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

#15.6 Homework
#Exercise 1:
#(a):
#(u,0) -> (2*u,u+0) = (2*u,u) -> y = (1/2)*x
#(0,v) -> (2*0,0+v) = (0,v) - > the y-axis
#ANSWER: The image of the u-axis is the line y = (1/2)*x, and the
#image of the v-axis is the y-axis
#(b):
#The four corners of the rectangle are (0,0), (5,0), (0,7) and (5,7).
#So, convert every single corner:
#(0,0) -> (2*0,0+0) = (0,0)
#(5,0) -> (2*5,5+0) = (10,5)
#(0,7) -> (2*0,0+7) = (0,7)
#(5,7) -> (2*5,5+7) = (10,12)
#ANSWER: The image is a parallelogram with vertices (0,0), (10,5), (0,7) and (10,12)
#(c):
#Convert the two endpoints of the line:
#(1,2) -> (2*1,1+2) = (2,3)
#(5,3) -> (2*5,5+3) = (10,8)
#ANSWER: The image is a line segment connecting (2,3) and (10,8)
#(d):
#Convert every vertex of the triangle:
#(0,1) -> (2*0,0+1) = (0,1)
#(1,0) -> (2*1,1+0) = (2,1)
#(1,1) -> (2*1,1+1) = (2,2)
#ANSWER: The image is a triangle with vertices (0,1), (2,1) and (2,2)

#Exercise 3:
#G(u,v) is not one-to-one because of the u^2 - the result for u and -u is the same.
#It is one-to-one for u<=0 and u>=0
#(a):
#(u,0) -> (u^2,0) -> x>=0
#(0,v) -> (0,v) -> the y-axis
#ANSWER: The image of the u-axis is the positive x-axis, and the image of
#the v-axis is the y-axis
#(b):
#The four corners of the rectangle are (-1,-1), (-1,1), (1,1) and (1,-1).
#We 
#So, convert every single corner:
#(-1,-1) -> ((-1)^2,-1) = (1,-1)
#(-1,1) -> ((-1)^2,1) = (1,1)
#(1,1) -> (1^2,1) = (1,1)
#(1,-1) -> (1^2,-1) = (1,-1)
#There are repeating points because of the x^2 as the first component.
#But, looking at the transformation itself, it could be inferred that
#the new rectangle would be the same rectangle, but only its right half
#ANSWER: The image is a rectangle with vertices at (0,-1), (0,1), (1,-1) and (1,1)
#(c):
#The line connecting (0,0) and (1,1) has a slope of 1, and can
#be represented as u=v, and u has a range of 0<=u<=u
#Looking at the conversion, we can see that u=sqrt(x), and v=y.
#So, plugging it into u=v:
#y=sqrt(x).
#ANSWER: The image is a curve y=sqrt(x) for 0<=x<=1
#(d):
#We have three line segments:
#From (0,0) to (0,1), or v=0,
#From (0,1) to (1,1), or u=1,
#From (1,1) to (0,0), or v=u
#Plug each of the lines into the conversion:
#(0,v) -> (0^2,v) = (0,v) -> y-axis
#(u,1) -> (u^2,1) -> x=1, on the positive side
#Because u^2 = x, and y=v, our last line segment is the curve y=sqrt(x) for 0<=x<=1
#Our vertices remain the same.
#ANSWER: The image is a triangle with vertices (0,0), (0,1) and (1,1), but
#the curve that connects (0,0) and (1,1) is y=sqrt(x) for 0<=x<=1

#Exercise 13:
#The formula for the Jacobian in 2D is:
#J = |x_u x_v|
#    |y_u y_v|
#So first, find all of the partial derivatives needed:
#x_u = 3
#x_v = 4
#y_u = 1
#y_v = -2
#Plug the into the matrix:
#J = |3  4|
#    |1 -2| = 3*(-2) - 4*1 = -6-4 = -10
#ANSWER: The Jacobian for this transformation is -10

#Exercise 15:
#The formula for the Jacobian in 2D is:
#J = |x_r x_t|
#    |y_r y_t|
#So first, find all of the partial derivatives needed:
#x_r = sin(t)
#x_t = r*cos(t)
#y_r = 1
#y_t = sin(t)
#Plug in the point (1,Pi) in each derivative:
#x_r(1,Pi) = sin(Pi) = 0
#x_t(1,Pi) = 1*cos(Pi) = -1
#y_r(1,Pi) = 1
#y_t(1,Pi) = sin(Pi) = 0
#Plug the into the matrix:
#J = |0 -1|
#    |1  0| = 0*0 - 1*(-1) = 0+1 = 1
#ANSWER: The Jacobian for this transformation is 1

#Exercise 17:
#The formula for the Jacobian in 2D is:
#J = |x_r x_theta|
#    |y_r y_theta|
#So first, find all of the partial derivatives needed:
#x_r = cos(theta)
#x_theta = (-1)*r*sin(theta)
#y_r = sin(theta)
#y_theta = r*cos(theta)
#Plug in the point (4,Pi/6) in each derivative:
#x_r(4,Pi/6) = cos(Pi/6) = sqrt(3)/2
#x_theta(4,Pi/6) = (-1)*4*sin(Pi/6) = (-4)*(1/2) = -2
#y_r(4,Pi/6) = sin(Pi/6) = 1/2
#y_theta(4,Pi/6) = 4*cos(Pi/6) = 2*sqrt(3)
#Plug the into the matrix:
#J = |sqrt(3)/2 -2       |
#    |1/2       2*sqrt(3)| = sqrt(3)*sqrt(3) - (1/2)*(-2) = 3 + 1 = 4
#ANSWER: The Jacobian for this transformation is 4

#Exercise 19:
#The four vertices of the original rectangle are (0,0), (0,1), (1,1) and (1,0)
#The four vertices of the resulting parallelogram are (0,0), (2,3), (6,4) and (4,1)
#We can use a system of equations to solve for x-component first:
#2 = a*0 + b*1
#6 = a*1 + b*1
#4 = a*1 + b*0
#So, a=4 and b=2.
#Do the same for the y-component:
#3 = a*0 + b*1
#4 = a*1 + b*1
#1 = a*1 + b*0
#So, a=1 and b=3.
#Our tranformation is:
#G(u,v) = (4*u+2*v,u+3*v)
#ANSWER: G(u,v) = (4*u+2*v,u+3*v)

#Exercise 23:
#To find the area of G(R), we take the area of the original region R
#and multiply it by the absolute value of the transformation's Jacobian.
#For that, we need to find the Jacobian first:
#J = |x_u x_v| = |3  1|
#    |y_u u_v|   |1 -2| = 3*(-2) - 1*1 = -6-1 = -7
#(a):
#The area of the original rectangle is:
#(3-0) * (5-0) = 3*5 = 15
#Calculate the new area:
#|-7|*15 = 7*15 = 105
#ANSWER: The area of G(R) is 105
#(b):
#The area of the original rectangle is:
#(5-2) * (7-1) = 3*6 = 18
#Calculate the new area:
#|-7|*18 = 7*18 = 126
#ANSWER: The area of G(R) is 126


#16.1 Homework
#Exercise 1:
#For P:
#(1,2) -> <1^2,1> = <1,1>
#For Q:
#(-1,1) -> <(-1)^2,-1> = <1,-1>
#ANSWER: In the sketch, there will be a vector <1,1> at the point (1,2),
#and a vector <1,-1> at the point (-1,1)

#Exercise 3:
#For P:
#(0,1,1) -> <0*1,1^2,0> = <0,1,0>
#For Q:
#(2,1,0) -> <2*1,0^2,2> = <2,0,2>
#ANSWER: In the sketch, there will be a vector <0,1,0> at the point (0,1,1),
#and a vector <2,0,2> at the point (2,1,0)

#Exercise 5:
#We can draw the vector field using Maple:
with(plots);
fieldplot(<1, 0>, x = -3 .. 3, y = -3 .. 3);
#ANSWER: From the drawing, we can see that at every point in the rectangle,
#all vectors are the i-vector.

#Exercise 7:
#We can draw the vector field using Maple:
with(plots);
fieldplot(<x, 0>, x = -3 .. 3, y = -3 .. 3);
#ANSWER: From the drawing, we can see that every vector points in the direction
#of its starting point's x-value

#Exercise 9:
#We can draw the vector field using Maple:
with(plots);
fieldplot(<0, x>, x = -3 .. 3, y = -3 .. 3);
#ANSWER: From the drawing, we can see that every vector points in the direction
#parallel to the y-axis, with the magnitude of its x-value. Its direction is determined
#by whether the x-value is negative or not.

#Exercise 11:
#We can draw the vector field using Maple:
with(plots);
fieldplot(<x/(x^2 + y^2), y/(x^2 + y^2)>, x = -3 .. 3, y = -3 .. 3);
#ANSWER: From the drawing, we can see that every vector points away from the origin,
#with the vectors decreasing in length the farther away they are from the origin

#Exercise 17:
#ANSWER: C, because all of the vectors are constant length and point in one direction

#Exercise 23:
#The formula for div(F) is:
#div(F) = delta . F = dFx/dx + dFy/dy + dFz/dz
#So first, we find each of the required partial derivatives:
#dFx/dx = y
#dFy/dy = z
#dFz/dz = 0
#Add the results:
#div(F) = y+z
#The formula for curl(F) is:
#curl(F) = delta x F = |i    j    k   |
#                      |d/dx d/dy d/dz|
#                      |Fx   Fy   Fz  | =
# = i*(dFz/dy - dFy/dz) - j*(dFz/dx - dFx/dz) + k*(dFy/dx - dFx/dy)
#Find all of the partial derivatives needed:
#dFz/dy = 2*y
#dFy/dz = y
#dFz/dx = -3*x^2
#dFx/dz = 0
#dFy/dx = 0
#dFx/dy = x
#Plug them into the formula:
#curl(F) = i*(2*y - y) - j*(-3*x^2 - 0) + k*(0 - x) =
# = (y)*i + (3*x^2)*j - (x)*k = <y, 3*x^2, -x>
#ANSWER: div(F) = y+z, curl(F) = <y, 3*x^2, -x>

#Exercise 25:
#The formula for div(F) is:
#div(F) = delta . F = dFx/dx + dFy/dy + dFz/dz
#So first, we find each of the required partial derivatives:
#dFx/dx = 1-4*z*x
#dFy/dy = -x
#dFz/dz = 2*z*x^2
#Add the results:
#div(F) = 1 - 4*z*x - x + 2*z*x^2
#The formula for curl(F) is:
#curl(F) = delta x F = |i    j    k   |
#                      |d/dx d/dy d/dz|
#                      |Fx   Fy   Fz  | =
# = i*(dFz/dy - dFy/dz) - j*(dFz/dx - dFx/dz) + k*(dFy/dx - dFx/dy)
#Find all of the partial derivatives needed:
#dFz/dy = 0
#dFy/dz = 1
#dFz/dx = 2*z^2*x
#dFx/dz = -2*x^2
#dFy/dx = -y
#dFx/dy = 0
#Plug them into the formula:
#curl(F) = i*(0 - 1) - j*(2*z^2*x + 2*x^2) + k*(-y - 0) =
# = (-1)*i - (2*z^2*x + 2*x^2)*j - (y)*k = <-1, 2*z^2*x + 2*x^2, -y>
#ANSWER: div(F) = 1 - 4*z*x - x + 2*z*x^2, curl(F) = <-1, 2*z^2*x + 2*x^2, -y>

#Exercise 27:
#The formula for div(F) is:
#div(F) = delta . F = dFx/dx + dFy/dy + dFz/dz
#So first, we find each of the required partial derivatives:
#dFx/dx = 0
#dFy/dy = 0
#dFz/dz = 0
#Add the results:
#div(F) = 0
#The formula for curl(F) is:
#curl(F) = delta x F = |i    j    k   |
#                      |d/dx d/dy d/dz|
#                      |Fx   Fy   Fz  | =
# = i*(dFz/dy - dFy/dz) - j*(dFz/dx - dFx/dz) + k*(dFy/dx - dFx/dy)
#Find all of the partial derivatives needed:
#dFz/dy = 1
#dFy/dz = 3*z^2
#dFz/dx = 2*x
#dFx/dz = 1
#dFy/dx = 1
#dFx/dy = -2*y
#Plug them into the formula:
#curl(F) = i*(1 - 3*z^2) - j*(2*x - 1) + k*(1 + 2*y) =
# = (1 - 3*z^2)*i - (2*x - 1)*j + (1 + 2*y)*k = <1 - 3*z^2, 1 - 2*x, 1 + 2*y>
#ANSWER: div(F) = 0, curl(F) = <1 - 3*z^2, 1 - 2*x, 1 + 2*y>