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Vector Calculus

https://twitter.com/bits_of_quantum/status/1575929782495109120

What are multivariable functions?

Multiple-number inputs. Multiple-number inputs. Vector-valued functions f:X⊂Rn→Y⊆Rmf:X \subset R^n \rightarrow Y \subseteq R^m.

Example: f:(x,y,z)→(xyz,sin(x))f:(x,y,z)\rightarrow (xyz,sin(x))

Single-number output. Single-number input. One variable f:X⊂R→Y⊆Rf:X \subset R \rightarrow Y \subseteq R.

Example: f(x)=xf(x)=x

Scalar-valued functions f:X⊂Rn→Y⊆Rf:X \subset R^n \rightarrow Y \subseteq R.

Example: f:(x,y,z)→xyz+sin(xyz)f:(x,y,z)\rightarrow xyz+sin(xyz)

Single-number input. Multiple-number output. f:X⊂R→Y⊆Rnf:X \subset R \rightarrow Y \subseteq R^n

Example: f:(t)→(t2,t+1)f:(t)\rightarrow (t^2,t+1)

Domain and Range of a Function

Graph function

Definition.

graphf={(x1,...,xn,f(x1,...xn))∈Rn+1∣(x1,...xn)∈U)graph_f=\{(x_1,...,x_n,f(x_1,...x_n))\in R^{n+1}|(x_1,...x_n)\in U)

The Method of Level curves

The Method of Section

Contour Map

Inline functions

Parametric functions

Parametric functions, two parameters

Transformations

Worked examples

Limits

Boundary

x+x0∈Rn,x−x0∈Ax+x_0 \in R^n,x-x_0\in A

Open sets

Assume r=min(∣∣b−x∣∣,∣∣x−b∣∣) and x0∈Dr(x)If r>∣∣b−x0∣∣>0and r>∣∣x0−b∣∣>0Then Dr(x)⊂A\text{Assume }r=min(||b-x||,||x-b||) \text{ and } x_0 \in D_r(x)\\ \text{If }r>||b-x_0||>0\\ \text{and }r>||x_0-b||>0\\ \text{Then } D_r(x) \sub A

Worked examples

  1. 𝐴={(𝑥,𝑦)∈R2∣x2+y2<1}𝐴 = \{(𝑥, 𝑦) \in R^2 | x^2+ y^2 < 1\}

  1. (x0,y0)∈A={(x,y)∈R2∣y>x2},(b,b2)∈B={(x,y)∈R2∣y=x2}(x_0,y_0)\in A = \{(x, y) \in R^2 | y>x^2 \},(b,b^2)\in B=\{(x,y)\in R^2| y=x^2\}
r=min∣∣(x0,y0)−(b,b2)∣∣r2=min{(x0−b)2+(y0−b2)2}min={solution to ddb((x0−b)2+(y0−b2)2)=0}min={solution to 2(x0−b)+2b(y0−b2)=0,b>0}r=min||(x_0,y_0)-(b,b^2)||\\ r^2=min\{(x_0-b)^2+(y_0-b^2)^2\}\\ min=\{\text{solution to }\dfrac{d}{db}((x_0-b)^2+(y_0-b^2)^2)=0\}\\ min=\{\text{solution to }2(x_0-b)+2b(y_0-b^2)=0,b>0\}

w=(x0,y0)−(b,b2),r=∣∣w∣∣=(x0−b)2+(y0−b2)w=(x_0,y_0)-(b,b^2),r=||w||=\sqrt{(x_0-b)^2+(y_0-b^2)}

TODO: Image to latex

https://www.geogebra.org/calculator/bycqw5nm

https://www.wolframalpha.com/input/?i=2(1-b)%2B4b(3-b^2)%3D0%2C+solution+for+b

Limits

Worked examples

  1. Let f:Rn→Rf:R^n \rightarrow R and suppose that lim(x,y)→(1,3)f(x,y)=5lim_{(x,y)\rightarrow(1,3)}f(x,y)=5. Analyze it.

f(x,y)f(x,y) approaches 55 as (x,y)(x,y)  approaches (1,3)(1,3), then his neighborhood exists. But We can't say f(1,3)f(1,3) exists by we've not known ff is continuous.

2. Let f:Rn→Rf:R^n \rightarrow R is continuous and suppose that lim(x,y)→(1,3)f(x,y)=5lim_{(x,y)\rightarrow(1,3)}f(x,y)=5. Analyze it.

f(x,y)f(x,y) approaches 55 as (x,y)(x,y)  approaches (1,3)(1,3), then his neighborhood exists. We can say f(1,3)f(1,3) exists by we've known ff is continuous.

3. lim(x,y)→(0,1)x3y=(limx→(0,1)x3)(limy→(0,1)y)=(0)3(1)=0lim_{(x,y)\rightarrow(0,1)}x^3y=(lim_{x\rightarrow(0,1)}x^3)(lim_{y\rightarrow(0,1)}y)=(0)^3(1)=0

4. limx→0cos(x)−1x2=limx→0−2sin2(x2)x2=limx→0−24(sin(x2)x/2)2=−12lim_{x\rightarrow0}\dfrac{cos(x)-1}{x^2}=lim_{x\rightarrow0}\dfrac{-2sin^2(\dfrac{x}{2})}{x^2}=lim_{x\rightarrow0}-\dfrac{2}{4}(\dfrac{sin(\dfrac{x}{2})}{x/2})^2=-\dfrac{1}{2}

5. limh→0eh−1h=limh→0((1+h)1/h)h−1h=limh→01+h−1h=1=ddxex∣0lim_{h \rightarrow 0}\dfrac{e^h-1}{h}=lim_{h \rightarrow 0}\dfrac{((1+h)^{1/h})^h-1}{h}=lim_{h \rightarrow 0}\dfrac{1+h-1}{h}=1=\dfrac{d}{dx}e^x\Big|_0

limx→0+1x=∞=limx→∞ylimx→∞1x=0=limy→0ylimx→∞(1x)x=0=limy→0+y1ylimx→∞(1+1x)x=e=limy→0+(1+y)1ylimx→x0limx→x0limx→x0...limx→x0f(x)=L,limx→x0=Llim_{x \rightarrow 0^+}\dfrac{1}{x}=\infty=lim_{x \rightarrow \infty}y\\ lim_{x \rightarrow \infty}\dfrac{1}{x}=0=lim_{y \rightarrow 0}y\\ lim_{x \rightarrow \infty}(\dfrac{1}{x})^x=0=lim_{y \rightarrow 0^+}y^\dfrac{1}{y}\\ lim_{x \rightarrow \infty}(1+\dfrac{1}{x})^x=e=lim_{y \rightarrow 0^+}(1+y)^\dfrac{1}{y}\\ lim_{x \rightarrow x_0}lim_{x \rightarrow x_0}lim_{x \rightarrow x_0}...lim_{x \rightarrow x_0}f(x)=L,lim_{x \rightarrow x_0}=L

6. lim(x,y)→(0,1)exy=(lim(x,y)→(0,1)ex)(lim(x,y)→(0,1)y)=(1)(1)=1lim_{(x,y)\rightarrow(0,1)}e^xy=(lim_{(x,y)\rightarrow(0,1)}e^x)(lim_{(x,y)\rightarrow(0,1)}y)=(1)(1)=1

7. limx→0sinn(x)x=limx→0sin(x)xlimx→0sinn(x)=(1)(0)=0,n>0lim_{x\rightarrow0}\dfrac{sin^n(x)}{x}=lim_{x\rightarrow0}\dfrac{sin(x)}{x}lim_{x\rightarrow0}sin^n(x)=(1)(0)=0,n>0

8. limx→0sinn(x)xn=(limx→0sin(x)x)n=1n=1,n>0lim_{x\rightarrow0}\dfrac{sin^n(x)}{x^n}=(lim_{x\rightarrow0}\dfrac{sin(x)}{x})^n=1^n=1,n>0

9. limx→ap(x)=p(a),p(x)=∑knaxklim_{x\rightarrow a}p(x)=p(a),p(x)=\sum_k^n ax^k

10. lim(x,y)→(0,0)exy−1sen(x)ln(1+y)=lim(x,y)→(0,0)1+xy+(xy)2/2!+(xy)3/3!+O(∣xy∣4)−1(x−x3/3!+x5/5!+O(∣x∣7))(y−y2/2+y3/3!+O(∣y∣4)=1lim_{(x,y) \rightarrow(0,0)} \dfrac{e^{xy}-1}{sen(x)ln(1+y)}=lim_{(x,y) \rightarrow(0,0)} \dfrac{1+xy+(xy)^2/2!+(xy)^3/3!+O(|xy|^{4})-1}{(x-x^3/3!+x^5/5!+O(|x|^{7}))(y-y^2/2+y^3/3!+O(|y|^{4})}=1\\ 

Gradient

direction and rate of fastest increase.

3. z−x=0z-x=0 and y−3x3+z2=0y-3x^3+z^2=0

https://www.geogebra.org/3d/bzcm27ut

https://www.geogebra.org/calculator/nvbvdvc4

High-Order Derivatives: Maxima and Minima

Iterated Partial Derivatives

Let f:Rn→Rf: \mathbb{R^n} \rarr \mathbb{R}  be of class CnC^n if ∂fn∂xn,∂fn∂yn,∂fn∂xn,...\dfrac{\partial f^n}{\partial x^n},\dfrac{\partial f^n}{\partial y^n},\dfrac{\partial f^n}{\partial x^n},... exists and are continous. How second-order derivatives are written:

∂2f∂x2=∂∂x(∂f∂x) Iterated partial derivatives∂2f∂x∂y=∂∂x(∂f∂y) Mixed partial derivatives\dfrac{\partial^2f}{\partial x^2}=\dfrac{\partial}{\partial x}(\dfrac{\partial f}{\partial x}) \text{ Iterated partial derivatives}\\ \dfrac{\partial^2f}{\partial x\partial y}=\dfrac{\partial}{\partial x}(\dfrac{\partial f}{\partial y})\text{ Mixed partial derivatives}
💡
The mixed partials are equal by Young's Theorem by recent contribuitor.
Euler → Alexis Claude Clairaut →H. A. Schwarz→W. H. Young

Young's theorem on equality of mixed partials

If f(x,y)f(x,y)  is of class C2C^2, then the mixed partial derivativates are equal; that is,

∂2f∂x∂y=∂2f∂y∂x\dfrac{\partial^2f}{\partial x\partial y}=\dfrac{\partial^2f}{\partial y\partial x}

proof. Key ideas Schwarz's theorem.

Taylor series

Big O

http://web.mit.edu/16.070/www/lecture/big_o.pdf

http://faculty.bard.edu/belk/math142af09/ApplicationsTaylorSeries.pdf

http://www.math.ubc.ca/~feldman/m120/taylorLimits.pdf

Polar functions

https://www.desmos.com/calculator/svnuhptykg?embed
XTX′=0X′=kn^,fn(t)=k≠0,∣∣n^∣∣=1You need to find t=fn∗(k).X^T X^{'}=0\\ X^{'}=k\hat{n},f_n(t)=k\neq0,||\hat{n}||=1\\ \text{You need to find } t=f_n^{*}(k).
Consider the polar curve r=f(θ)Therefore the equation of our tangent line is:y−f(θ)sin(θ)=dydx∣θ(x−f(θ)cos(θ))\text{Consider the polar curve }r=f(\theta)\\ \text{Therefore the equation of our tangent line is:}\\ y-f(\theta)sin(\theta)=\dfrac{dy}{dx}\biggr\rvert _\theta (x-f(\theta)cos(\theta))

Worked examples

  1. r=1−sin(θ)r=1-sin(\theta), At which values of θ does the graph of r have a horizontal tangent line?
XTX′=[(1−sin(θ))cos(θ)(1−sin(θ))sin(θ)][(sin(θ)−1)sin(θ)−cos2(θ)(1−2sin(θ))cos(θ)]=0X^TX^{'}=\begin{bmatrix} (1-sin(\theta))cos(\theta)\\ (1-sin(\theta))sin(\theta) \end{bmatrix} \begin{bmatrix} (sin(\theta)-1)sin(\theta)-cos^2(\theta) & (1-2sin(\theta))cos(\theta) \end{bmatrix}=0
Horizontal tangent line: X′=ki^f0′(θ)=−sin(θ)−cos(2θ)≠0f1′(θ)=(1−2sin(θ))cos(θ)=0θ=...\text{Horizontal tangent line: }X^{'}=k\hat{i}\\ f_0^{'}(\theta)=-sin(\theta)-cos(2\theta)\neq0\\ f_1^{'}(\theta)=(1-2sin(\theta))cos(\theta)=0\\ \\ \theta=...
https://www.desmos.com/calculator/akueesxruu

2. r=4sin(3θ)r=4sin(3\theta), What is the slope of the tangent line to the curve rr when θ=π3\theta=\dfrac{\pi}{3}?

https://www.desmos.com/calculator/yeaoaleswf
dydx=4(3cos⁡(3a)sin⁡(a)+cos⁡(a)sin⁡(3a))12cos⁡(a)cos⁡(3a)−4sin⁡(a)sin⁡(3a)dydx∣pi/3=3\dfrac{dy}{dx}=\frac{4(3\cos(3a)\sin(a)+\cos(a)\sin(3a))}{12\cos(a)\cos(3a)-4\sin(a)\sin(3a)}\\ \dfrac{dy}{dx}\biggr\rvert_{pi/3}=\sqrt{3}

Double integral

Worked examples

∫0a∫0a2−x2(a2−y2)dydx∫0a∫0a2−y2(a2−y2)dxdy=∫0a(a2−y2)32=3πa416\int_0^a\int_0^{\sqrt{a^2-x^2}}(a^2-y^2)dydx\\\int_0^a\int_0^{\sqrt{a^2-y^2}}(a^2-y^2)dxdy=\int_0^a\left(a^2-y^2\right)^\frac{3}{2}=\dfrac{3{\pi}a^4}{16}\\

Polar coordinates

∣∂(x,y)∂(r,θ)∣=r|\dfrac{\partial (x,y)}{\partial(r,\theta)}|=r

x=rcosθx=rcos\theta, y=rsinθy=rsin\theta

Substitute

∫0π/2∫0ar(a2−r2sin2θ)drdθ=3πa416\int^{\pi/2}_0\int_0^ar(a^2-r^2sin^2\theta)drd\theta=\dfrac{3{\pi}a^4}{16}