248 lines
8.7 KiB
Markdown
248 lines
8.7 KiB
Markdown
# MATH 119: Calculus 2
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## Multivariable functions
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!!! definition
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- A **multivariable function** accepts more than one independent variable, e.g., $f(x, y)$.
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The signature of multivariable functions is indicated in the form *[identifier]*: *[input type]* → *[return type]*. Where $n$ is the number of inputs:
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$$f: \mathbb R^n \to \mathbb R$$
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!!! example
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The following function is in the form $f: \mathbb R^2\to\mathbb R$ and maps two variables into one called $z$ via function $f$.
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$$(x,y)\longmapsto z=f(x,y)$$
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### Sketching multivariable functions
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!!! definition
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- In a **scalar field**, each point in space is assigned a number. For example, topography or altitude maps are scalar fields.
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- 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.
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- A **contour plot** is a graph obtained by substituting a constant for $k$ in a level curve.
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Please see [level set](https://en.wikipedia.org/wiki/Level_set) and [contour line](https://en.wikipedia.org/wiki/Contour_line) for example images.
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In order to create a sketch for a multivariable function, this site does not have enough pictures so you should watch a YouTube video.
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!!! example
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For the function $z=x^2+y^2$:
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For each $x, y, z$:
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- Set $k$ equal to the variable and substitute it into the equation
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- 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
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Combine the three **contour plots** in a three-dimensional plane to form the full sketch.
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A **hyperbola** is formed when the difference between two points is constant. Where $r$ is the x-intercept:
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$$x^2-y^2=r^2$$
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<img src="/resources/images/hyperbola.svg" width=600 />
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If $r^2$ is negative, the hyperbola is is bounded by functions of $x$, instead.
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## Limits of two-variable functions
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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:
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$$\text{continuous}\iff \lim_{(x, y)\to(x_0, y_0)}f(x, y) = L$$
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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.
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!!! example
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For the function $\lim_{(x, y)\to (0,0)}\frac{x^2}{x^2+y^2}$:
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Along $y=0$:
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$$\lim_{(x,0)\to(0, 0)} ... = 1$$
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Along $x=0$:
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$$\lim_{(0, y)\to(0, 0)} ... = 0$$
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Therefore the limit does not exist.
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## Partial derivatives
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Partial derivatives have multiple different symbols that all mean the same thing:
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$$\frac{\partial f}{\partial x}=\partial_x f=f_x$$
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For two-input-variable equations, setting one of the input variables to a constant will return the derivative of the slice at that constant.
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By definition, the **partial** derivative of $f$ with respect to $x$ (in the x-direction) at point $(a, B)$ is:
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$$\frac{\partial f}{\partial x}(a, B)=\lim_{h\to 0}\frac{f(a+h, B)-f(a, B)}{h}$$
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Effectively:
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- if finding $f_x$, $y$ should be treated as constant.
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- if finding $f_y$, $x$ should be treated as constant.
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!!! example
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With the function $f(x,y)=x^2\sqrt{y}+\cos\pi y$:
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\begin{align*}
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f_x(1,1)&=\lim_{h\to 0}\frac{f(1+h,1)-f(1,1)} h \\
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\tag*{$f(1,1)=1+\cos\pi=0$}&=\lim_{h\to 0}\frac{(1+h)^2-1} h \\
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&=\lim_{h\to 0}\frac{h^2+2h} h \\
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&= 2 \\
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\end{align*}
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### Higher order derivatives
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!!! definition
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- **wrt.** is short for "with respect to".
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$$\frac{\partial^2f}{\partial x^2}=\partial_{xx}f=f_{xx}$$
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Derivatives of different variables can be combined:
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$$f_{xy}=\frac{\partial}{\partial y}\frac{\partial f}{\partial x}=\frac{\partial^2 f}{\partial xy}$$
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The order of the variables matter: $f_{xy}$ is the derivative of f wrt. x *and then* wrt. y.
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**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.
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!!! warning
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In multivariable calculus, **differentiability does not imply continuity**.
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### Linear approximations
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A **tangent plane** represents all possible partial derivatives at a point of a function.
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For two-dimensional functions, the differential could be used to extrapolate points ahead or behind a point on a curve.
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$$
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\Delta f=f'(a)\Delta d \\
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\boxed{y=f(a)+f'(a)(x-a)}
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$$
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The equations of the two unit direction vectors in $x$ and $y$ can be used to find the normal of the tangent plane:
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$$
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\vec n=\vec d_1\times\vec d_2 \\
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\begin{bmatrix}-f_x(a,b) \\ -f_y(a,b) \\ 1\end{bmatrix} = \begin{bmatrix}1\\0\\f_x(a,b)\end{bmatrix}
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\begin{bmatrix}0\\1\\f_y(a,b)\end{bmatrix}
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$$
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Therefore, the general expression of a plane is equivalent to:
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$$
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z=C+A(x-a)+B(x-b) \\
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\boxed{z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)}
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$$
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??? tip "Proof"
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The general formula for a plane is $c_1(x-a)+c_2(y-b)+c_3(z-c)=0$.
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If $y$ is constant such that $y=b$:
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$$z=C+A(x-a)$$
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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 $f_y(a,b)$.
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If both $x=a$ and $y=b$ are constant:
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$$z=C$$
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where $C$ must be the $z$-point.
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Usually, functions can be approximated via the **tangent at $x=a$.**
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$$f(x)\simeq L(x)$$
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!!! warning
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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.
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!!! example
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Given the function $f(x,y)=\ln(\sqrt[3]{x}+\sqrt[4]{y}-1)$, $f(1.03, 0.98)$ can be linearly approximated.
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$$
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L(x=1.03, y=0.98)=f(1,1)=f_x(1,1)(x-1)+f_y(1,1)(y-1) \\
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f(1.03,0.98)\simeq L(1.03,0.98)=0.005
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$$
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### Differentials
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Linear approximations can be used with the help of differentials. Please see [MATH 117#Differentials](/1a/math117/#differentials) for more information.
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$\Delta f$ can be assumed to be equivalent to $df$.
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$$\Delta f=f_x(a,b)\Delta x+f_y(a,b)\Delta y$$
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Alternatively, it can be expanded in Leibniz notation in the form of a **total differential**:
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$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$
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??? tip "Proof"
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The general formula for a plane in three dimensions can be expressed as a tangent plane if the differential is small enough:
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$$f(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(x-b)$$
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As $\Delta f=f(x,y)-f(a,b)$, $\Delta x=x-a$, and $\Delta y=y-b$, it can be assumed that $\Delta x=dx,\Delta y=dy, \Delta f\simeq df$.
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$$\boxed{\Delta f\simeq df=f_x(a,b)dx+f_y(a,b)dy}$$
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### Related rates
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Please see [SL Math - Analysis and Approaches 1](/g11/mhf4u7/#related-rates) for more information.
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!!! example
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For the gas law $pV=nRT$, if $T$ increases by 1% and $V$ increases by 3%:
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\begin{align*}
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pV&=nRT \\
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\ln p&=\ln nR + \ln T - \ln V \\
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\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) \\
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\frac{dp}{p} &=\frac{dT}{T}-\frac{dV}{V} \\
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&=0.01-0.03 \\
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&=-2\%
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\end{align*}
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### Parametric curves
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Because of the existence of the parameter $t$, these expressions have some advantages over scalar equations:
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- the direction of $x$ and $y$ can be determined as $t$ increases, and
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- the rate of change of $x$ and $y$ relative to $t$ as well as each other is clearer
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$$
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\begin{align*}
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f(x,y,z)&=\begin{bmatrix}x(t) \\ y(t) \\ z(t)\end{bmatrix} \\
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&=(x(t), y(t), z(t))
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\end{align*}
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$$
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The **derivative** of a parametric function is equal to the vector sum of the derivative of its components:
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$$\frac{df}{dt}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}$$
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Sometimes, the **chain rule for multivariable functions** creates a new branch in a tree for each independent variable.
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For two-variable functions, if $z=f(x,y)$:
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$$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$$
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Sample tree diagram:
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<img src="/resources/images/two-var-tree.jpg" width=300>(Source: LibreTexts)</img>
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!!! example
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This can be extended for multiple functions — for the function $z=f(x,y)$, where $x=g(u,v)$ and $y=h(u,v)$:
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<img src="/resources/images/many-var-tree.jpg" width=300>(Source: LibreTexts)</img>
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Determining the partial derivatives with respect to $u$ or $v$ can be done by only following the branches that end with those terms.
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$$
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\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u} \\
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$$
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!!! warning
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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.
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