eifueo/docs/1b/math119.md

MATH 119: Calculus 2

Multivariable functions

!!! definition - A multivariable function accepts more than one independent variable, e.g., \(f(x, y)\).

The signature of multivariable functions is indicated in the form [identifier]: [input type] → [return type]. Where \(n\) is the number of inputs:

\[f: \mathbb R^n \to \mathbb R\]

!!! example The following function is in the form \(f: \mathbb R^2\to\mathbb R\) and maps two variables into one called \(z\) via function \(f\).

$$(x,y)\longmapsto z=f(x,y)$$

Sketching multivariable functions

!!! definition - In a scalar field, each point in space is assigned a number. For example, topography or altitude maps are scalar fields. - A level curve is a slice of a three-dimensional graph by setting to a general variable \(f(x, y)=k\). It is effectively a series of contour plots set in a three-dimensional plane. - A contour plot is a graph obtained by substituting a constant for \(k\) in a level curve.

Please see level set and contour line for example images.

In order to create a sketch for a multivariable function, this site does not have enough pictures so you should watch a YouTube video.

!!! example For the function \(z=x^2+y^2\):

For each $x, y, z$:

- Set $k$ equal to the variable and substitute it into the equation
- Sketch a two-dimensional graph with constant values of $k$ (e.g., $k=-2, -1, 0, 1, 2$) using the other two variables as axes

Combine the three **contour plots** in a three-dimensional plane to form the full sketch.

A hyperbola is formed when the difference between two points is constant. Where \(r\) is the x-intercept:

\[x^2-y^2=r^2\]

If \(r^2\) is negative, the hyperbola is is bounded by functions of \(x\), instead.

Limits of two-variable functions

A function is continuous at \((x, y)\) if and only if all possible lines through \((x, y)\) have the same limit. Or, where \(L\) is a constant:

\[\text{continuous}\iff \lim_{(x, y)\to(x_0, y_0)}f(x, y) = L\]

In practice, this means that if any two paths result in different limits, the limit is undefined. Substituting \(x|y=0\) or \(y=mx\) or \(x=my\) are common solutions.

!!! example For the function \(\lim_{(x, y)\to (0,0)}\frac{x^2}{x^2+y^2}\):

Along $y=0$:

$$\lim_{(x,0)\to(0, 0)} ... = 1$$

Along $x=0$:

$$\lim_{(0, y)\to(0, 0)} ... = 0$$

Therefore the limit does not exist.

Partial derivatives

Partial derivatives have multiple different symbols that all mean the same thing:

\[\frac{\partial f}{\partial x}=\partial_x f=f_x\]

For two-input-variable equations, setting one of the input variables to a constant will return the derivative of the slice at that constant.

By definition, the partial derivative of \(f\) with respect to \(x\) (in the x-direction) at point \((a, B)\) is:

\[\frac{\partial f}{\partial x}(a, B)=\lim_{h\to 0}\frac{f(a+h, B)-f(a, B)}{h}\]

Effectively:

• if finding \(f_x\), \(y\) should be treated as constant.
• if finding \(f_y\), \(x\) should be treated as constant.

!!! example With the function \(f(x,y)=x^2\sqrt{y}+\cos\pi y\):

\begin{align*}
f_x(1,1)&=\lim_{h\to 0}\frac{f(1+h,1)-f(1,1)} h \\
\tag*{$f(1,1)=1+\cos\pi=0$}&=\lim_{h\to 0}\frac{(1+h)^2-1} h \\
&=\lim_{h\to 0}\frac{h^2+2h} h \\
&= 2 \\
\end{align*}

Higher order derivatives

!!! definition - wrt. is short for “with respect to”.

\[\frac{\partial^2f}{\partial x^2}=\partial_{xx}f=f_{xx}\]

Derivatives of different variables can be combined:

\[f_{xy}=\frac{\partial}{\partial y}\frac{\partial f}{\partial x}=\frac{\partial^2 f}{\partial xy}\]

The order of the variables matter: \(f_{xy}\) is the derivative of f wrt. x and then wrt. y.

Clairaut’s theorem states that if \(f_x, f_y\), and \(f_{xy}\) all exist near \((a, b)\) and \(f_{yx}\) is continuous at \((a,b)\), \(f_{yx}(a,b)=f_{x,y}(a,b)\) and exists.

!!! warning In multivariable calculus, differentiability does not imply continuity.

Linear approximations

A tangent plane represents all possible partial derivatives at a point of a function.

For two-dimensional functions, the differential could be used to extrapolate points ahead or behind a point on a curve.

\[ \Delta f=f'(a)\Delta d \\ \boxed{y=f(a)+f'(a)(x-a)} \]

The equations of the two unit direction vectors in \(x\) and \(y\) can be used to find the normal of the tangent plane:

\[ \vec n=\vec d_1\times\vec d_2 \\ \begin{bmatrix}-f_x(a,b) \\ -f_y(a,b) \\ 1\end{bmatrix} = \begin{bmatrix}1\\0\\f_x(a,b)\end{bmatrix} \begin{bmatrix}0\\1\\f_y(a,b)\end{bmatrix} \]

Therefore, the general expression of a plane is equivalent to:

\[ z=C+A(x-a)+B(x-b) \\ \boxed{z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)} \]

??? tip “Proof” The general formula for a plane is \(c_1(x-a)+c_2(y-b)+c_3(z-c)=0\).

If $y$ is constant such that $y=b$:

$$z=C+A(x-a)$$

which must represent in the x-direction as an equation in the form $y=b+mx$. It follows that $A=f_x(a,b)$. A similar concept exists for $f_y(a,b)$.

If both $x=a$ and $y=b$ are constant:

$$z=C$$

where $C$ must be the $z$-point.

Usually, functions can be approximated via the tangent at \(x=a\).

\[f(x)\simeq L(x)\]

!!! warning Approximations are less accurate the stronger the curve and the farther the point is away from \(f(a,b)\). A greater \(|f''(a)|\) indicates a stronger curve.

!!! example Given the function \(f(x,y)=\ln(\sqrt[3]{x}+\sqrt[4]{y}-1)\), \(f(1.03, 0.98)\) can be linearly approximated.

$$
L(x=1.03, y=0.98)=f(1,1)=f_x(1,1)(x-1)+f_y(1,1)(y-1) \\
f(1.03,0.98)\simeq L(1.03,0.98)=0.005
$$

Differentials

Linear approximations can be used with the help of differentials. Please see MATH 117#Differentials for more information.

\(\Delta f\) can be assumed to be equivalent to \(df\).

\[\Delta f=f_x(a,b)\Delta x+f_y(a,b)\Delta y\]

Alternatively, it can be expanded in Leibniz notation in the form of a total differential:

\[df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy\]

??? tip “Proof” The general formula for a plane in three dimensions can be expressed as a tangent plane if the differential is small enough:

$$f(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(x-b)$$

As $\Delta f=f(x,y)-f(a,b)$, $\Delta x=x-a$, and $\Delta y=y-b$, it can be assumed that $\Delta x=dx,\Delta y=dy, \Delta f\simeq df$.

$$\boxed{\Delta f\simeq df=f_x(a,b)dx+f_y(a,b)dy}$$

Related rates

Please see SL Math - Analysis and Approaches 1 for more information.

!!! example For the gas law \(pV=nRT\), if \(T\) increases by 1% and \(V\) increases by 3%:

\begin{align*}
pV&=nRT \\
\ln p&=\ln nR + \ln T - \ln V \\
\tag{multiply both sides by $d$}\frac{d}{dp}\ln p(dp)&=0 + \frac{d}{dT}\ln T(dt)-\frac{d}{dV}\ln V(dV) \\
\frac{dp}{p} &=\frac{dT}{T}-\frac{dV}{V} \\
&=0.01-0.03 \\
&=-2\%
\end{align*}

Parametric curves

Because of the existence of the parameter \(t\), these expressions have some advantages over scalar equations:

• the direction of \(x\) and \(y\) can be determined as \(t\) increases, and
• the rate of change of \(x\) and \(y\) relative to \(t\) as well as each other is clearer

\[ \begin{align*} f(x,y,z)&=\begin{bmatrix}x(t) \\ y(t) \\ z(t)\end{bmatrix} \\ &=(x(t), y(t), z(t)) \end{align*} \]

The derivative of a parametric function is equal to the vector sum of the derivative of its components:

\[\frac{df}{dt}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}\]

Sometimes, the chain rule for multivariable functions creates a new branch in a tree for each independent variable.

For two-variable functions, if \(z=f(x,y)\):

\[\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\]

Sample tree diagram:

(Source: LibreTexts)

!!! example This can be extended for multiple functions — for the function \(z=f(x,y)\), where \(x=g(u,v)\) and \(y=h(u,v)\):

<img src="/resources/images/many-var-tree.jpg" width=300>(Source: LibreTexts)</img>

Determining the partial derivatives with respect to $u$ or $v$ can be done by only following the branches that end with those terms.

$$
\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u} \\
$$

!!! warning If the function only depends on one variable, \(\frac{d}{dx}\) is used. Multivariable functions must use \(\frac{\partial}{\partial x}\) to treat the other variables as constant.

Gradient vectors

The gradient vector is the vector of the partial derivatives of a function with respect to its independent variables. For \(f(x,y)\):

\[\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)\]

This allows for the the following replacements to appear more like single-variable calculus. Where \(\vec r=(x,y)\) is a desired point, \(\vec a=(a,b)\) is the initial point, and all vector multiplications are dot products:

Linear approximations are simplified to:

\[f(\vec r)=f(\vec a)+\nabla f(\vec a)\bullet(\vec r-\vec a)\]

The chain rule is also simplified to:

\[\frac{dz}{dt}=\nabla f(\vec r(t))\bullet\vec r'(t)\]

A directional derivative is any of the infinite derivatives at a certain point with the length of a unit vector. Specifically, in the unit vector direction \(\vec u\) at point \(\vec a=(a,b)\):

\[D_{\vec u}f(a_b)=\lim_{h\to 0}\frac{f(\vec a+h\vec u)\bullet f(\vec a)}{h}\]

This reduces down by taking only \(h\) as variable to:

\[D_{\vec u}f(a,b)=\nabla f(a,b)\bullet\vec u\]

Cartesian and polar coordinates can be easily converted between:

• \(x=r\sin\theta\cos\phi\)
• \(y=r\sin\theta\sin\phi\)
• \(z=r\cos\theta\)

Optimisation

Local maxima / minima exist at points where all points in a disk-like area around it do not pass that point. Practically, they must have \(\nabla f=0\).

Critical points are any point at which \(\nabla f=0|undef\). A critical point that is not a local extrema is a saddle point.

Local maxima tend to be concave down while local minima are concave up. This can be determined via the second derivative test. For the critical point \(P_0\) of \(f(x,y)\):

1. Calculate \(D(x,y)= f_{xx}f_{yy}-(f_{xy})^2\)
2. If it greater than zero, the point is an extremum
   1. If \(f_{xx}(P_0)<0\), the point is a maximum — otherwise it is a minimum
3. If it is less than zero, it is a saddle point — otherwise the test is inconclusive and you must use your eyeballs

Optimisation with constraints

If there is a limitation in optimising for \(f(x,y)\) in the form \(g(x,y)=K\), new critical points can be found by setting them equal to each other, where \(\lambda\) is the Lagrange multiplier that determines the rate of increase of \(f\) with respect to \(g\):

\[\nabla f = \lambda\nabla g, g(x,y)=K\]

The largest/smallest values of \(f(x,y)\) from the critical points return the maxima/minima. If possible, \(\nabla g=\vec 0, g(x,y)=K\) should also be tested afterward.

!!! example If \(A(x,y)=xy\), \(g(x,y)=K: x+2y=400\), and \(A(x,y)\) should be maximised:

\begin{align*}
\nabla f &= \left<y, x\right> \\
\nabla g &= \left<1, 2\right> \\
\left<y, x\right> &= \lambda \left<1, 2\right> \\
\begin{cases}
y &= \lambda \\
x &= 2\lambda \\
x + 2y &= 400 \\
\end{cases}
\\
\\
\therefore y=100,x=200,A=20\ 000
\end{align*}

??? example If \(f(x,y)=y^2-x^2\) and the constraint \(\frac{x^2}{4} + y^2=1\) must be satisfied:

\begin{align*}
\nabla f &=\left<-2x, 2y\right> \\
\nabla g &=\left<\frac{1}{2} x,2y\right> \\
\tag{$\left<0,0\right>$ does not satisfy constraints} \left<-2x,2y\right>&=\lambda\left<-\frac 1 2 x,2y\right> \\
&\begin{cases}
-2x &= \frac 1 2\lambda x \\
2y &= \lambda2y \\
\frac{x^2}{4} + y^2&= 1
\end{cases} \\
\\
2y(1-\lambda)&=0\implies y=0,\lambda=1 \\
&\begin{cases}
y=0&\implies x=\pm 2\implies\left<\pm2, 0\right> \\
\lambda=1&\implies \left<0,\pm 1\right>
\end{cases}
\\
\tag{by substitution} \max&=(2,0), (-2, 0) \\
\min&=(0, -1), (0, 1)
\end{align*}

??? example If \(f(x, y)=x^2+xy+y^2\) and the constraint \(x^2+y^2=4\) must be satisfied:

\begin{align*}
\tag{domain: bounded at $-2\leq x\leq 2$}y=\pm\sqrt{4-x^2} \\
f(x,\pm\sqrt{4-x^2}) &= x^2+(\pm\sqrt{4-x^2})x + 4-x^2 \\
\frac{df}{dx} &=\pm(\sqrt{4-x^2}-\frac{1}{2}\frac{1}{\sqrt{4-x^2}}2x(x)) \\
\tag{$f'(x)=0$} 0 &=4-x^2-x^2 \\
x &=\pm\sqrt{2} \\
\\
2+y^2 &= 4 \\
y &=\pm\sqrt{2} \\
\therefore f(x,y) &= 2, 6
\end{align*}

Alternatively, trigonometric substitution may be used to solve the system parametrically.

\begin{align*}
x^2+y^2&=4\implies &x=2\cos t \\
& &y=2\sin t \\
\therefore f(x,y) &= 4+2\sin(2t),0\leq t\leq 2\pi \\
\tag{include endpoints $0,2\pi$}t &= \frac\pi 4,\frac{3\pi}{4},\frac{5\pi}{4} \\
\end{align*}

!!! warning Terms cannot be directly cancelled out in case they are zero.

This applies equally to higher dimensions and constraints by adding a new term for each constraint. Given \(f(x,y,z)\) with constraints \(g(x,y,z)=K\) and \(h(x,y,z)=M\):

\[\nabla f=\lambda_1\nabla g + \lambda_2\nabla h\]

Absolute extrema

• If end points exist, those should be added
• If no endpoints exist and the limits go to \(\pm\infty\), there are no absolute extrema