# MIT OCW

## 18.02 Multivariable Calculus

### Non-independent Variables

#### Review of Implicit Differentiation

Like the implicit differentiation in single variable calculus, there are times we cannot solve a variable as a function dependent on others (especially the powers are larger than $2$).
e.g. Given the implicit function $y^3+xy+5=0$, what is $\frac{\mathrm{d}y}{\mathrm{d}x}$?
Solution Apply derivatives at both sides we get $$3y^2\cdot\frac{\mathrm{d}y}{\mathrm{d}x}+\left(1\cdot y+x\cdot\frac{\mathrm{d}y}{\mathrm{d}x}\right)=0,$$ and the answer is $$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{y}{3y^2+x}.$$

#### Multivariable Function

When there are not only two variables in a formula, and it's not easy to solve a variable as a function dependent on others, is there similar tricks? The answer is to use the total differential.
e.g. Given an equation $x^3+yz+z^3=8$, find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ at point $(2,3,1)$.
Solution Consider a function $f(x,y,z)=x^2+yz+z^3=8$ and its total differential $$\mathrm{d}f=2x\mathrm{d}x+z\mathrm{d}y+(y+3z^2)\mathrm{d}z,$$ since $f(x,y,z)=8$ so $$\mathrm{d}f=2x\mathrm{d}x+z\mathrm{d}y+(y+3z^2)\mathrm{d}z=0.$$
And let's think about the actual meaning of partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$, it keeps a variable active and others constant, so to get $\frac{\partial z}{\partial x}$, we can set $\mathrm{d}y=0$, since $y$ is considered constant and get $$3x^2\mathrm{d}x+(y+3z^2)\mathrm{d}z=0,$$ so plug the point into it and we have $$4\mathrm{d}x+6\mathrm{d}z=0,$$ so $$\frac{\partial z}{\partial x}=-\frac23,$$ and similarly $$\frac{\partial z}{\partial y}=-\frac16.$$

##### Relationship with Implicit Differentiation

When I carefully reviewed the equation above, I was ignited by the meaning of the partial derivative and find a bridge between the method and implicit differentiation.
When we are computing $\frac{\partial z}{\partial x}$, we only care about the variation of $x$ and $z$, so we can treat anything else as constants and the problem is again solved by implicit differentiation that is $$\frac{\mathrm{d}}{\mathrm{d}x}(x^2+yz+z^3)=\frac{\mathrm{d}}{\mathrm{d}x}8,$$ and here $y$ is a number not a variable, we can get $$2x+yz'+3z^2z'=0,$$ so $$z'=-\frac{2x}{3z^2+y},$$ namely $$\frac{\partial z}{\partial x}=-\frac{2x}{3z^2+y}.$$

#### Notation Traps

Consider a function $f(x,y)=x+y$ and set $x=u,y=u+v$, so $f(u,v)=2u+v$. Compute $\frac{\partial f}{\partial x}=1,\frac{\partial f}{\partial u}=2$, but $x=u$, what is the contradiction?
Actually it's a trap from notation, because when we write $\frac{\partial f}{\partial x}$, we are assuming that $x$ is active and $y$ is constant. On the other hand, when we write $\frac{\partial f}{\partial u}$, we are assuming that $u$ is active and $v$ is constant. The common variation of $x=u$ is the same, but what we keep constant is different in these two notations. So here are another notation to clearify what we keep constant: $\left(\frac{\partial f}{\partial x}\right)_y$ indicates that $x$ is active and $y$ is constant.

#### Application

e.g. Given a right triangle with area $A$, like the following:

find $\frac{\partial A}{\partial \theta}$.
Solution We can write the expression $$A=\frac12ab\sin(\theta),$$ the key point here is that we have a constraint: a right triangle. So here exist the notation traps.

• Treat $a,b$ are independent variables
Then the answer is just the partial derivative which keeps both $a$ and $b$ constant: $\left(\frac{\partial A}{\partial \theta}\right)_{a,b}=\frac12ab\cos(\theta).$
• Treat $a,b$ are non-independent
The constraint here is exactly $a=b\cos(\theta)$, and there are another two ways: either $a$ or $b$ is constant.
• Treat $a$ constant
Method 1 (Total Differential) Now we compute the differential \begin{aligned}\mathrm{d}A&=\frac{\partial A}{\partial a}\mathrm{d}a+\frac{\partial A}{\partial b}\mathrm{d}b+\frac{\partial A}{\partial \theta}\mathrm{d}\theta\\&=\frac12b\sin(\theta)\mathrm{d}a+\frac12a\sin(\theta)\mathrm{d}b+\frac12ab\cos(\theta)\mathrm{d}\theta,\end{aligned} and another differential derived from the constraint \begin{aligned}\mathrm{d}a&=\frac{\partial a}{\partial b}\mathrm{d}b+\frac{\partial a}{\partial \theta}\mathrm{d}\theta\\&=\cos(\theta)\mathrm{d}b-b\sin(\theta)\mathrm{d}\theta.\end{aligned}
Here we're keeping $a$ constant, so $\mathrm{d}a=0$, and get the differential relationship between $b$ and $\theta$ that is $$\cos(\theta)\mathrm{d}b-b\sin(\theta)\mathrm{d}\theta=0,$$ namely $$\mathrm{d}b=b\tan(\theta)\mathrm{d}\theta.$$
Substitute $\mathrm{d}b$ and $\mathrm{d}a=0$ into the previous differential, we can get $$\mathrm{d}A=\frac12ab\sin(\theta)\tan(\theta)\mathrm{d}\theta+\frac12ab\cos(\theta)\mathrm{d}\theta,$$
so we have $$\mathrm{d}A=\frac12ab\left(\sin(\theta)\tan(\theta)+\cos(\theta)\right)\mathrm{d}\theta,$$ therefore $$\left(\frac{\partial A}{\partial \theta}\right)_{a}=\frac12ab\sec(\theta).$$
Method 2 (Chain Rule) We can have \begin{aligned}\left(\frac{\partial A}{\partial \theta}\right)_{a}&=\left(\frac{\partial A}{\partial \theta}\right)_{a}\left(\frac{\partial \theta}{\partial \theta}\right)_{a}+\left(\frac{\partial A}{\partial a}\right)_{a}\left(\frac{\partial a}{\partial \theta}\right)_{a}+\left(\frac{\partial A}{\partial b}\right)_{a}\left(\frac{\partial b}{\partial \theta}\right)_{a}\\&=\frac{\partial A}{\partial \theta}+\frac{\partial A}{\partial b}\frac{\partial b}{\partial \theta}=\frac12ab\cos(\theta)+\left(\frac12a\sin(\theta)\right)\cdot\left(a\tan(\theta)\sec(\theta)\right)\\&=\frac12ab\cos(\theta)+\frac12a\sec(\theta)\cdot\sin(\theta)\tan(\theta)\\&=\frac12ab\sec(\theta).\end{aligned}
• Treat $b$ constant
Omitted.

### Partial Differential Equations

Phenomenon in reality is generally governed by partial differential equations, like heat equation in 3D space $$\frac {\partial u}{\partial t}=\alpha \left({\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}+{\frac {\partial ^{2}u}{\partial z^{2}}}\right).$$