# 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](https://en.wikipedia.org/wiki/Level_set) and [contour line](https://en.wikipedia.org/wiki/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](/1a/math117/#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](/g11/mhf4u7/#related-rates) 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 $y=h(u,v)$: (Source: LibreTexts) 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 a. 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 $g$: $$\nabla f = \lambda\nabla g, g(x,y)=K$$ If possible, $\nabla g=\vec 0, g(x,y)=K$ should also be tested. !!! example If $A(x,y)=xy$, $g(x,y)=K: x+2y=400$, and $A(x,y)$ should be maximised: \begin{align*} \nabla f &= (y, x) \\ \nabla g &= (1, 2) \\ (y, x) &= \lambda (1, 2) \\ \begin{cases} y &= \lambda \\ x &= 2\lambda \\ x + 2y &= 400 \\ \end{cases} \\ \\ \therefore y=100,x=200,A=20\ 000 \end{align*} 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