dfrac{\partial A}{\partial\theta}\right)_b$ , which variable is the dependent one?

✅ Correct!

a

is the letter missing from the symbol: it changes as

a = b\cos\theta

forces it to.

❌ Not quite.

\theta

is the variable we change on purpose — it sits in the denominator, so it is independent.

❌ Not quite. The subscript names the variable held fixed, so

b

is constant here.

❌ Not quite. One relation among three variables leaves only two independent; the constraint makes one of them follow.

Show solution

The symbol $\left(\partial A/\partial\theta\right)_b$ says: change $\theta$ , hold $b$ fixed. The constraint $a = b\cos\theta$ is one relation among $a$ , $b$ , $\theta$ , so exactly two of them are independent — here $\theta$ and $b$ .

The remaining variable, $a$ , is dependent: $a = b\cos\theta$ shrinks as $\theta$ grows with the hypotenuse fixed. It is the letter that appears nowhere in the symbol.

Problem 2 · Same Variable, Different Rate

Given: $f(x, y) = 2x + 5y$ and new variables with $x = u$ , $y = u + v$ . Find $\left(\dfrac{\partial f}{\partial u}\right)_v$ .

✅ Correct! Holding

v

fixed forces

y

to grow with

x

, so

f

picks up

2 + 5 = 7

❌ Close, but…

2

\left(\partial f/\partial x\right)_y

. Even though

x = u

, holding

v

fixed is not holding

y

fixed:

y = u + v

moves with

u

❌ Not quite.

5

\left(\partial f/\partial v\right)_u

, the rate in the other new variable.

❌ Not quite. Substitute first:

f = 2u + 5(u + v)

, then differentiate in

u

with

v

constant.

Show solution

Substitute $x = u$ , $y = u + v$ :

f = 2u + 5(u + v) = 7u + 5v

\left(\frac{\partial f}{\partial u}\right)_v = 7

Compare $\left(\partial f/\partial x\right)_y = 2$ . Since $x = u$ , also $\left(\partial f/\partial x\right)_v = 7$ : the two symbols differ only in what they hold fixed, $v = y - x$ versus $y$ .

Problem 3 · Two Rates at One Point

Given: $f(x, y) = x^2 y$ with $x = u$ , $y = u + v$ , at the point $(x, y) = (1, 2)$ , that is $(u, v) = (1, 1)$ . Find both rates.

What is $\left(\partial f/\partial x\right)_y$ ?

What is $\left(\partial f/\partial u\right)_v$ ?

✅ Correct!

\left(\partial f/\partial x\right)_y = 4

but

\left(\partial f/\partial u\right)_v = \left(\partial f/\partial x\right)_v = 5

, even though

x = u

❌ Check the first rate.

\partial(x^2 y)/\partial x = 2xy

— the factor

y = 2

stays.

❌ Check the first rate.

1 = x^2

is the rate in

y

, not in

x

❌ Check the first rate.

5

is the rate with

v

held fixed; with

y

fixed, only

2xy

counts.

❌ Check the second rate.

4

is the

y

-fixed rate. With

v

fixed,

y = u + v

grows with

u

, adding

x^2 = 1

❌ Check the second rate.

f = u^3 + u^2 v

: the term

u^2 v

also depends on

u

and contributes

2uv = 2

❌ Check the second rate.

1 = u^2

\left(\partial f/\partial v\right)_u

, the rate in

v

Show solution

$y$ held fixed:

\left(\frac{\partial f}{\partial x}\right)_y = 2xy = 2(1)(2) = 4

$v$ held fixed: substitute $x = u$ , $y = u + v$ :

f = u^2(u + v) = u^3 + u^2 v

\left(\frac{\partial f}{\partial u}\right)_v = 3u^2 + 2uv = 3 + 2 = 5

Check by the chain rule: with $v$ fixed, $y$ grows one-for-one with $x$ , so $\left(\partial f/\partial x\right)_v = f_x + f_y \cdot 1 = 2xy + x^2 = 4 + 1 = 5$ .

Problem 4 · A Different Dependent Variable

Given: the surface $x^2 + yz + z^3 = 8$ at the point $(2, 3, 1)$ , where the previous video found $\left(\partial z/\partial x\right)_y = -\tfrac{2}{3}$ and $\left(\partial z/\partial y\right)_x = -\tfrac{1}{6}$ . Now treat $y$ and $z$ as independent. Find $\left(\dfrac{\partial x}{\partial y}\right)_z$ .

✅ Correct! With

dz = 0

, the differential gives

4\,dx + dy = 0

, so

dx/dy = -\tfrac{1}{4}

❌ Not quite.

-4

\left(\partial y/\partial x\right)_z

, the reciprocal. Here

x

is the dependent variable, so solve for

dx

❌ Not quite.

-\tfrac{1}{6}

\left(\partial z/\partial y\right)_x

, where

z

was dependent. Now

z

is held fixed and

x

moves.

❌ Not quite.

-\tfrac{2}{3}

\left(\partial z/\partial x\right)_y

from the previous video — a different symbol with a different held-fixed variable.

Show solution

Differentiate the relation:

2x\,dx + z\,dy + y\,dz + 3z^2\,dz = 0

At $(2, 3, 1)$ :

4\,dx + dy + 6\,dz = 0

Holding $z$ fixed means $dz = 0$ , so $4\,dx = -dy$ :

\left(\frac{\partial x}{\partial y}\right)_z = -\frac{1}{4}

The subscript $z$ is held fixed, $y$ is changed, and $x$ — the letter missing — is the dependent variable.

Solved: 0 / 4

What Is Held Constant? Subscript Notation for Partial Derivatives and the Right-Triangle Area

IKey Formulas

IIVisualization 1 — Holding yyy Fixed vs Holding v=y−xv = y - xv=y−x Fixed

IIIVisualization 2 — Three Motions of the Triangle, Three Rates

IVQuiz Questions

IIVisualization 1 — Holding $y$ Fixed vs Holding $v = y - x$ Fixed