MAT021C: Calculus: Partial Derivatives and Series

Incomplete

My notes for MAT21C is incomplete since I regularly skip classes and only take notes on problems on the homework that I don’t know or forgot how to do from AP Calculus AB/BC.

Series

N: common Maclaurin series

common Maclaurin series

$$\begin{array}{rlrl}\frac{1}{1-x}& =\sum _{n=0}^{\infty }{x}^n& & =1+x+x^2+x^3+\dots \\ e^x& =\sum _{n=0}^{\infty }\frac{x^n}{n!}& & =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \\ \cos x& =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n)!}& & =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \\ \sin x& =\sum _{n=1}^{\infty }\frac{(-1{)}^{n-1}{x}^{2n-1}}{(2n-1)!}& & =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots \\ \ln (1+x)& =\sum _{n=1}^{\infty }\frac{(-1{)}^{n-1}{x}^n}{n}& & =x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots \\ \arctan x& =\sum _{n=1}^{\infty }\frac{(-1{)}^{n-1}{x}^{2n-1}}{2n-1}& & =x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dots \end{array}$$Link to original

P: binomial series expansion

binomial function Taylor series

We have a binomial function $f_m(x)=(1+x{)}^m$ where $m\in \mathbb{R}$.

If $m$ is a positive integer, then the binomial function $f_m$ is a polynomial, and its Taylor series would be the same polynomial.

If $m$ is not a positive integer, then its Taylor series has infinitely many non-zero terms. The Taylor series converges for $|x|<1$ and is given by:

$$T(x)=1+\sum _{n=1}^{\infty }\left(\genfrac{}{}{0ex}{}{m}{n}\right)x^n$$

with binomial coefficients $\left(\genfrac{}{}{0ex}{}{m}{1}\right)=m$, $\left(\genfrac{}{}{0ex}{}{m}{2}\right)=\frac{m(m-1)}{2!}$, and so on.

Link to original

Algebra in 3D

P: angle between vectors

angle between vectors

The angle between vectors can be determined with dot product:

$$\cos \theta =\frac{\vec{u}\cdot \vec{v}}{|\vec{u}|\cdot |\vec{v}|}$$

or alternatively using cross product:

$$\sin \theta =\frac{\vec{u}\times \vec{v}}{|\vec{u}|\cdot |\vec{v}|}$$Link to original

P: Two vectors orthogonal?

Two vectors are orthogonal if $\vec{u}\cdot \vec{v}=0$.

dot product

The dot product or scalar product of $\vec{u}$ and $\vec{v}$ is $\vec{u}\cdot \vec{v}$. The resulting value is a scalar. The dot product is 0 when the vectors are orthogonal to each other.

$\vec{A}\cdot \vec{B}=AB\cos \varphi =|\vec{A}||\vec{B}|\cos \varphi$

• length of A (or B) times the length of A (or B) in the direction of B (or A), a.k.a the length of the other side of the triangle
• $\varphi$ is the acute or flat angle between $\vec{A}$ and $\vec{B}$

The dot product is equal to the sum of the product of the corresponding components. For example, if they are two dimensional, then: $\vec{u}\cdot \vec{v}=u_x\cdot v_x+u_y\cdot v_y$

# assume u, v are two tuples of arbitrary dimensions
assert len(u) == len(v)
u_dot_v = sum([u_i * v_i for u_i, v_i in zip(u, v)])

Link to original

P: Three vectors in same plane?

Yes if the triple scalar product is 0.

triple scalar product

The triple scalar product of $\vec{u}$, $\vec{v}$, and $\vec{w}$ is the determinant of the vertically concatenated matrix:

$$\left|\begin{array}{ccc}{u}_x& u_y& v_z\\ v_x& v_y& v_z\\ w_x& w_y& v_z\end{array}\right|$$

or in short:

$$\left|\begin{array}{c}\vec{u}\\ \vec{v}\\ \vec{w}\end{array}\right|$$

The above is equivalent to $\vec{u}\cdot (\vec{v}\times \vec{w})$

Triple scalar product of zero

The triple scalar product of three vectors is 0 if they are in the same plane.

Link to original

P: two plane intersection (parametric line eqn)

find a parametric equation for the line where two planes intersect

1. Find normal vectors for the two plane. ![[normal vector#find—vecn-of-a-plane|Find $vecn$ of a plane]]
2. The intersecting line has the same direction as $\overrightarrow{n_1}\times \overrightarrow{n_2}$. Find the cross product.
   • NOTE: You don’t have to find the unit vector (direction) of the cross product, since the magnitude doesn’t matter, only the direction and where the initial point of the vector do (which we will determine later.
3. Find a point $P$ shared by both planes where we can place the direction vector of the line.

   find a point shared by two planes

   • Case 1: the point is apparent in the plane equations.
     • $3(x+2)-2(y-1)+2(z+1)=0$
     • $(x+2)+2(y-1)-3(z+1)=0$
     • Common point: $(-2,1,-1)$
   • Case 2: try setting one variable to zero, plug it into the plane equations, then solve the resulting system of equations (two equations, two variables). See this post.

   Link to original

4. Create a parametric equation now that we have $\overrightarrow{n_1}\times \overrightarrow{n_2}$ and point $P$.

   parametric equation of line from vector and point

   Given vector $\vec{u}$ and point $(x_0,y_0,z_0)$, a parametric equation for a line can be found:

   $$\begin{array}{rl}x& =x_0+u_x\cdot t\\ y& =y_0+u_y\cdot t\\ z& =z_0+u_z\cdot t\end{array}$$Link to original

Link to original

P: parametric equation from vector and point

parametric equation of line from vector and point

Given vector $\vec{u}$ and point $(x_0,y_0,z_0)$, a parametric equation for a line can be found:

$$\begin{array}{rl}x& =x_0+u_x\cdot t\\ y& =y_0+u_y\cdot t\\ z& =z_0+u_z\cdot t\end{array}$$Link to original

P: normal vector to surface at point

normal vector to surface at point

The normal vector to surface $f$ is given by the gradient $\nabla f$. Evaluate the function at point to obtain the normal vector tangent to surface at point.

Link to original

P: tangent plane to surface

tangent plane to surface at point

To find the plane, first get its normal vector.

normal vector to surface at point

The normal vector to surface $f$ is given by the gradient $\nabla f$. Evaluate the function at point to obtain the normal vector tangent to surface at point.

Link to original

Given the point and a normal vector, it is easy to make a plane.

plane from point and normal vector

Given a point $(x_0,y_0,z_0)$ and normal vector $\vec{n}$, a plane can be defined as follows:

$$\begin{array}{rl}{n}_x(x-x_0)+n_y(y-y_0)+n_z(z-z_0)& =0\\ n_x\cdot x+n_y\cdot y+n_z\cdot z& =n_x\cdot x_0+n_y\cdot y_0+n_z\cdot z_0\end{array}$$Link to original

Link to original

P: area of parallelogram given three vertices

area of parallelogram given three vertices

Problem

Given three vertices of the parallelogram $P$, $Q$, and $R$, calculate the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.

Solution

The area of the parallelogram is $\overrightarrow{PQ}\times \overrightarrow{PR}$, their cross product.

Link to original

P: area of triangle given three vertices

area of triangle given three vertices

Problem

Given three vertices of the triangle $P$, $Q$, and $R$, calculate the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.

Solution

The area of the parallelogram is $\overrightarrow{PQ}\times \overrightarrow{PR}$, their cross product. The corresponding triangle is half the area.

Link to original

P: volume of parallelopiped given four vertices

volume of parallelopiped given four vertices

Assume we are given four vertices $A$ to $D$. Treat the $A$ as the origin and calculate three vectors $\vec{P}=\overrightarrow{AB}$, $\vec{Q}=\overrightarrow{AC}$, and $\vec{R}=\overrightarrow{AD}$. These three vectors are the three edges ending on the same vertex. The volume is the triple scalar product of $\vec{P}$, $\vec{Q}$, and $\vec{R}$, i.e., $\vec{P}\cdot (\vec{Q}\times \vec{R})$, or find the determinant of their matrix from vertical concatenation).

Link to original

P: intersection of lines or minimum distance between lines

intersection of lines or minimum distance between lines

Without already knowing if the lines intersect, we have to follow through these steps one by one.

Parallel - distance

First check whether the lines are parallel. The lines are parallel if the cross product of the direction vectors is zero. To find distance between parallel lines: Find one point on each line. Let’s call them $P$ and $Q$. Calculate the vector $\vec{u}=\overrightarrow{PQ}$

The distance $d$ between the lines is $d=\frac{|\vec{u}\times \vec{v}|}{|\vec{v}|}$ where $\vec{v}$ is the direction vector for any one of the two lines.

Intersection?

If the lines aren’t parallel, we need to determine whether the lines intersect or are skew. Treat the two parametric equations as a system of equations, and solve for both parameters. If the solutions are valid (check with all equations), then we have an intersection, otherwise the lines are skew. Detailed steps below:

• Set the first pair of parametric equations (both $x=\cdots$) equal to each other. Solve for one of the parameter (parametric variable).
• Set the second pair of equations equal to each other. Plug the solved parameter in, and solve for the other parameter.
• Set the third pair equal to each other. Plug the first parameter into it and solve for the other one. Check if the solution for the parametric variable is consistent with step 2. If not, then the lines are skew. If it is, then plug the one parameter into a line equation to obtain the intersection coordinate.

Skew!

To calculate distance between skew lines:

• Find the cross product $\vec{N}$ between the two direction vectors. If the two lines were moved to the same plane, $\vec{N}$ would be its normal vector.
• Then find any point on each line ($P$ and $Q$) and calculate the vector $\overrightarrow{PQ}$. Note that the $\vec{N}$-component of $\overrightarrow{PQ}$ is the basically distance between these two lines. We can calculate that using projection.
• The distance between the two lines is the projection of $\overrightarrow{PQ}$ onto $\vec{N}$: $d=\left|{\mathrm{proj}}_{\vec{N}}\overrightarrow{PQ}\right|=\frac{\overrightarrow{PQ}\cdot \vec{N}}{|\vec{N}|}$

Link to original

P: distance between point and line

distance between point and line

• Find a point $Q$ on the line.
• The distance can be calculated from:

$$d=\frac{|\overrightarrow{PQ}\times \vec{v}|}{|\vec{v}|}$$

where $\vec{v}$ is the direction vector.

Link to original

P: Projection vector

projection vector

The projection vector of $\vec{u}$ onto $\vec{v}$ is a vector in $\vec{v}$ direction with a magnitude of $\vec{u}$‘s $\vec{v}$-component.

Its magnitude is

$$\left|{\mathrm{proj}}_{\vec{v}}\vec{u}\right|=\frac{\vec{u}\cdot \vec{v}}{\left|\vec{v}\right|}=\frac{\vec{u}\cdot \vec{v}}{\vec{v}\cdot \vec{v}}$$

To make it a vector in $\vec{v}$‘s direction, multiply it by $\frac{\vec{v}}{|\vec{v}|}$, which is $\vec{v}$‘s direction vector.

Link to original

Multivariate calculus

P: partial differentiation

partial differentiation

The partial differentiation of an equation with respect to $x$ ($\frac{\partial }{\partial x}$) can be obtained by differentiating the equation with respect to $x$ and treating other variables as if they were constants. For example, $\frac{\partial }{\partial x}(3x+2xy)=3+2y$ .

For partial differentiation involving underlying variables, use chain rule and chain rule dependency diagram (e.g. $\frac{\partial y}{\partial t}$ when $y$ is defined not in terms of $t$, but some other variables that can be defined with $t$)

Link to original

P: chain rule dependency diagram

chain rule dependency diagram

A chain rule dependency diagram is useful for calculating more involved partial differentiations using chain rule.

To draw a diagram for $\frac{\partial z}{\partial t}$, start from the initial function/variable, say $z$, which is defined in terms of undesired variables (say $x$ and $y$). The diagram starts with $z$ as the first node. Add the variables as children to the current variable’s node. Do this repeatedly to all the newly added child nodes (leaves) until they reach at the desired variable ($t$ in this case). Each edge in the diagram represents a partial derivative ($\frac{\partial p}{\partial c}$ where $p$ is the parent variable and $c$ is child variable). For each possible path from the initial to the desired variable ($z$ to $t$), multiply all the edges (e.g. $z$-$x$-$t$ becomes $\frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial t}$). Sum all the paths together to get the desired partial differentiation (e.g. $\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial t}$).

Link to original

P: gradient

gradient

Gradient of a multivariate function at a point is a vector pointing at the direction of steepest increase in value, with a magnitude of the function’s directional derivative in the pointed direction. each partial derivative. Its components consist of partial derivatives of the corresponding variables. For instance, $\nabla f(x,y,z)=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}+\frac{\partial f}{\partial z}\mathbf{k}$

Link to original

P: find direction of maximum growth and decrease of function at point

find direction of maximum growth and decrease of function at point

At a point $(a,b,c)$, the maximum growth is in the direction of the gradient ($\frac{\nabla }{|\nabla |}$), and the maximum decrease is in the opposite direction ($-\frac{\nabla }{|\nabla |}$). This can be used for optimization (e.g. stochastic gradient descent).

Link to original

P: directional derivative of f in direction of u

directional derivative

The directional derivative of $f$ in the direction $\vec{u}$, assuming that $\vec{u}$ is a direction vector.

$$\begin{array}{rl}{D}_{\vec{u}}f(x,y,z)& =f_x\cdot u_x+f_y\cdot u_y+f_z\cdot u_z\\ & =\nabla f(x,y,z)\cdot \frac{\vec{u}}{|\vec{u}|}\\ & =⟨f_x,f_y,f_z⟩\cdot \frac{\vec{u}}{|\vec{u}|}\end{array}$$

where $f_x$ and so on are the partial derivatives with respect to the subscripted variable.

If $\vec{u}$ is the direction vector ($|\vec{u}|=1$) of $\nabla f$ at point $P$, then:

$$D_{\vec{u}}f(x,y,z)=\nabla f\cdot \vec{u}$$Link to original

P: critical points of multivariate function

find critical points of multivariate function

Problem

Assume the function $z=f$ is defined in terms of $x$ and $y$, i.e. $z=f(x,y)$. Find the critical points (any minimum, maximum, or saddle point).

At critical points, $f_x=0$ and $f_y=0$, where $f_{(\cdot )}$ is the partial derivative with respect to the corresponding variable. Solve the system of equations for $x$ and $y$. Check the coordinates for extraneous solutions (not sure if this can happen but it might).

second derivative test

For a univariate function $f(x)$, a second derivative test tells:

• if $f(x)$ is concave up (if $f^{\prime \prime }>0$)
• if $f(x)$ is concave down (if $f^{\prime \prime }<0$)
• where inflection points are (where concavity changes)

If $f$ is a multivariate function, to determine whether $z=f(x,y)$ has a minimum, maximum, or saddle point at the critical point $P=(x_0,y_0)$, we can use the second derivative test.

First find $f_{xx}$, $f_{yy}$, and $f_{xy}$ ($f_{xx}=\frac{{\partial }^{2}z}{\partial x^2}$, $f_{xy}=\frac{{\partial }^{2}f}{\partial y^2}$, $f_{xy}=\frac{{\partial }^{2}f}{\partial x\partial y}$). Then calculate the determinant $D$ as below:

$$D=f_{xx}(x_0,y_0)\cdot f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0){)}^2$$

Then:

1. If $D>0$ and $f_{xx}(x_0,y_0)>0$, then $f$ has a local minimum at $P$.
2. If $D>0$ and $f_{xx}(x_0,y_0)<0$, then $f$ has a local maximum at $P$.
3. If $D<0$, then $f$ has saddle point at $P$.
4. If $D=0$ then test is inconclusive.

Link to original

Link to original

P: evaluate a multivariate limit (or prove it doesn’t exist)

evaluate a multivariate limit

Try each step and move on to the next if indeterminate form is encountered:

Step 1 - Simple Substitution

Try just substituting the coordinate of the limit $(x,y,\dots )$ into the expression.

Step 2 - Algebraic Manipulation

Factoring

If the expression is a fraction, try factoring the numerator and denominator and see if anything cancels. Then evaluate the limit again.

Multiply by conjugate over conjugate

If the fraction contains a root in numerator/denominator (e.g. in the form $a+\sqrt{b}$), try getting rid of it by multiplying the fraction by the conjugate over conjugate. The conjugate for $a+\sqrt{b}$ is $a-\sqrt{b}$. This might get rid of indeterminate forms if you are lucky.

Try to find other ways

See if there’s any other things you can try, like trignometry identities, to make the expression evaluatable.

Step 3 - Evaluate along paths to prove limit DNE

It’s likely that the limit does not exist, but we need to prove it. Try approaching the limit along different paths, like $x=0$ (y-axis), $y=0$ (x-axis), $y=x$, $y=x^3$, etc. Substitute the value into the limit expression then evaluate the limit. If any two limits evaluated along different paths are not equal to each other, then the limit does not exist. Note that if two limits evaluated along different paths end up being equal, this number is not necessarily the limit of the function.

Link to original

P: level curve

level curve

Level curves are like contour lines but for 3D functions.

To sketch level curves for a function, determine a range of $z=f(x,y)$ values and step size (e.g. from -2 to 2 with step size 1, meaning $c\in \left\{-2,-1,0,1,2\right\}$). Then for each value of $c$: set $c=f(x,y)$, solve for $y$, and draw the resulting curve.

Link to original

P: multivariate function domain

multivariate function domain

For a (real) multivariate function (let’s say $z=f(x,y)$), its domain is a region $R\in {\mathbb{R}}^n$ where $n$ is $f$‘s arity.

The boundary of $R$ is a curve that separates points internal to the domain from the points outside the domain. Formally, a region in a plane is bounded if it lies inside a disk of finite radius. Note that boundary points can be either in or out of the domain, and two boundary points don’t have to all be in or out of the domain.

• The domain is closed if $R$ contains all boundary points, if any.
• The domain is open if $R$ contains no boundary points.
• In cases where there is no boundary (e.g. when the domain is the entire plane), then the domain is both open and closed.
• The domain can be neither open nor closed if $R$ contains some but not all boundary points.

Link to original

P: piecewise multivariate partial derivative

piecewise multivariate partial derivative

Problem

Given a piecewise function $f$: $f(x,y)=\{\begin{array}{ll}g(x)& (x,y)\ne (0,0)\\ h(x)& (x,y)=(0,0)\end{array}$ Find its partial derivatives at every point $(x,y)$

For any point $(x,y)\ne (0,0)$,

$$\begin{array}{rl}\frac{\partial f}{\partial x}(x,y)& =\frac{\partial g}{\partial x}(x,y)\\ \frac{\partial f}{\partial y}(x,y)& =\frac{\partial g}{\partial y}(x,y)\end{array}$$

Plug the $y=0$ into $\frac{\partial f}{\partial x}$, and evaluate. We need to check if the second derivative exists at this point. If not, then the partial derivative doesn’t exist at $(0,0)$. Do the same for $\frac{\partial f}{\partial y}$.

If the second partial derivative does exist at $(0,0)$ (piecewise is OK), and the left and right side first derivatives evaluate to the same value at $(0,0)$, then the partial derivative exist at $(0,0)$.

Link to original