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

#16.2 Homework
#Exercise 3:
#a)
#The path C is represented parametrically as:
#x = t, y = 1/t
#To find F(r(t)), we plug in the equations above into x and y variables of F:
#F(r(t)) = <(1/t)^2, t^2> = <1/t^2, t^2>
#And to find dr, we take the derivative with respect to t of each of its components
#and multiply them by dt:
#dr = <1, -1/t^2>*dt
#ANSWER: F(r(t)) = <1/t^2, t^2>, dr = <1, -1/t^2>*dt
#b)
#Our t-bounds for the curve would be 1<=t<=2, making it also be the line
#integral's bounds
#We first find the dot product of the results above:
#F(r(t)) . dr = <1/t^2, t^2> . <1, -1/t^2>*dt = (1/t^2 - t^2/t^2)*dt = (1/t^2 - 1)*dt
#Plug this and the t-bounds into the integral:
#int(1/t^2 - 1,t=1..2) = -1/t - t,t=1..2 = -1/2 - 2 + 1/1 + 1 = -1/2
#ANSWER: The value of the line integral is -1/2

#Exercise 9:
#First, we need the parametric representation of the curve:
#x = t, y = t^3, for 0<=t<=1
#The formula for the line integral of a function f(x,y) is:
#intC(f(x,y)*ds), where ds = sqrt(x_t^2 + y_t^2)
#First, let's find ds. Find all of the derivatives needed:
#x_t = 1
#y_t = 3*t^2
#And plug them into the ds formula:
#sqrt(1^2 + (3*t^2)^2) = sqrt(1 + 9*t^4)
#Now, represent the function f using the parametric equations above:
#f(t) = sqrt(1 + 9*t*t^3) = sqrt(1 + 9*t^4)
#Plug both results into the integral, with the appropriate t-bounds:
#int(sqrt(1 + 9*t^4)*sqrt(1 + 9*t^4),t=0..1) = 
# = int(1 + 9*t^4,t=0..1) = t + (9/5)*t^5,t=0..1 = 1 + (9/5) = 14/5 = 2.8
#ANSWER: The value of the line integral is 2/8

#Exercise 11:
#The parametric representation of the curve is:
#x = 2*t, y = 3*t, z=4*t, for 0<=t<=2
#The formula for the line integral of a function f(x,y,z) is:
#intC(f(x,y,z)*ds), where ds = sqrt(x_t^2 + y_t^2 + z_t^2)
#First, let's find ds. Find all of the derivatives needed:
#x_t = 2
#y_t = 3
#z_t = 4
#And plug them into the ds formula:
#sqrt(2^2 + 3^2 + 4^2) = sqrt(4 + 9 + 16) = sqrt(29)
#Now, represent the function f using the parametric equations above:
#f(t) = (4*t)^2 = 16*t^2
#Plug both results into the integral, with the appropriate t-bounds:
#int(sqrt(29)*16*t^2,t=0..2) = (sqrt(29)*16/3)*t^3,t=0..2 = (sqrt(29)*16/3)*8 ≈ 229.8
#ANSWER: The value of the line integral is about 229.8

#Exercise 13:
#The piecewise path can be represented parametrically in two parts:
#Part 1: x=0, y=2*t, z=1-t, for 0<=t<=1
#Part 2: x=t, y=1-t, z=t,   for 0<=t<=1
#The formula for the line integral of a function f(x,y,z) is:
#intC(f(x,y,z)*ds), where ds = sqrt(x_t^2 + y_t^2 + z_t^2)
#First, let's find ds for each path, starting with path 1:
#x_t = 0, y_t = 2, z_t = -1
#ds1 = sqrt(0 + 2^2 + (-1)^2) = sqrt(4+1) = sqrt(5)
#And the same for part 2:
#x_t = 1, y_t = -1, z_t = 1
#ds2 = sqrt(1^2 + (-1)^2 + 1^2) = sqrt(3)
#Now, represent the function f using the parametric equations above for part 1:
#f(t) = 0*exp((1-t)^2) = 0
#And the same for part 2:
#f(t) = t*exp(t^2)
#Using the results above, make two line integrals (one for each part) and add them
#together:
#int(0*sqrt(5),t=0..1) + int(t*exp(t^2)*sqrt(3),t=0..1) = int(t*exp(t^2)*sqrt(3),t=0..1) ->
# -> substitute u = t^2, du = 2*t*dt, dt = du/(2*t) -> int((sqrt(3)/2)*exp(u),u=0..1) = 
#sqrt(3)/2)*exp(u),u=0..1 = (sqrt(3)/2)*exp(1) - sqrt(3)/2 ≈ 1.488
#ANSWER: The value of the line integral is about 1.488

#Exercise 17:
#The parametric representation of the curve is:
#x = 4*t, y = -3*t, z=12*t, for 2<=t<=5
#The formula for the line integral of a function f(x,y,z) is:
#intC(f(x,y,z)*ds), where ds = sqrt(x_t^2 + y_t^2 + z_t^2)
#First, let's find ds. Find all of the derivatives needed:
#x_t = 4
#y_t = -3
#z_t = 12
#And plug them into the ds formula:
#sqrt(4^2 + (-3)^2 + 12^2) = sqrt(16 + 9 + 144) = sqrt(169) = 13
#The function stays the same, since it is just a constant 1.
#Plug both results into the integral, with the appropriate t-bounds:
#int(13,t=2..5) = 13*t,t=2..5 = 13*5 - 13*2 = 65 - 26 = 39
#ANSWER: The value of the line integral is 39, and it represents the magnitude of
#the line from t=2 to t=5

#Exercise 27:
#First, we need the parametric representation of the curve:
#x = t, y = t^2, for 0<=t<=2
#To find dx and dy, we need to find dx/dt and dy/dt and multiply those by dt:
#dx = dt
#dy = 2*t*dt
#Represent the functions of x and y inside the integral as a function of t:
#P = y = t^2
#Q = x = t
#Plug the results into the integral, with the appropriate t-bounds:
#int(t^2 - 2*t*t,t=0..2) = int(t^2 - 2*t^2,t=0..2) = int((-1)*t^2,t=0..2) = 
# = (-1/3)*t^3,t=0..2 = -8/3
#ANSWER: The value of the line integral is -8/3

#Exercise 29:
#First, we need the parametric representation of the curve:
#x = t, y = 4*t, z=4*t for 0<=t<=1
#To find dx, dy and dz, we need to find dx/dt, dy/dt and dz/dt and multiply those by dt:
#dx = dt
#dy = 4*dt
#dz = 4*dt
#Represent the functions of x, y and z inside the integral as a function of t:
#P = (x-y) = t-4*t = -3*t
#Q = (y-z) = 4*t-4*t = 0
#R = z = 4*t
#Plug the results into the integral, with the appropriate t-bounds:
#int(-3*t + 0 + 4*t*4,t=0..1) = int(-3*t + 16*t,t=0..1) = int(13*t,t=0..1) = 
# = (13/2)*t^2,t=0..1 = 13/2
#ANSWER: The value of the line integral is 13/2

#Exercise 31:
#First, we need the parametric representation of the curve:
#x = r*cos(t), y = r*sin(t), for 0<=t<=Pi/2
#The radius of the path is 1. So, our final representation is:
#x = cos(t), y = sin(t), for 0<=t<=Pi/2
#To find dx and dy, we need to find dx/dt and dy/dt and multiply those by dt:
#dx = -sin(t)*dt
#dy = cos(t)*dt
#Represent the functions of x and y inside the integral as a function of t:
#P = -y = -sin(t)
#Q = x = cos(t)
#Plug the results into the integral, with the appropriate t-bounds:
#int((sin(t)^2 + cos(t)^2)/(sin(t)^2 + cos(t)^2),t=0..Pi/2) = int(1,t=0..Pi/2) = 
# = t,t=0..Pi/2 = Pi/2
#ANSWER: The value of the line integral is Pi/2

#Exercise 35:
#There are 3 parts to that path that can be represented parametrically:
#Part 1: x=0, y=0, z=t, for 0<=t<=1
#Part 2: x=0, y=t, z=1, for 0<=t<=t
#Part 3: x=-t, y=1, z=1, for 0<=t<=1
#Make a line integral for part 1 by finding its dx, dy, and dz and representing the F
#field as a field of t:
#dx = 0, dy = 0, dz = dt
#F(t) = <exp(t), exp(0), exp(0)> = (exp(t), 1, 1)
#So, the part 1 integral would be:
#int(exp(t)*0 + 1*0 + 1*1,t=0..1) = int(1,t=0..1) = t,t=0..1 = 1
#Next, make a line integral for part 2 by finding its dx, dy, and dz and representing the F
#field as a field of t:
#dx = 0, dy = dt, dz = 0
#F(t) = <exp(1), exp(-t), exp(t)>
#So, the part 1 integral would be:
#int(exp(1)*0 + exp(-t)*1 + exp(t)*0,t=0..1) = int(exp(-t),t=0..1) = (-1)*exp(-t),t=0..1 = 
# = (-1)*exp(-1) + exp(0) = 1 - exp(-1)
#Lastly, make a line integral for part 2 by finding its dx, dy, and dz and representing 
#the F field as a field of t:
#dx = -dt, dy = 0, dz = 0
#F(t) = <exp(1), exp(-t-1), exp(1)>
#So, the part 1 integral would be:
#int(exp(1)*(-1) + exp(-t-1)*0 + exp(1)*0,t=0..1) = int(exp(1)*(-1),t=0..1) = 
# = (-1)*exp(1)*t,t=0..1 = (-1)*exp(1)
#Finally, add all of the line integrals together:
# 1 + 1 - exp(-1) - exp(1) = 2 - exp(-1) - exp(1)
#ANSWER: The value of the line integral is 2 - exp(-1) - exp(1)


#16.3 Homework
#Exercise 1:
#First, let's find F, if we know that F=grad(f):
#F = <f_x, f_y, f_z> = <y*sin(y*z), x*sin(y*z) + x*y*z*cos(y*z), x*y^2*cos(y*z)>
#We need to check that the function is path-independent, or that dP/dy = dQ/dx:
#dP/dy = sin(y*z) + y*z*cos(y*z)
#dQ/dx = sin(y*z) + y*z*cos(y*z)
#dP/dy = dQ/dx, so, the function is path-independent, meaning we can taky any path from
#the starting point to the ending point and still get the same value for the line integral.
#We can pick a path and take its parametric representation:
#x = t, y = t, z = Pi*t, for 0<=t<=1
#To find F(r(t)), we plug in the equations above into x and y variables of F:
#F(r(t)) = <t*sin(Pi*t^2), t*sin(Pi*t^2) + Pi*t^3*cos(Pi*t^2), t^3*cos(Pi*t^2)>
#And to find dr, we take the derivative with respect to t of each of its components
#and multiply them by dt:
#dr = <1, 1, Pi>*dt
#We first find the dot product of the results above:
#F(r(t)) . dr = <t*sin(Pi*t^2), t*sin(Pi*t^2) + Pi*t^3*cos(Pi*t^2), t^3*cos(Pi*t^2)> . 
# . <1, 1, Pi>*dt = (t*sin(Pi*t^2) + t*sin(Pi*t^2) + Pi*t^3*cos(Pi*t^2) + Pi*t^3*cos(Pi*t^2))*dt 
#Plug this and the t-bounds into the integral, and use Maple to find the answer:
int(t*sin(Pi*t^2) + t*sin(Pi*t^2) + Pi*t^3*cos(Pi*t^2) + Pi*t^3*cos(Pi*t^2),t=0..1);
#Result: 0
#ANSWER: The value of the line integral is 0

#Exercise 3:
#We can check if F=grad(f) by taking the partial derivatives of f and see if they are equal
#to the components of F:
#f_x = 3 == 3
#f_y = 6*y == 6*y
#Our parametric representation is:
#x = t, y = 2/t, for 1<=t<=4
#To find F(r(t)), we plug in the equations above into x and y variables of F:
#F(r(t)) = <3, 6*2/t> = <3, 12/t>
#And to find dr, we take the derivative with respect to t of each of its components
#and multiply them by dt:
#dr = <1, (-2)/t^2>*dt
#Our t-bounds for the curve would be 1<=t<=4, making it also be the line
#integral's bounds
#We first find the dot product of the results above:
#F(r(t)) . dr = <3, 12/t> . <1, (-2)/t^2>*dt = (3 - 24/t^3)*dt
#Plug this and the t-bounds into the integral:
#int(3 - 24/t^3,t=1..4) = 3*t + 12/t^2,t=1..4 = 3*4 + 12/16 - 3*1 - 12 = 12 + 3/4 - 3 - 12 = 
# = (3-12)/4 = -9/4
#ANSWER: The value of the line integral is -9/4

#Exercise 5:
#We can check if F=grad(f) by taking the partial derivatives of f and see if they are equal
#to the components of F:
#f_x = y*exp(z) == y*exp(z)
#f_y = x*exp(z) == x*exp(z)
#f_z = x*y*exp(z) == x*y*exp(z)
#Our parametric representation is:
#x = t^2, y = t^3, z=t-1, for 1<=t<=2
#To find F(r(t)), we plug in the equations above into x and y variables of F:
#F(r(t)) = <t^3*exp(t-1), t^2*exp(t-1), t^5*exp(t-1)>
#And to find dr, we take the derivative with respect to t of each of its components
#and multiply them by dt:
#dr = <2*t, 3*t^2, 1>*dt
#Our t-bounds for the curve would be 1<=t<=2, making it also be the line
#integral's bounds
#We first find the dot product of the results above:
#F(r(t)) . dr = <t^3*exp(t-1), t^2*exp(t-1), t^5*exp(t-1)> . <2*t, 3*t^2, 1>*dt = 
# = (2*t^4*exp(t-1) + 3*t^4*exp(t-1) + t^5*exp(t-1))*dt
#Plug this and the t-bounds into the integral, and use Maple to find the answer:
int(2*t^4*exp(t-1) + 3*t^4*exp(t-1) + t^5*exp(t-1),t=1..2);
#Result: -1 + 32*exp(1)
#ANSWER: The value of the line integral is 32*exp(1) - 1

#Exercise 9:
#We can check if F is conservative if its dP/dy is equal to dQ/dx:
#dP/dy = 2*y
#dQ/dx = 2*y
#dP/dy = dQ/dx, so, F is conservative and we can find its potential function.
#Since P = f_x, we can integrate it with respect to x to find f:
#int(y^2,x) = y^2*x + g(y,z), where g(y,z) is like an arbitrary constant.
#To find g(y,z), we can take the derivative with respect to y of the result, and then 
#integrate the resulting g_y:
#(y^2*x + g(y,z))_y = 2*x*y + exp(z)
#2*x*y + g_y = 2*x*y + exp(z)
#g_y = exp(z) -> g(y,z) = exp(z)*y
#So, our updated function f is:
#y^2*x + exp(z)*y + h(z)
#To find h(z), we can take the derivative with respect to z of the result, and then
#integrate the resulting h_z:
#(y^2*x + exp(z)*y + h(z))_z = y*exp(z)
#exp(z)*y + h_z = exp(z)*y
#h_z = 0 -> h(z) = 0
#So, our final potential function is:
#y^2*x + exp(z)*y
#ANSWER: The potential function is f(x,y,z) = y^2*x + exp(z)*y

#Exercise 13:
#We can check if F is conservative if its dP/dy is equal to dQ/dx:
#dP/dy = 0
#dQ/dx = 0
#dP/dy = dQ/dx, so, F is conservative and we can find its potential function.
#Since Q = f_y, we can integrate it with respect to y to find f:
#int(z,y) = z*y + g(x,z), where g(x,z) is like an arbitrary constant.
#To find g(x,z), we can take the derivative with respect to z of the result, and then 
#integrate the resulting g_z:
#(z*y + g(x,z))_z = y + tan(x)
#y + g_z = y + tan(x)
#g_z = tan(x) -> g(x,z) = tan(x)*z
#So, our updated function f is:
#z*y + tan(x)*z + h(x)
#To find h(x), we can take the derivative with respect to x of the result, and then
#integrate the resulting h_x:
#(z*y + tan(x)*z + h(x))_x = z*sec(x)^2
#z*sec(x)^2 + h_x = z*sec(x)^2
#h_x = 0  -> h(x) = 0
#So, our final potential function is:
#z*y + tan(x)*z
#ANSWER: The potential function is f(x,y,z) = z*y + tan(x)*z

#Exercise 15:
#We can check if F is conservative if its dP/dy is equal to dQ/dx:
#dP/dy = 2*x
#dQ/dx = 2*x
#dP/dy = dQ/dx, so, F is conservative and we can find its potential function.
#Since P = f_x, we can integrate it with respect to x to find f:
#int(z,y) = x^2*y + 5*x + g(y,z), where g(y,z) is like an arbitrary constant.
#To find g(y,z), we can take the derivative with respect to y of the result, and then 
#integrate the resulting g_y:
#(x^2*y + 5*x + g(y,z))_y = x^2 - 4*z
#x^2 + g_y = x^2 - 4*z
#g_y = -4*z -> g(x,z) = -4*y*z
#So, our updated function f is:
#x^2*y + 5*x - 4*y*z + h(z)
#To find h(z), we can take the derivative with respect to z of the result, and then
#integrate the resulting h_z:
#(x^2*y + 5*x - 4*y*z + h(z))_z = -4*y
#-4*y + h_z = -4*y
#h_z = 0  -> h(z) = 0
#So, our final potential function is:
#x^2*y + 5*x - 4*y*z
#ANSWER: The potential function is f(x,y,z) = x^2*y + 5*x - 4*y*z

#Exercise 17:
#If we prove that F is path-independent, we can pick a parametric representation
#that is easier to integrate. To prove this, we need to check that F's dP/dy is equal
#to its dQ/dx:
#dP/dy = 2*x*z
#dQ/dx = 2*x*z
#They are equal, so, we can make a different parametric representation:
#r(0) = <0,sin(0),exp(0)> = <0,0,1>
#r(2) = <4,sin(Pi/2),exp(0)> = <4,1,1>
#Our new parametric equations are:
#x=4*t, y=t, z=1, for 0<=t<=1
#To find dx, dy and dz, we need to find dx/dt, dy/dt and dz/dt and multiply those by dt:
#dx = 4*dt
#dy = dt
#dz = 0
#Represent the functions of x, y and z inside the integral as a function of t:
#P = 2*x*y*z = 2*4*t*t*1 = 8*t^2
#Q = x^2*z = 16*t^2*1 = 16*t^2
#R = x^2*y = 16*t^2*t = 16*t^3
#Plug the results into the integral, with the appropriate t-bounds:
#int(32*t^2 + 16*t^2 + 0,t=0..1) = int(48*t^2,t=0..1) = 16*t^3,t=0..1 = 16
#ANSWER: The value of the line integral is 16

#Exercise 19:
#F = graf(f), so:
#F = <f_x, f_y, f_z> = <2*x*y, x^2, -1>
#First path:::::
#Our parametric equations are:
#x=t, y=t, z=0, for 0<=t<=1
#To find dx, dy and dz, we need to find dx/dt, dy/dt and dz/dt and multiply those by dt:
#dx = dt
#dy = dt
#dz = 0
#Represent the functions of x, y and z inside the integral as a function of t:
#P = 2*x*y = 2*t^2
#Q = x^2 = t^2
#R = -1
#Plug the results into the integral, with the appropriate t-bounds:
#int(2*t^2 + t^2 + 0,t=0..1) = int(3*t^2,t=0..1) = t^3,t=0..1 = 1
#Second path:::::
#Our parametric equations are:
#x=t, y=t^2, z=0, for 0<=t<=1
#To find dx, dy and dz, we need to find dx/dt, dy/dt and dz/dt and multiply those by dt:
#dx = dt
#dy = 2*t*dt
#dz = 0
#Represent the functions of x, y and z inside the integral as a function of t:
#P = 2*x*y = 2*t^3
#Q = x^2 = t^2
#R = -1
#Plug the results into the integral, with the appropriate t-bounds:
#int(2*t^3 + 2*t^3 + 0,t=0..1) = int(4*t^3,t=0..1) = t^4,t=0..1 = 1
#ANSWER: Both line integrals are equal to 1