Multivariable-Calculus · Unit 13 · Video 3 · Interactive Practice

Computing Constrained Partial Derivatives: Total Differentials vs. the Chain Rule

IKey Formulas

Formula	Name	What it does
$A = \tfrac{1}{2}ab\sin\theta,\quad a = b\cos\theta$	Setup	Area of the right triangle; $a$ is held fixed and $b$ is the dependent variable
$b = a\sec\theta \;\Rightarrow\; A = \tfrac{1}{2}a^2\tan\theta \;\Rightarrow\; \left(\dfrac{\partial A}{\partial\theta}\right)_a = \tfrac{1}{2}a^2\sec^2\theta$	Method 0: solve and substitute	Works only when the constraint can be solved for the dependent variable
$dA = A_\theta\,d\theta + A_a\,da + A_b\,db$ , with $da = 0$ and $da = \cos\theta\,db - b\sin\theta\,d\theta$	Method 1: total differentials	The constraint gives $db = b\tan\theta\,d\theta$ ; substitute and read off the coefficient of $d\theta$
$\left(\dfrac{\partial A}{\partial\theta}\right)_a = A_\theta + A_b\left(\dfrac{\partial b}{\partial\theta}\right)_a = \tfrac{1}{2}ab\sec\theta$	Method 2: chain rule	$\left(\frac{\partial\theta}{\partial\theta}\right)_a = 1$ , $\left(\frac{\partial a}{\partial\theta}\right)_a = 0$ , and $\left(\frac{\partial b}{\partial\theta}\right)_a = b\tan\theta$ from the constraint

Key Insight: Divide $dA = A_\theta\,d\theta + A_a\,da + A_b\,db$ by $d\theta$ and put the subscript $a$ on every rate: that is the chain-rule formula. Both methods differentiate the constraint instead of solving it, and both agree with method 0, since $\tfrac{1}{2}ab\sec\theta = \tfrac{1}{2}a^2\sec^2\theta$ when $b = a\sec\theta$ .

IIWhy the Answer Is Not Just $A_\theta$

Holding $a$ fixed, the apex moves in two pieces: rotate with $b$ frozen ( $A_\theta\,d\theta$ ), then stretch $b$ ( $A_b\,db$ ).

Angle θ 35° Step Δθ 8°

💡 With $b$ frozen the shape is a triangle with sides $a$ , $b$ and included angle $\theta$ , so $A = \tfrac{1}{2}ab\sin\theta$ still gives its area, but it no longer satisfies $a = b\cos\theta$ . The $A_b\,db$ term is what brings the apex back onto the constraint.

IIIOne Computation, Written Two Ways

Each step of the total-differential method has a matching chain-rule step; dividing by $d\theta$ turns one into the other.

Step 1

1 · Total differentials

Write $dA$ with the formal partials:

\begin{aligned} dA &= A_\theta\,d\theta + A_a\,da + A_b\,db \\ &= \tfrac{1}{2}ab\cos\theta\,d\theta + \tfrac{1}{2}b\sin\theta\,da + \tfrac{1}{2}a\sin\theta\,db \end{aligned}

2 · Chain rule

Every rate is taken with $a$ held fixed:

\begin{aligned} \left(\frac{\partial A}{\partial\theta}\right)_a = {} & A_\theta\left(\frac{\partial\theta}{\partial\theta}\right)_a + A_a\left(\frac{\partial a}{\partial\theta}\right)_a \\ & + A_b\left(\frac{\partial b}{\partial\theta}\right)_a \end{aligned}

IVQuiz Questions

Problem 1 · Evaluate the Constrained Rate

Given: the right triangle with $A = \tfrac{1}{2}ab\sin\theta$ and $a = b\cos\theta$ , at $a = 2$ and $\theta = \tfrac{\pi}{4}$ — find $\left(\dfrac{\partial A}{\partial\theta}\right)_a$ .

✅ Correct!

\tfrac{1}{2}a^2\sec^2\theta = \tfrac{1}{2}\cdot 4\cdot 2 = 4

, and

\tfrac{1}{2}ab\sec\theta

with

b = 2\sqrt{2}

gives the same value.

❌ Not quite.

2

A_\theta = \tfrac{1}{2}ab\cos\theta

, the rate with

b

frozen. Holding

a

fixed makes

b

grow, and that adds

A_b\left(\frac{\partial b}{\partial\theta}\right)_a = 2

❌ Close, but… the derivative of

\tan\theta

\sec^2\theta

, not

\sec\theta

. With

\sec\frac{\pi}{4} = \sqrt{2}

, squaring gives

2

❌ Not quite. The constant

\tfrac{1}{2}a^2

comes out of the derivative intact:

\tfrac{1}{2}\cdot 4\cdot\sec^2\tfrac{\pi}{4}

, not

4\cdot\sec^2\tfrac{\pi}{4}

Show solution

Method 0 applies, since the constraint solves for $b$ : $b = a\sec\theta$ , so $A = \tfrac{1}{2}a\cdot a\sec\theta\cdot\sin\theta = \tfrac{1}{2}a^2\tan\theta$ . With $a$ fixed this is an ordinary derivative:

\left(\frac{\partial A}{\partial\theta}\right)_a = \tfrac{1}{2}a^2\sec^2\theta = \tfrac{1}{2}\cdot 2^2\cdot\left(\sqrt{2}\right)^2 = 4.

Check with methods 1 and 2: $b = 2\sec\tfrac{\pi}{4} = 2\sqrt{2}$ , and $\tfrac{1}{2}ab\sec\theta = \tfrac{1}{2}\cdot 2\cdot 2\sqrt{2}\cdot\sqrt{2} = 4$ .

Problem 2 · The Missing Term

Given: a student differentiates $A = \tfrac{1}{2}ab\sin\theta$ with respect to $\theta$ , treating $a$ and $b$ as constants, and reports $\left(\dfrac{\partial A}{\partial\theta}\right)_a = \tfrac{1}{2}ab\cos\theta$ — find the term the student left out.

✅ Correct!

b

is the dependent variable, so it changes with

\theta

A_b\left(\frac{\partial b}{\partial\theta}\right)_a = \tfrac{1}{2}a\sin\theta\cdot b\tan\theta

❌ Not quite.

a

is the variable held fixed, so

\left(\frac{\partial a}{\partial\theta}\right)_a = 0

and the

A_a

term vanishes.

❌ Close, but…

A_b

is only a coefficient. It must multiply the rate

\left(\frac{\partial b}{\partial\theta}\right)_a = b\tan\theta

❌ Not quite. Treating

b

as constant while

a

is fixed breaks

a = b\cos\theta

: opening the angle with both sides frozen loses the right angle.

Show solution

The chain rule with $a$ held fixed gives

\left(\frac{\partial A}{\partial\theta}\right)_a = A_\theta\cdot 1 + A_a\cdot 0 + A_b\left(\frac{\partial b}{\partial\theta}\right)_a.

The student kept only $A_\theta = \tfrac{1}{2}ab\cos\theta$ . Differentiating the constraint gives $\left(\frac{\partial b}{\partial\theta}\right)_a = b\tan\theta$ , so the missing term is $\tfrac{1}{2}a\sin\theta\cdot b\tan\theta = \tfrac{1}{2}ab\sin\theta\tan\theta$ . Together:

\tfrac{1}{2}ab\left(\cos\theta + \frac{\sin^2\theta}{\cos\theta}\right) = \tfrac{1}{2}ab\cdot\frac{\cos^2\theta + \sin^2\theta}{\cos\theta} = \tfrac{1}{2}ab\sec\theta.

Problem 3 · Differentials on a Constraint You Cannot Solve

Given: $w = xy + z^2$ , where $x$ , $y$ , $z$ satisfy $x^2 + yz + z^3 = 8$ (a cubic in $z$ ). At the point $(x, y, z) = (2, 3, 1)$ — find $\left(\dfrac{\partial w}{\partial x}\right)_y$ .

✅ Correct!

dz = -\tfrac{2}{3}\,dx

on the constraint with

dy = 0

, so

dw = 3\,dx + 2\left(-\tfrac{2}{3}\right)dx = \tfrac{5}{3}\,dx

❌ Not quite.

3

is the formal partial

w_x = y

. Here

z

is the dependent variable, so

w_z\,dz

contributes too.

❌ Close, but check the sign of $dz$ . From

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

dz = -\tfrac{2x}{y + 3z^2}\,dx

, which is negative.

❌ Close, but…

dz

enters

dw

multiplied by its coefficient

w_z = 2z = 2

Show solution

The point is on the constraint: $4 + 3 + 1 = 8$ . Step 1: $dw = y\,dx + x\,dy + 2z\,dz$ . Step 2: $y$ is fixed, so $dy = 0$ .

Step 3: differentiate the constraint: $2x\,dx + z\,dy + (y + 3z^2)\,dz = 0$ . With $dy = 0$ at $(2, 3, 1)$ : $4\,dx + 6\,dz = 0$ , so $dz = -\tfrac{2}{3}\,dx$ .

Step 4: substitute:

dw = 3\,dx + 2\cdot 1\cdot\left(-\tfrac{2}{3}\right)dx = \tfrac{5}{3}\,dx \quad\Rightarrow\quad \left(\frac{\partial w}{\partial x}\right)_y = \tfrac{5}{3}.

Method 0 would need $z$ solved from a cubic; the differential never solves the constraint.

Problem 4 · Hold the Hypotenuse Instead

Given: the same triangle, $A = \tfrac{1}{2}ab\sin\theta$ with $a = b\cos\theta$ , but now the hypotenuse $b$ is held fixed and $a$ is dependent — find $\left(\dfrac{\partial A}{\partial\theta}\right)_b$ .

✅ Correct!

A_\theta + A_a\left(\frac{\partial a}{\partial\theta}\right)_b = \tfrac{1}{2}b^2\cos^2\theta - \tfrac{1}{2}b^2\sin^2\theta = \tfrac{1}{2}b^2\cos 2\theta

; it vanishes at

\theta = \tfrac{\pi}{4}

, where the area is largest.

❌ Not quite. That is the rate with

a

fixed. With

b

fixed the dependent variable is

a

, and

\left(\frac{\partial a}{\partial\theta}\right)_b = -b\sin\theta

changes the second term.

❌ Not quite.

\tfrac{1}{2}ab\cos\theta

is only

A_\theta

. The leg

a = b\cos\theta

shrinks as

\theta

opens, which adds

A_a\left(\frac{\partial a}{\partial\theta}\right)_b

❌ Close, but… that is only the dependent-variable term

A_a\left(\frac{\partial a}{\partial\theta}\right)_b

. The

A_\theta\left(\frac{\partial\theta}{\partial\theta}\right)_b = A_\theta

term stays.

Show solution

Chain rule with $b$ held fixed: $\left(\frac{\partial b}{\partial\theta}\right)_b = 0$ , so

\left(\frac{\partial A}{\partial\theta}\right)_b = A_\theta + A_a\left(\frac{\partial a}{\partial\theta}\right)_b.

Differentiate $a = b\cos\theta$ with $b$ fixed: $\left(\frac{\partial a}{\partial\theta}\right)_b = -b\sin\theta$ . With $A_\theta = \tfrac{1}{2}ab\cos\theta$ and $A_a = \tfrac{1}{2}b\sin\theta$ :

\left(\frac{\partial A}{\partial\theta}\right)_b = \tfrac{1}{2}ab\cos\theta - \tfrac{1}{2}b^2\sin^2\theta = \tfrac{1}{2}b^2\cos^2\theta - \tfrac{1}{2}b^2\sin^2\theta = \tfrac{1}{2}b^2\cos 2\theta,

using $a = b\cos\theta$ . Method 0 agrees: $A = \tfrac{1}{2}b^2\sin\theta\cos\theta = \tfrac{1}{4}b^2\sin 2\theta$ , whose $\theta$ -derivative is $\tfrac{1}{2}b^2\cos 2\theta$ .

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Computing Constrained Partial Derivatives: Total Differentials vs. the Chain Rule

IKey Formulas

IIWhy the Answer Is Not Just AθA_\thetaAθ​

IIIOne Computation, Written Two Ways

IVQuiz Questions

IIWhy the Answer Is Not Just $A_\theta$