#Irina Mukhametzhanova, Section 24
#This homework is OK to post

#14.6 Homework
#Exercise 1:
#a)
#Taking the partial derivatives with respect means treating other variables as constants
#So:
#df/dx = 2*x*y^3
#df/dy = 3*x^2*y^2
#df/dz = 4*z^3
#b)
#dx/ds = 2*s
#dy/ds = t^2
#dz/ds = 2*s*t
#c)
#df/ds = (df/dx)*(dx/ds) + (df/dy)*(dy/ds) + (df/dz)*(dz/ds) =
# = (2*x*y^3)*(2*s) + (3*x^2*y^2)*(t^2) + (4*z^3)*(2*s*t) = 
# = (2*s^2*(s*t^2)^3)*(2*s) + (3*s^4*s^2*t^4)*(t^2) + (4*s^6*t^3)*(2*s*t) = 
# = 4*s^2*s^3*t^6*s + 3*s^6*t^6 + 8*s^7*t^4 = 4*s^6*t^6 + 3*s^6*t^6 + 8*s^7*t^4 = 
# = 7*s^6*t^6 + 8*s^7*t^4
#ANSWER: df/ds = 7*s^6*t^6 + 8*s^7*t^4

#Exercise 3:
#The formula for the chain rule to find df/ds is:
#df/ds = (df/dx)*(dx/ds) + (df/dy)*(dy/ds) + (df/dz)*(dz/ds)
#And to find df/dr is:
#df/dr = (df/dx)*(dx/dr) + (df/dy)*(dy/dr) + (df/dz)*(dz/dr)
#So first, we find the first partial derivatives of f(x,y,z):
#df/dx = y
#df/dy = x
#df/dz = 2*z
#Then, find partial derivatives of x, y, and z with respect to s:
#dx/ds = 2*s
#dy/ds = 2*r
#dz/ds = 0
#And with respect to r:
#dx/dr = 0
#dy/dr = 2*s
#dz/dr = 2*r
#Now, find df/ds and df/dr using the chain rule:
#df/ds = (y)*(2*s) + (x)*(2*r) + (2*z)*(0) = (2*r*s)*(2*s) + (s^2)*(2*r) = 
# = 4*r*s^2 + 2*r*s^2 = 6*r*s^2
#df/dr = (y)*(0) + (x)*(2*s) + (2*z)*(2*r) = (s^2)*(2*s) + (2*r^2)*(2*r) = 
# = 2*s^3 + 4*r^3
#ANSWER: df/ds = 6*r*s^2, df/dr = 2*s^3 + 4*r^3

#Exercise 5:
#The formula for the chain rule to find dg/du is:
#dg/du = (dg/dx)*(dx/du) + (dg/dy)*(dy/du)
#And to find dg/dv is:
#dg/dv = (dg/dx)*(dx/dv) + (dg/dy)*(dy/dv)
#So first, we find the first partial derivatives of g(x,y):
#dg/dx = -sin(x-y)
#dg/dy = sin(x-y)
#Then, find partial derivatives of x and y with respect to u:
#dx/du = 3
#dy/du = -7
#And with respect to v:
#dx/dv = -5
#dy/dv = 15
#Now, find dg/du and dg/dv using the chain rule:
#dg/du = (-sin(x-y))*(3) + (sin(x-y))*(-7) =
# = (-3)*sin(3*u - 5*v + 7*u - 15*v) - 7*sin(3*u - 5*v + 7*u - 15*v) = 
# = (-3)*sin(10*u - 20*v) - 7*sin(10*u - 20*v) = (-10)*sin(10*u - 20*v)
#dg/dv = (-sin(x-y))*(-5) + (sin(x-y))*(15) = 5*sin(10*u - 20*v) + 15*sin(10*u - 20*v) = 
# = 20*sin(10*u - 20*v)
#ANSWER: dg/du = (-10)*sin(10*u - 20*v), dg/dv = 20*sin(10*u - 20*v)

#Exercise 7:
#The formula for the chain rule to find dF/dy is:
#dF/dy = (dF/du)*(du/dy) + (dF/dv)*(dv/dy)
#So first, we find the first partial derivatives of F(u,v):
#dF/du = exp(u+v)
#dF/dv = exp(u+v)
#Then, find partial derivatives of u and v with respect to y:
#du/dy = 0
#dv/dy = x
#Now, find dF/dy using the chain rule:
#dF/dy = (exp(u+v))*(0) + (exp(u+v))*(x) = (exp(x^2+x*y))*x
#ANSWER: dF/dy = (exp(x^2+x*y))*x

#Exercise 15:
#First, find all first partial derivatives of g(x,y):
#dg/dx = 2*x
#dg/dy = (-2)*y
#And partial derivatives of x and y with respect to u:
#dx/du = exp(u)*cos(v)
#dy/du = exp(u)*sin(v)
#Now, find dg/du using the chain rule:
#dg/du = (2*x)*(exp(u)*cos(v)) + ((-2)*y)*(exp(u)*sin(v)) 
#At (0,1):
#x = exp(0)*cos(1) = cos(1)
#y = exp(0)*sin(1) = sin(1)
#Plug (0,1) and the resulting x and y into the chain rule:
#dg/du(0,1) = (2*cos(1))*(exp(0)*cos(1)) + ((-2)*sin(1))*(exp(0)*sin(1)) = 
# = 2*cos(1)^2 - 2*sin(1)^2 = 2*(cos(1)^2 - sin(1)^2) = 2*cos(2)
#ANSWER: dg/du = 2*cos(2) 

#Exercise 23:
#Let's expand the right side of the equation:
#(df/ds)*(df/dt) = ((df/dx)*(dx/ds) + (df/dy)*(dy/ds)) * ((df/dx)*(dx/dt) + (df/dy)*(dy/dt))
#Calculate dx/ds, dy/ds, dx/dt, and dy/dt:
#dx/ds = 1
#dy/ds = 1
#dx/dt = 1
#dy/dt = -1
#Plug them back into the equation above:
#((df/dx)*(1) + (df/dy)*(1)) * ((df/dx)*(1) + (df/dy)*(-1)) = 
# = ((df/dx) + (df/dy)) * ((df/dx) - (df/dy)) = 
# = (df/dx)^2 + (df/dx)*(df/dy) - (df/dx)*(df/dy) + (df/dy)^2 = 
# = (df/dx)^2 - (df/dy)^2 -> Checks out

#Exercise 27:
#Using implicit differentiation means that to find dz/dx, we take the
#derivative with respect to x on both sides. So:
#x' = 1, y' = 0, and z' = dz/dx
#Using that:
#(x^2*y)' + (y^2*z)' + (x*z^2)' = (10)'
#(2*x*y) + (y^2*z') + (z^2 + x*2*z*z') = 0
#Collect z' on one side:
#(y^2*z') + (x*2*z*z') = (-2)*x*y - z^2
#Factor it out:
#z'*(y^2 + 2*x*z) = (-2)*x*y - z^2
#z' = ((-2)*x*y - z^2)/(y^2 + 2*x*z)
#ANSWER: dz/dx = ((-2)*x*y - z^2)/(y^2 + 2*x*z)

#Exercise 29:
#To find dz/dy, we take the derivative with respect to y on both sides. So:
#x' = 0, y' = 1, and z' =  dz/dy
#Using that:
#(exp(x*y))' + (sin(x*z))' + (y)' = 0
#x*exp(x*y) + cos(x*z)*x*z' + 1 = 0
#Collect z' on one side:
#cos(x*z)*x*z' = (-1)*x*exp(x*y) - 1
#Factor it out:
#z' = ((-1)*x*exp(x*y) - 1)/(cos(x*z)*x)
#ANSWER: ((-1)*x*exp(x*y) - 1)/(cos(x*z)*x)

#Exercise 31:
#First, we check that the point satisfies the equation:
#((1)/(1^2 + 1^2)) + ((1)/(1^2 + 1^2) = (1/2) + (1/2) = 1 -> Checks out
#To find dw/dy, we take the derivative with respect to y on both sides. So:
#x' = 0, y' = 1, and w' = dw/dy
#Using that:
#((1)/(w^2 + x^2))' + ((1)/(w^2 + y^2)' = 0
#((-1)/(w^2 + x^2)^2)*(2*w*w') + (((-1)*(w^2 + y^2)')/(w^2 + y^2)^2) = 0
#(((-2)*w*w')/(w^2 + x^2)^2) + (((-1)*(2*w*w' + 2*y))/(w^2 + y^2)^2) = 0
#Collect w' on one side:
#(((-4)*w*w')/(w^2 + x^2)^2) = ((2*y)/(w^2 + x^2)^2)
#Factor it out:
#w' = ((2*y)/(w^2 + x^2)^2) / (((-4)*w)/(w^2 + x^2)^2)
#w' = (2*y)/((-4)*w)
#Plug in (1,1,1):
#w' = (2*1)/((-4)*w) = 2/(-4) = -0.5
#ANSWER: dw/dy = (2*y)/((-4)*w), and dw/dy(1,1,1) = -0.5


#14.7 Homework
#Exercise 1:
#a)
#f_x = 2*x - 4*y = 0
#2*x = 4*y
#So, a = 2*b
#f_y = 4*y^3 - 4*x = 0
#4*y^3 = 4*x
#y^3 = x
#Check the three points:
#(0)^3 = 0 - Checks out
#(sqrt(2))^3 = 2*sqrt(2) - Checks out
#((-1)*sqrt(2))^3 = (-2)*sqrt(2) - Checks out
#b)
#Local minima are at (2*sqrt(2), sqrt(2)) and ((-2)*sqrt(2)), 
#and the saddle point is at (0,0)
#The absolute minimum value of f is:
#f(2*sqrt(2), sqrt(2)) = 4*2 + 4 - 8*2 = 8 + 4 - 16 = -4

#Exercise 3:
#To find the critical point, we first find the first order derivatives
#and set them to 0:
#f_x = 2*x + y = 0
#y = (-2)*x
#f_y = 32*y^3 + x - 6*y - 3*y^2 = 0
#32*(-2*x)^3 + x - 6*(-2)*x - 3*((-2)*x)^2 = 0
#(-256)*x^3 - 12*x^2 + 13*x = 0
#x*(256*x^2 + 12*x - 13) = 0
#x = (12 +- sqrt(144 - 4*256*(-13)))/512 = (12 +- 116)/512
#x = 0, 13/64, (-1)/4
#y = 0, -26/64, 1/2
#Looking at the figure:
#ANSWER: (0,0) is a saddle point, (13/64,-13/32) is a local minimum,
#and (-1/4,1/2) is also a local minimum

#Exercise 5:
#a)
#f_x = y^2 - 2*y*x + y = y*(y - 2*x + 1) = 0
#f_y = 2*y*x - x^2 + x = x*(2*y - x + 1) = 0
#b)
#At x = 0:
#f_x(0,y) = y*(y - 2*0 + 1) = y*(y+1) = 0
#f_y(0,y) = 0*(2*y - 0 + 1) = 0
#So, for both equations, if x is 0, y can either be 0 or a non-zero number (-1)
#And at y = 0:
#f_x(x,0) = 0*(0 - 2*x + 1) = 0
#f_y(x,0) = x*(2*0 - x + 1) = x*(1 - x) = 0
#So, for both equations, if y is 0, x can either be 0 or a non-zero number (1)
#We can also use substitution:
#y - 2*x + 1 = 0
#y = 2*x - 1
#Into the second equation:
#x*(2*(2*x - 1) - x + 1) = 0
#x*(4*x - 2 - x + 1) = 0
#x*(3*x - 1) = 0
#3*x - 1 = 0
#3*x = 1
#x = 1/3
#y = 2/3 - 1 = (-1)/3
#So, our critical points are (0,0), (0,-1), (1,0), and (1/3, (-1)/3)
#c)
#Find the second partial derivatives:
#f_xx = (-2)*y
#f_xy = 2*y - 2*x + 1
#f_yy = 2*x
#To see if a point is a minimum, maximum, or a saddle point, we can
#use the determinant:
#D = (f_xx)*(f_yy) - (f_xy)^2
#If D > 0 and f_xx > 0, the point is a min
#If D > 0 and f_xx < 0, the point is a max
#If D < 0, the point is a saddle point
#If D = 0, we do not know
#So test each of the 4 points:
#D(0,0) = (f_xx(0,0))*(f_yy(0,0)) - (f_xy(0,0))^2 = 
# = ((-2)*0)*(2*0) - (2*0 - 2*0 + 1)^2 = -1
#Because D(0,0) < 0, (0,0) is a saddle point
#D(0,-1) = (f_xx(0,-1))*(f_yy(0,-1)) - (f_xy(0,-1))^2 = 
# = ((-2)*(-1))*(2*0) - (2*(-1) - 2*0 + 1)^2 = 0 - (-2 + 1)^2 = -1
#Because D(0,-1) < 0, (0,-1) is a saddle point
#D(1,0) = (f_xx(1,0))*(f_yy(1,0)) - (f_xy(1,0))^2 = 
# = ((-2)*0)*(2*1) - (2*0 - 2*1 + 1)^2 = 0 - (-2 + 1)^2 = -1
#Because D(1,0) < 0, (1,0) is a saddle point
#D(1/3, (-1)/3) = (f_xx(1/3, (-1)/3))*(f_yy(1/3, (-1)/3)) - (f_xy(1/3, (-1)/3))^2 =
# = ((-2)*(-1)/3)*(2*1/3) - (2*(-1)/3 - 2*1/3 + 1)^2 = 
# = (2/3)*(2/3) - ((-2)/3 - 2/3 + 1)^2 = 3/9 = 1/3
#Because D(1/3, (-1)/3) > 0 and f_xx(1/3, (-1)/3) > 0, it is a local min
#ANSWER: (0,0), (0,-1), and (1,0) are sadle points,
#while (1/3, (-1)/3) is a local minimum

#Exercise 7:
#First, find the first partial derivatives of f:
#f_x = 2*x - y + 1
#f_y = 2*y - x
#Set both equations to 0 to find the critical points:
#2*y - x = 0
#x = 2*y
#Substitute:
#2*(2*y) - y + 1 = 0
#4*y - y = -1
#3*y = -1
#y = (-1)/3
#x = (-2)/3
#The critical point is ((-2)/3, (-1)/3)
#Next, find the second partial derivatives of f:
#f_xx = 2
#f_xy = -1
#f_yy = 2
#Find the determinant at the critical point:
#D((-2)/3, (-1)/3) = (f_xx)*(f_yy) - (f_xy)^2 = 
# = (2)*(2) - (-1)^2 = 4 - 1 = 3
#Because D((-2)/3, (-1)/3) > 0 and f_xx((-1)/3, (-2)/3) > 2,
#the point is a local minimum
#ANSWER: ((-2)/3, (-1)/3) is a local minimum

#Exercise 11:
#First, find the first partial derivatives of f:
#f_x = 4 - 9*x^2 - 2*y^2
#f_y = (-4)*x*y
#Set both equations to 0 to find the critical points:
#(-4)*x*y = 0
#x or y is 0. So, let's test every option
#At x = 0:
#4 - 0 - 2*y^2 = 0
#2*y^2 = 4
#y = +-sqrt(2)
#At y = 0:
#4 - 9*x^2 - 0 = 0
#9*x^2 = 4
#x = +-(2/3)
#Our critical points are: (0, sqrt(2)), (0, (-1)*sqrt(2)),
#(2/3, 0), and ((-2)/3, 0)
#Next, find the second partial derivatives of f:
#f_xx = (-18)*x
#f_xy = (-4)*y
#f_yy = (-4)*x
#Test every point with the Second Derivative Test:
#D(0, sqrt(2)) = ((-18)*0)*((-4)*0) - ((-4)*sqrt(2))^2 = 
# = -32
#Because D(0,sqrt(2)) < 0, it is a saddle point
#D(0, (-1)*sqrt(2)) = ((-18)*0)*((-4)*0) - ((-4)*(-1)*sqrt(2))^2 = 
# = -32
#Because D(0,(-1)*sqrt(2)) < 0, it is a saddle point
#D(2/3,0) = ((-18)*(2/3))*((-4)*(2/3)) - ((-4)*0)^2 = 
# = (-12)*((-8)/3) - 0 = 32
#Because D(2/3,0) > 32 and f_xx(2/3,0) < 0, it is a local max
#D((-2)/3,0) = ((-18)*((-2)/3))*((-4)*((-2)/3)) - ((-4)*0)^2 = 
# = (12)*((8)/3) - 0 = 32
#Because D((-2)/3,0) > 32 and f_xx((-2)/3,0) > 0, it is a local min
#ANSWER: (0, +-sqrt(2)) are saddle points, (2/3,0) is a local maximum,
#and ((-2)/3,0) is a local minimum

#Exercise 13:
#First, find the first partial derivatives of f:
#f_x = 4*x^3 - 4*y
#f_y = 4*y^3 - 4*x
#Set both equations to 0 to find the critical points:
#4*x^3 = 4*y
#x^3 = y
#4*y^3 = 4*x
#y^3 = x
#So, our points are (0,0), (1,1) and (-1,-1)
#Next, find the second partial derivatives of f:
#f_xx = 12*x^2
#f_xy = -4
#f_yy = 12*y^2
#Test every point with the Second Derivative Test:
#D(0,0) = (12*0^2)*(12*0^2) - (-4)^2 = 0 - 16 = -16
#Because D(0,0) < 0, it is a saddle point
#D(1,1) = (12*1^2)*(12*1^2) - (-4)^2 = 144 - 16 = 128
#Because D(1,1) > 0 anf f_xx(1,1) > 0, it is a local min
#D(-1,-1) = (12*(-1)^2)*(12*(-1)^2) - (-4)^2 = 144 - 16 = 128
#Because D(-1,-1) > 0 and f_xx(-1,-1) > 0, it is a local min
#ANSWER: (0,0) is a saddle point, and (1,1) and (-1,-1) are local minima

#Exercise 19:
#First, find the first partial derivatives of f:
#f_x = (1/x) - 1
#f_y = (2/y) - 4
#Set both equations to 0 to find the critical points:
#1/x = 1
#x = 1
#2/y = 4
#y = 0.5
#Our critical point is (1,0.5)
#Next, find the second partial derivatives of f:
#f_xx = (-1)/(x^2)
#f_xy = 0
#f_yy = (-2)/(y^2)
#Test the critical point with the Second Derivative Test:
#D(1,0.5) = ((-1)/(1^2))*((-2)/(0.5^2)) - 0^2 = (-1)*(-0.5) = 0.5
#Because D(1,0.5) > 0 and f_xx(1,0.5) < 0, it is a local max
#ANSWER: (1,0.5) is a local maximum

#Exercise 21:
#First, find the first partial derivatives of f:
#f_x = 1 - (1/(x + y))
#f_y = (-2)*y - (1/(x + y))
#Set both equations to 0 to find the critical points:
#1 - (1/(x + y)) = 0
#(1/(x + y)) = 1
#x + y = 1
#(-2)*y - (1/(x + y)) = 0
#(-2)*y - 1 = 0
#(-2)*y = 1
#y = -0.5
#x = 1.5
#Our critical point is (1.5,-0.5)
#Next, find the second partial derivatives of f:
#f_xx = 1/(x + y)^2
#f_xy = 1/(x + y)^2
#f_yy = -2 + (1/(x + y)^2)
#Test the critical point with the Second Derivative Test:
#D(1.5,-0.5) = (1/(1.5 - 0.5)^2)*(-2 + (1/(1.5 - 0.5)^2)) - (1/(1.5 - 0.5)^2)^2 = 
# = (1)*(-1) - 1 = -2
#Because D(1.5,-0.5) < 0, it is a saddle point
#ANSWER: (1.5,-0.5) is a saddle point

#Exercise 23:
#First, find the first partial derivatives of f:
#f_x = (x + 3*y)*((-2)*x*exp(y - x^2)) + exp(y - x^2) 
#f_y = (x + 3*y)*exp(y - x^2) + 3*exp(y - x^2)
#Set both equations to 0 to find the critical points:
#(x + 3*y)*exp(y - x^2) + 3*exp(y - x^2) = 0
#(x + 3*y + 3)*exp(y - x^2) = 0
#x + 3*y + 3 = 0
#x + 3*y = -3
#(x + 3*y)*((-2)*x*exp(y - x^2)) + exp(y - x^2) = 0
#(-2)*x^2 - 6*x*y + 1 = 0
#(-2)*x*(x + 3*y) + 1 = 0
#Substitute:
#(-2)*x*(-3) + 1 = 0
#6*x = -1
#x = (-1)/6
#y = (-17)/18
#Our critical point is ((-1)/6, (-17)/18)
#Next, find the second partial derivatives of f:
#f_xx = (2*x^3 + 6*x^2 - 3*x - 3*y)*2*exp(y - x^2)
#f_xy = (1 - 6*x*y - 2*x^2 - 6*x)*exp(y - x^2)
#f_yy = (6 + x + 3*y)*exp(y - x^2)
#Test the critical point with the Second Derivative Test:
#D((-1)/6, (-17)/18) = used a calculator = 2.57
#Because D((-1)/6, (-17)/18) > 0, and f_xx((-1)/6, (-17)/18) > 0,
#it is a local minimum
#ANSWER: The point ((-1)/6, (-17)/18) is a local minimum

#Exercise 29:
#Absolute min = 0 and is at (0,0) and absolute max = 2 and is at (1,1)

#Exercise 35:
#a)
#First, find the first partial derivatives of f:
#f_x = 1 - 2*x - y
#f_y = 1 - 2*y - x
#Set both equations to 0 to find the critical point:
#1 - 2*x - y = 0
#y = 1 - 2*x
#Substitute:
#1 - 2*(1 - 2*x) - x = 0
#1 - 2 + 4*x - x = 0
#3*x = 1
#x = 1/3
#y = 1/3
#The critical point is at (1/3, 1/3)
#f(1/3, 1/3) = (1/3) + (1/3) - (1/3)^2 - (1/3)^2 - (1/3)(1/3) = 
# = 2/3 - 1/9 - 1/9 - 1/9 = (6 - 1 - 1 - 1)/9 = 1/3
#ANSWER: At the critical point (1/3,1/3), the function equals to 1/3
#b)
#Find the derivative of the resulting function:
#F_x = 1 - 2*x
#Set it equal to 0:
#1 - 2*x = 0
#2*x = 1
#x = 0.5
#f(0.5,0) = 0.5 - 0.5^2 = 0.25
#Next, find the resulting function at the edgepoints:
#f(0,0) = 0
#f(2,0) = 2 - 2^2 = -2
#At the bottom edge, the absolute min is -2, and the absolute max is 0.25
#c)
#To find the extreme values of the top edge, we set y equal to 2:
#f(x,2) = x + 2 - x^2 - 2^2 - 2*x
#f(x,2) = -x^2 - x - 2
#Find the derivative of the resulting function and set it to 0:
#F_x = (-2)*x - 1 = 0
#(-2)*x = 1
#x = -0.5
#f(-0.5,2) = -(-0.5)^2 - (-0.5) - 2 = -(0.25) + 0.5 - 2 = -1.75
#Next, find the resulting function at the edgepoints:
#f(0,2) = -2
#f(2,2) = -(2)^2 - 2 - 2 = -4 - 2 - 2 = -8
#At the top edge, the absolute min is -8, and the absolute max is -1.75
#To find the extreme values of the left edge, we set x equal to 0:
#f(0,y) = y - y^2
#Find the derivative of the resulting function and set it to 0:
#F_y = 1 - 2*y = 0
#2*y = 1
#y = 0.5
#f(0,0.5) = 0.5 - 0.5^2 = 0.25
#Next, find the resulting function at the edgepoints:
#f(0,0) = 0
#f(0,2) = 2 - 2^2 = -2
#At the left egde, the absolute min is -2, and the absolute max is 0.25
#To find the extreme values of the right edge, we set x equal to 2:
#f(2,y) = 2 + y - 2^2 - y^2 - 2*y
#f(2,y) = -y^2 - y - 2
#Find the derivative of the resulting function and set it to 0:
#F_y = (-2)*y - 1 = 0
#(-2)*y = 1
#y = -0.5
#f(2,-0.5) = -(-0.5)^2 - (-0.5) - 2 = -0.25 + 0.5 - 2 = -1.75
#Next, find the resulting function at the edgepoints:
#f(2,0) = -2
#f(2,2) = -(2)^2 - 2 - 2 = -4 - 2 - 2 = -8
#At the right edge, the absolute min is -8, and the absolute max is -1.75
#d)
#ANSWER: The maximum value of the function in that region is 1/3