From 54ca67bc4b75fe2ad6d2f68aa4eff61f781d2af8 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Mon, 1 Nov 2021 16:18:06 -0400
Subject: updates for math and physics homeworks

---
 jarrio/week09.bc          | 161 ++++++++++++++++++++++++++++++
 jarrio/week10.bc          | 245 ++++++++++++++++++++++++++++++++++++++++++++++
 jarrio/week10resistors.py |  66 +++++++++++++
 jarrio/week9practice.bc   | 162 ------------------------------
 li/hw6.tex                |  14 ++-
 5 files changed, 481 insertions(+), 167 deletions(-)
 create mode 100644 jarrio/week09.bc
 create mode 100644 jarrio/week10.bc
 create mode 100644 jarrio/week10resistors.py
 delete mode 100644 jarrio/week9practice.bc

diff --git a/jarrio/week09.bc b/jarrio/week09.bc
new file mode 100644
index 0000000..b0357f8
--- /dev/null
+++ b/jarrio/week09.bc
@@ -0,0 +1,161 @@
+print "Weekly Homework 9\n"
+pi = 4*a(1)
+epsilon = 8.85 * 10^(-12)
+epercoul = 6.24 * 10^18
+# Conductivity constants ( resistitvity = 1/sigma)
+sigmairon = 1*10^7
+sigmacopper = 6*10^7
+sigmasilver = 6.2*10^7
+sigmagold = 4.1*10^7
+sigmaaluminum = 3.5*10^7
+sigmatungsten = 1.8*10^7
+sigmanichrome = 6.7*10^5
+sigmacarbon = 2.9*10^4
+
+print "Question 1\n"
+#params
+i = 1.9/1000 # A
+diam = 5/1000 # m
+
+r = diam/2
+"Protons Per Second "
+i*epercoul
+# i = \int_0^R J_edge (r/R) 2\pi r dr
+#   = J_edge/R [ 2\pi r^3 / 3 ]_0^R
+#   = 2\pi J_edge R^2/3
+jedge = 3*i/(2*pi*r^2)
+"J_edge (A/m^2) "
+jedge
+
+print "Question 2\n"
+# Equilibrium: eta/2epsilon + EF_I2 = EF_I1
+# EF_I = V/L = IR/L = I(L/sigma/A)/L = I/(sigma*A)
+# eta = 2epsilon * ( I/(sigma1*A)-I/(sigma2*A) )
+# Corrections from pearson: apparently it should be eta/epsilon, and the
+# sign was opposite, so correct is
+print "eta = epsilon * ( I/(sigma2*A)-I/(sigma1*A))\n"
+#params
+diam = 1/1000 # m
+i = 3 # A
+
+a = pi*r^2 # m^2
+r = diam/2 # m
+"Charge on the boundary "
+# eta * a
+epsilon * i * (1/(sigmairon*a) - 1/(sigmacopper*a)) * a
+
+print "Question 3\n"
+#params
+diam = .5 # mm
+i = 20 # mA
+
+diam = diam/1000 # m
+r = diam/2 # m
+i = i/1000 # A
+
+a = pi*r^2
+j = i/a
+# j = ef*sigma
+ef = j/sigmasilver
+"Electric Field (V/m) "
+ef
+
+"Electron drift speed (m/s) " # idk lol
+ve = 5.8 * 10^28 # number/m^3
+ie = i*epercoul # number/s
+ie/a/ve
+
+print "Question 4\n"
+diam = 1 # mm
+diam = diam/1000 # m
+r = diam/2
+a = pi*r^2
+# efnichrome = efaluminum
+# jnichrome/sigmanichrome = jaluminum/sigmaaluminum
+# i/pi*(diam_nichrome/2)^2/sigmanichrome = i/a/sigmaaluminum
+# pi*(diam_nichrome/2)^2*sigmanichrome = a*sigmaaluminum
+"Diameter of nichrome (m) "
+2*sqrt(a*sigmaaluminum/sigmanichrome/pi)
+
+print "Question 5\n"
+#params
+l = 10 # cm
+innerdiam = 2.8 # mm
+outerdiam = 3   # mm
+v = 3 # V
+
+l = l/100 # m
+innerdiam = innerdiam/1000 # m
+innerr = innerdiam/2
+outerdiam = outerdiam/1000 # m
+outerr = outerdiam/2
+a = pi*outerr^2 - pi*innerr^2
+ef = v/l
+j = ef*sigmanichrome
+i = j*a
+"Current (A) "
+i
+
+print "Question 6\n"
+print "60W means higher resistance, so it has a higher voltage drop\n"
+print "and takes more energy\n"
+
+v = 120 # V
+w1 = 60 # W (wattage of lightbulb 1 @ 120V)
+# w1 = i*v = v*v/r1
+r1 = v^2/w1
+r1
+w2 = 100 # W
+r2 = v^2/w2
+i = v/(r1+r2)
+print "Power by the first bulb (W) "
+i^2*r1
+print "Power by the second bulb (W) "
+i^2*r2
+
+print "Question 7\n"
+#params
+r = 70 # Ohm
+
+v = 9 # V
+i = v/r
+
+"Magnitude of Current (A) "
+i
+
+print "Direction is left to right (positive to negative)\n"
+
+print "Question 8\n"
+#params
+w = 1100 # W
+
+v = 120 # V
+"Resistance (Ohm) "
+r = v^2/w
+r
+
+"Current (A) "
+v/r
+
+print "Question 9\n"
+#params
+m = 1 # g aluminum
+p = 8 # W
+v = 1.5 # V
+
+m = m/1000 # kg
+r = v^2/p # Ohm
+densityaluminum = 2.7 # g/cm^3
+densityaluminum = densityaluminum*1000 #kg/m^3
+vol = m/densityaluminum # m^3
+rhoaluminum = 1/sigmaaluminum # Ohm*m
+# R = rhoaluminum * l / A = rhoaluminum * l / (vol/l) =
+# rhoaluminum*l^2 / vol
+l = sqrt(r/rhoaluminum*vol) # m
+# Cross-sectional area
+a = vol/l # m^2
+"Diameter (mm) "
+2*sqrt(a/pi) * 1000
+
+"Length (m) "
+l
diff --git a/jarrio/week10.bc b/jarrio/week10.bc
new file mode 100644
index 0000000..80a8fee
--- /dev/null
+++ b/jarrio/week10.bc
@@ -0,0 +1,245 @@
+print "Daily Homework 10a\n"
+
+print "Question 4\n"
+#params
+vbattery = 3.05
+vchange = 2.9
+ammeter = 1.7
+
+bigr = vchange/ammeter
+"emf "
+vbattery
+"internal resistance "
+rtot = vbattery/ammeter
+rtot - bigr
+"circuit resistance "
+bigr
+
+print "\nDaily Homework 10b\n"
+
+print "Question 1\n"
+#params
+r1 = 2
+r2 = 5
+r3 = 1
+r4 = 7
+r5 = 3
+r6 = 3
+r7 = 3
+
+"RA "
+r4+r3+1/(1/r1+1/r2)
+"RB (open) "
+r5 + r4 + r3 + 1/(1/(r6+r1) + 1/r2)
+"RB (closed) "
+r5 + 1/(1/r4+1/r7) + r3 + 1/(1/(r6+r1) + 1/r2)
+
+print "Question 2\n"
+print "Rlong^2\n"
+
+print "Question 3\n"
+# params
+ratio = 10 # Ip = 10Is
+# Without loss of generality, let Is = 1, and R2 = 1, so r = R1.
+# IR = V
+# 1(1+r) = V
+# 10/(1/1 + 1/r) = V
+# 1 + r = 10/(1 + 1/r)
+# (1+r)(1+1/r) = 10
+# r^2 + 2r + 1 = 10r
+# r^2 - 8r + 1 = 0
+b = 2 - ratio
+(-b - sqrt(b^2 - 4*1*1))/2
+
+print "Question 4\n"
+#params
+ammeter = 9
+
+# clockwise loop rule on top loop
+# ammeter - 3*i1 + 2*ammeter = 0
+i1 = ammeter
+print "I1 (A)\n"
+i1
+
+# branch rule
+i2 = i1+ammeter
+print "I2 (A)\n"
+i2
+
+# clockwise loop rule on bottom loop
+# -2*ammeter + electromotive - 1*i2 = 0
+electromotive = 2*ammeter + i2
+print "Electromotive (V)\n"
+electromotive
+
+print "\nDaily Homework 10c\n"
+print "Question 2\n"
+print "C > A = E > B > D\n"
+
+print "Question 3\n"
+#params
+c = 10 # uF
+q = 30 # uC
+r = 1.8 # kOhm
+qtarget = 15 # uC
+
+# c = q(t)/v(t) --> v(t) = q(t)/c
+# dq(t)/dt = -I = -v(t)/r
+# dq(t)/dt = -q(t)/cr
+# q(t) = Ce^{-t/cr}
+# C = q
+# q(t) = qe^{-t/cr} = qtarget => -t/cr = ln(qtarget/q)
+
+r = r*1000 # Ohm
+qtarget = qtarget/10^6 # C
+c = c/10^6 # F
+q = q/10^6 # C
+
+t = -c*r*l(qtarget/q)
+print "Time (s)\n"
+t
+
+print "\nWeekly Homework 10\n"
+print "Question 1\n"
+#params
+r1 = 2.5 # kOhm
+r2 = 4.0 # kOhm
+r3 = 5.0 # kOhm
+v = 100 # V
+# v = ir
+# p = iv = v^2/r
+
+pparallel = v^2 / (1/(1/r1 + 1/r2 + 1/r3))
+pseries = v^2 / (r1 + r2 + r3)
+pparallel/pseries
+
+print "Question 2\n"
+# I = ef/(r+R)
+# P = I^2R = ef^2*R/(r+R)^2
+# dP/dR = 0 = d/dR (ef^2*R/(r+R)^2)
+# d/dR (R/(r+R)^2) = 0
+# (r+R)/(r+R)^3 - 2R/(r+R)^3 = (r-R)/(r+R)^3 = 0
+# r = R
+print "Part A: R = r\n"
+
+print "Part B\n"
+# params
+ef = 5 # V
+r = 1.8 # Ohm
+i = ef/(r+r)
+print "Power (W)\n"
+i^2*r
+
+print "Question 3\n"
+#params
+bulbr = 6 # Ohm
+r = .8 # Ohm
+v = 1.5 # V
+
+iopen = v/(r+bulbr)
+print "I_open (A)\n"
+iopen
+
+iclosed = v/(r+bulbr/2) / 2 # because current is even between a and b
+print "I_closed (A)\n"
+iclosed
+
+print "pct change\n"
+100*(iclosed-iopen)/iopen
+
+print "Question 4\n"
+v = 24 # V
+# No params. Effective resistance w/ open is 6Ohm and 6Ohm in parallel,
+# giving 3Ohm
+print "Part A - Ibat (A)\n"
+v/3
+
+print "Part B - dV (V)\n"
+v/3
+
+# the two pairs of parallel resistors = two effective resistors in
+# parallel
+print "Part C - Ibat (A)\n"
+v / (1/(1/3 + 1/5) + 1/(1/3 + 1))
+
+print "Part D - dV (V)\n"
+0
+
+print "Question 5\n"
+v = 24 # V
+i6 = v/10 # effective resistance calculation
+i5 = v/15 # same
+i10 = i5
+i4 = i6
+
+print "Currents (i6, i5, i10, i4)\n"
+i6
+i5
+i10
+i4
+
+v6 = i6*6 # V = IR
+v5 = i5*5
+v10 = i10*10
+v4 = i4*4
+print "Voltages (v6, v5, v10, v4)\n"
+v6
+v5
+v10
+v4
+
+print "Question 6\n"
+#params
+v = 150
+rtot = 2 + 1/(1/20 + 1/5) + 4
+itot = v/rtot
+vparallel = itot / (1/20 + 1/5) # v through the parallel resistors
+i20 = vparallel/20
+
+print "I20 (A)\n"
+i20
+
+print "Question 7\n"
+
+# 1/(1/4 + 1/12) = 3.
+# 3 + 5 = 8
+# 1/(1/8 + 1/24) = 6.
+# Assume current goes clockwise.
+# Loop: 12 - I*6 - 3 - I*3 = 0
+# 9 = I*9 => I = 1A.
+# Then, week10resistors.py has the solution
+print "See source\n"
+
+print "Question 8\n"
+#params
+tconstant = 7 # ms
+
+print "Charge reduced to half (ms)\n"
+-l(1/2)*tconstant
+
+print "Energy reduced to half (ms)\n"
+-l(1/sqrt(2))*tconstant
+
+print "Question 9\n"
+#params
+c = 40 # uF
+# The graph shows a reduction from 30V to 10V in 4ms, giving a time
+# constant of
+tconstant = 4/l(30/10) # ms
+# tconstant = cr => r = tconstant/c
+c = c/10^6 # F
+tconstant = tconstant/1000 # s
+r = tconstant/c
+print "Resistance\n"
+r
+
+print "Question 10\n"
+#params
+c = .25 # uF
+v = 70 # V
+r1 = 25 # Ohm (the one the question asks about)
+r2 = 150 # Ohm
+q = c*v # uC
+j = q*v/2
+print "Energy dissipated (uJ)\n"
+j*r1/(r1+r2)
diff --git a/jarrio/week10resistors.py b/jarrio/week10resistors.py
new file mode 100644
index 0000000..f7041c7
--- /dev/null
+++ b/jarrio/week10resistors.py
@@ -0,0 +1,66 @@
+class resistor:
+    def __init__(self, resistance: float):
+        self._resistance = resistance;
+        self.has_current(0)
+
+    def has_current(self, current: float):
+        self._current = current
+        self._voltage = current*self._resistance
+
+    def has_voltage(self, voltage: float):
+        self._voltage = voltage
+        self._current = voltage/self._resistance
+
+    def __repr__(self):
+        return f"Resistor with resistance {self._resistance}: current {self._current} through voltage {self._voltage}"
+
+class parallel(resistor):
+    def __init__(self, r1: resistor, r2: resistor):
+        self._r1 = r1
+        self._r2 = r2
+        self._resistance = 1/(1/r1._resistance + 1/r2._resistance)
+
+    def _update_children(self):
+        self._r1.has_voltage(self._voltage)
+        self._r2.has_voltage(self._voltage)
+
+    def has_current(self, current: float):
+        self._current = current
+        self._voltage = current*self._resistance
+        self._update_children()
+
+    def has_voltage(self, voltage: float):
+        self._voltage = voltage
+        self._current = voltage/self._resistance
+        self._update_children()
+
+class series(resistor):
+    def __init__(self, r1: resistor, r2: resistor):
+        self._r1 = r1
+        self._r2 = r2
+        self._resistance = r1._resistance + r2._resistance
+
+    def _update_children(self):
+        self._r1.has_current(self._current)
+        self._r2.has_current(self._current)
+
+    def has_current(self, current: float):
+        self._current = current
+        self._voltage = current*self._resistance
+        self._update_children()
+
+    def has_voltage(self, voltage: float):
+        self._voltage = voltage
+        self._current = voltage/self._resistance
+        self._update_children()
+
+r24 = resistor(24)
+r4 = resistor(4)
+r12 = resistor(12)
+r5 = resistor(5)
+rtot = parallel( r24, series( parallel(r4,r12), r5 ) )
+rtot.has_current(1)
+print(r24)
+print(r5)
+print(r4)
+print(r12)
diff --git a/jarrio/week9practice.bc b/jarrio/week9practice.bc
deleted file mode 100644
index 611dc8e..0000000
--- a/jarrio/week9practice.bc
+++ /dev/null
@@ -1,162 +0,0 @@
-########################################################################
-print "Weekly Homework 9\n"
-pi = 4*a(1)
-epsilon = 8.85 * 10^(-12)
-epercoul = 6.24 * 10^18
-# Conductivity constants ( resistitvity = 1/sigma)
-sigmairon = 1*10^7
-sigmacopper = 6*10^7
-sigmasilver = 6.2*10^7
-sigmagold = 4.1*10^7
-sigmaaluminum = 3.5*10^7
-sigmatungsten = 1.8*10^7
-sigmanichrome = 6.7*10^5
-sigmacarbon = 2.9*10^4
-
-print "Question 1\n"
-#params
-i = 1.9/1000 # A
-diam = 5/1000 # m
-
-r = diam/2
-"Protons Per Second "
-i*epercoul
-# i = \int_0^R J_edge (r/R) 2\pi r dr
-#   = J_edge/R [ 2\pi r^3 / 3 ]_0^R
-#   = 2\pi J_edge R^2/3
-jedge = 3*i/(2*pi*r^2)
-"J_edge (A/m^2) "
-jedge
-
-print "Question 2\n"
-# Equilibrium: eta/2epsilon + EF_I2 = EF_I1
-# EF_I = V/L = IR/L = I(L/sigma/A)/L = I/(sigma*A)
-# eta = 2epsilon * ( I/(sigma1*A)-I/(sigma2*A) )
-# Corrections from pearson: apparently it should be eta/epsilon, and the
-# sign was opposite, so correct is
-print "eta = epsilon * ( I/(sigma2*A)-I/(sigma1*A))\n"
-#params
-diam = 1/1000 # m
-i = 3 # A
-
-a = pi*r^2 # m^2
-r = diam/2 # m
-"Charge on the boundary "
-# eta * a
-epsilon * i * (1/(sigmairon*a) - 1/(sigmacopper*a)) * a
-
-print "Question 3\n"
-#params
-diam = .5 # mm
-i = 20 # mA
-
-diam = diam/1000 # m
-r = diam/2 # m
-i = i/1000 # A
-
-a = pi*r^2
-j = i/a
-# j = ef*sigma
-ef = j/sigmasilver
-"Electric Field (V/m) "
-ef
-
-"Electron drift speed (m/s) " # idk lol
-ve = 5.8 * 10^28 # number/m^3
-ie = i*epercoul # number/s
-ie/a/ve
-
-print "Question 4\n"
-diam = 1 # mm
-diam = diam/1000 # m
-r = diam/2
-a = pi*r^2
-# efnichrome = efaluminum
-# jnichrome/sigmanichrome = jaluminum/sigmaaluminum
-# i/pi*(diam_nichrome/2)^2/sigmanichrome = i/a/sigmaaluminum
-# pi*(diam_nichrome/2)^2*sigmanichrome = a*sigmaaluminum
-"Diameter of nichrome (m) "
-2*sqrt(a*sigmaaluminum/sigmanichrome/pi)
-
-print "Question 5\n"
-#params
-l = 10 # cm
-innerdiam = 2.8 # mm
-outerdiam = 3   # mm
-v = 3 # V
-
-l = l/100 # m
-innerdiam = innerdiam/1000 # m
-innerr = innerdiam/2
-outerdiam = outerdiam/1000 # m
-outerr = outerdiam/2
-a = pi*outerr^2 - pi*innerr^2
-ef = v/l
-j = ef*sigmanichrome
-i = j*a
-"Current (A) "
-i
-
-print "Question 6\n"
-print "60W means higher resistance, so it has a higher voltage drop\n"
-print "and takes more energy\n"
-
-v = 120 # V
-w1 = 60 # W (wattage of lightbulb 1 @ 120V)
-# w1 = i*v = v*v/r1
-r1 = v^2/w1
-r1
-w2 = 100 # W
-r2 = v^2/w2
-i = v/(r1+r2)
-print "Power by the first bulb (W) "
-i^2*r1
-print "Power by the second bulb (W) "
-i^2*r2
-
-print "Question 7\n"
-#params
-r = 70 # Ohm
-
-v = 9 # V
-i = v/r
-
-"Magnitude of Current (A) "
-i
-
-print "Direction is left to right (positive to negative)\n"
-
-print "Question 8\n"
-#params
-w = 1100 # W
-
-v = 120 # V
-"Resistance (Ohm) "
-r = v^2/w
-r
-
-"Current (A) "
-v/r
-
-print "Question 9\n"
-#params
-m = 1 # g aluminum
-p = 8 # W
-v = 1.5 # V
-
-m = m/1000 # kg
-r = v^2/p # Ohm
-densityaluminum = 2.7 # g/cm^3
-densityaluminum = densityaluminum*1000 #kg/m^3
-vol = m/densityaluminum # m^3
-rhoaluminum = 1/sigmaaluminum # Ohm*m
-# R = rhoaluminum * l / A = rhoaluminum * l / (vol/l) =
-# rhoaluminum*l^2 / vol
-l = sqrt(r/rhoaluminum*vol) # m
-# Cross-sectional area
-a = vol/l # m^2
-"Diameter (mm) "
-2*sqrt(a/pi) * 1000
-
-"Length (m) "
-l
diff --git a/li/hw6.tex b/li/hw6.tex
index e0ff219..7acccb6 100644
--- a/li/hw6.tex
+++ b/li/hw6.tex
@@ -12,8 +12,12 @@ $$P = A(A^TA)^{-1}A^T = \bmatrix{1&1\cr 1&-1\cr -2&4}
 \fr1{22}\bmatrix{9&4\cr4&3}\bmatrix{1&1&-2\cr1&-1&4}
 = \fr1{22}\bmatrix{20&6&2\cr 6&4&-6\cr2&-6&20}.$$
 
-$$Pb = \fr1{22}\bmatrix{20&6&2\cr 6&4&-6\cr2&-6&20}\bmatrix{1\cr2\cr7}
-= \fr1{22}\bmatrix{46\cr-28\cr130}$$
+$$p = Pb = \fr1{22}\bmatrix{20&6&2\cr
+6&4&-6\cr2&-6&20}\bmatrix{1\cr2\cr7} =
+\fr1{22}\bmatrix{46\cr-28\cr130}$$
+With $b = p + q,$ and $p$ known, $q = b-p,$ and $q$ is in the left null
+space of $A$ by the definition of orthogonality with the column space
+which contains $p.$
 
 \noindent{\bf 7.}
 
@@ -47,12 +51,12 @@ longer distance.
 
 \noindent{\bf 16.}
 
-A is a $3\times2$ matrix, $Q$ is also a $3\times2$ matrix, and $R$ is a
-$2\times2$ matrix.
-
 $$A = \bmatrix{1&1\cr2&3\cr2&1} = \bmatrix{1/3&0\cr
 2/3&1/\sqrt2\cr2/3&-1/\sqrt2}\bmatrix{3&3\cr0&\sqrt2} = QR.$$
 
+Generally, $A$ is an $m\times n$ matrix, $Q$ is an $m\times r$ (where
+$r$ is the rank of $A$) matrix, and $R$ is an $r\times r$ matrix.
+
 \noindent{\bf 17.}
 
 $$Pb = QQ^Tb = \bmatrix{1/9&2/9&2/9\cr 2/9&17/18&-1/18\cr
-- 
cgit