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


#14.3 Homework
#Exercise 3:
#If f(x) = g(x)/h(x), then the Quotient rule states that:
#f'(x) = (g'(x)*h(x) - h'(x)*g(x))/(h(x))^2
#So, following this rule:
#(d/dy)(y/(x+y)) = (1*(x+y) - 1*y)/(x+y)^2 = (x)/(x+y)^2
#ANSWER: The derivative with respect to y is (x)/(x+y)^2

#Exercise 5:
#First, we find the partial derivative with respect to z, df/dz
#For that, we treat x and y as constants:
#df/dz = (x*y*z)' = x*y
#Now, we plug in the values x=2, y=3 and z=1
#df/dz(2,3,1) = 2*3 = 6
#ANSWER: At point (2,3,1), df/dz = 6

#Exercise 17:
#To find partial derivative dz/dx, we treat y as a constant,
#and to find partial derivative dz/dy, we treat x as a constant:
#dz/dx = (x/y)' = 1/y
#dz/dy = (x/y)' = (-1)*x/y^2
#ANSWER: dz/dx = 1/y, dz/dy = (-1)*x/y^2

#Exercise 19:
#To find partial derivative dz/dx, we treat y as a constant,
#and to find partial derivative dz/dy, we treat x as a constant:
#dz/dx = (sqrt(9 - x^2 - y^2))' = 1/(2*sqrt(9 - x^2 - y^2)) * -2*x
#dz/dx = (-1)*x/sqrt(9 - x^2 - y^2)
#dz/dy = (sqrt(9 - x^2 - y^2))' = 1/(2*sqrt(9 - x^2 - y^2)) * -2*y
#dz/dy = (-1)*y/sqrt(9 - x^2 - y^2)
#ANSWER: dz/dx = (-1)*x/sqrt(9 - x^2 - y^2), dz/dy = (-1)*y/sqrt(9 - x^2 - y^2)

#Exercise 21:
#To find partial derivative dz/dx, we treat y as a constant,
#and to find partial derivative dz/dy, we treat x as a constant:
#dz/dx = (sin(x)*sin(y))' = sin(y)*cos(x)
#dz/dy = (sin(x)*sin(y))' = sin(x)*cos(y)
#ANSWER: dz/dx = sin(y)*cos(x), dz/dy = sin(x)*cos(y)

#Exercise 27:
#To find partial derivative dW/dr, we treat s as a constant,
#and to find partial derivative dW/ds, we treat r as a constant:
#dW/dr = (exp(r+s))' = exp(r+s) * 1 = exp(r+s)
#dW/ds = (exp(r+s))' = exp(r+s) * 1 = exp(r+s)
#ANSWER: dW/dr = exp(r+s), dW/ds = exp(r+s)

#Exercise 31:
#To find partial derivative dz/dx, we treat y as a constant,
#and to find partial derivative dz/dy, we treat x as a constant:
#dz/dx = exp((-1)*x^2 - y^2)' = exp((-1)*x^2 - y^2) * (-2)*x = -2*x*exp((-1)*x^2 - y^2)
#dz/dy = exp((-1)*x^2 - y^2)' = exp((-1)*x^2 - y^2) * (-2)*y = -2*y*exp((-1)*x^2 - y^2)
#ANSWER: dz/dx = -2*x*exp((-1)*x^2 - y^2), dz/dy = -2*y*exp((-1)*x^2 - y^2)

#Exercise 39:
#To find partial derivative dQ/dL, we treat M and t as constants,
#to find partial derivative dQ/dM, we treat L and t as constants,
#and to find partial derivative dQ/dt, we treat M and L as constants:
#dQ/dL = ((L/M)*exp(-L*t/M))' = (L/M)'*(exp(-L*t/M) + (L/M)*(exp(-L*t/M))' =
# = exp(-L*t/M)/M + (L/M)*exp(-L*t/M)*(-t/M) = exp(-L*t/M)/M + (-t*L/M^2*exp(-L*t/M) =
# = (M - t*L)*exp(-L*t/M)/M^2
#dQ/dM = ((L/M)*exp(-L*t/M))' = (L/M)'*(exp(-L*t/M) + (L/M)*(exp(-L*t/M))' = 
# = (-L/M^2)*(exp(-L*t/M) + (L/M)*exp(-L*t/M)*(L*t/M^2) = (-L/M^2)*(exp(-L*t/M) + (L^2*t/M^3)*exp(-L*t/M) = 
# = (-L*M + L^2*t)*exp(-L*t/M)/M^3 = L*(L*t - M)*exp(-L*t/M)/M^3
#dQ/dt = ((L/M)*exp(-L*t/M))' = (L/M)*exp(-L*t/M)*(-L/M) = (-L^2/M^2)*exp(-L*t/M)
#ANSWER: dQ/dL = (M - t*L)*exp(-L*t/M)/M^2, dQ/dM = L*(L*t - M)*exp(-L*t/M)/M^3, dQ/dt = (-L^2/M^2)*exp(-L*t/M)

#Exercise 47:
#a)
HeatIndex := (T, H) -> 45.33 + 0.6845*T + 5.758*H + (-1)*0.00365*T^2 + (-1)*0.1565*H*T + 0.001*H*T^2;
Answer := HeatIndex(95,50);
#ANSWER: At 95 degrees Fahrenheit and 50% humidity, the heat index is 73.19125
#b) 
#The increase in I per degree increase in T can also be written as dI/dT:
HeatIndexPerDegree := (T, H) -> 0.6845 + (-1)*0.00365*2*T + (-1)*0.1565*H + 0.002*H*T;
Answer := HeatIndexPerDegree(95,50);
#ANSWER: The increase in I per degree increase in T at 95 degrees Fahrenheit and
#50% humidity is 1.66600


#14.4 Homework
#Exercise 3:
#To find the equation of the tangent plane, we first need to find
#the vector normal to the surface at point (2,1)
#For this, we find its gradient vector:
#∇f = <df/dx, df/dy>
#So first, we find the first partial derivatives of f(x,y):
#df/dx = 2*x*y + y^3
#df/dy = x^2 + 3*x*y^2
#Plug in the point (2,1) into each partial derivative:
#df/dx(2,1) = 2*2*1 + 1^3 = 4 + 1 = 5
#df/dy(2,1) = 2^2 + 3*2*1^2 = 4 + 6 = 10
#So, the vector normal to the surface is:
#∇f = <5,10>
#We also need to find f(2,1), for the plane equation:
#f(2,1) = 2^2*1 + 2*1^3 = 4 + 2 = 6
#If our normal vector is <n1,n2> and our point on the plane is (x0,y0,z0),
#the formula for the tangent plane is:
#n1*(x-x0) + n2*(y-y0) = z-z0
#Plug in the normal vector <5,10> and the point (2,1,6):
#5*(x-2) + 10*(y-1) = z-6
#5*x - 10 + 10*y - 10 = z-6
#5*x + 10*y = z+14
#ANSWER: The equation for the tangent plane at point (2,1) is z = 5*x + 10*y - 14

#Exercise 5:
#To find the equation of the tangent plane, we first need to find
#the vector normal to the surface at point (4,1)
#For this, we find its gradient vector:
#∇f = <df/dx, df/dy>
#So first, we find the first partial derivatives of f(x,y):
#df/dx = 2*x
#df/dy = -2*y^(-3)
#Plug in the point (4,1) into each partial derivative:
#df/dx(2,1) = 2*4 = 8
#df/dy(2,1) = -2*1^(-3) = -2
#So, the vector normal to the surface is:
#∇f = <8,-2>
#We also need to find f(4,1), for the plane equation:
#f(4,1) = 4^2 + 1^(-2) = 16 + 1 = 17
#If our normal vector is <n1,n2> and our point on the plane is (x0,y0,z0),
#the formula for the tangent plane is:
#n1*(x-x0) + n2*(y-y0) = z-z0
#Plug in the normal vector <8,-2> and the point (4,1,17):
#8*(x-4) + (-2)*(y-1) = z-17
#8*x - 32 - 2*y + 2 = z-17
#8*x - 2*y = z+13
#ANSWER: The equation for the tangent plane at point (4,1) is z = 8*x - 2*y - 13

#Exercise 7:
#To find the equation of the tangent plane, we first need to find
#the vector normal to the surface at point (2,1)
#For this, we find its gradient vector:
#∇f = <df/dr, df/ds>
#So first, we find the first partial derivatives of f(r,s):
#df/dr = 2*r*s^(-0.5) = 2*r/sqrt(s)
#df/ds = (-1)*r^2*s^(-3/2)/2 - 3*s^(-4)
#Plug in the point (2,1) into each partial derivative:
#df/dr(2,1) = 2*2/sqrt(1) = 4/1 = 4
#df/ds(2,1) = (-1)*2^2*1^(-3/2)/2 - 3*1^(-4)= (-1)*4/2 - 3= -5
#So, the vector normal to the surface is:
#∇f = <4,-5>
#We also need to find f(2,1), for the plane equation:
#f(2,1) = 2^2*1^(-0.5) + 1^(-3) = 4 + 1 = 5
#If our normal vector is <n1,n2> and our point on the plane is (r0,s0,z0),
#the formula for the tangent plane is:
#n1*(r-r0) + n2*(s-s0) = z-z0
#Plug in the normal vector <8,-11> and the point (2,1,5):
#4*(r-2) + (-5)*(s-1) = z-5
#4*r - 8 - 5*s + 5 = z-5
#4*r - 5*s = z-2
#ANSWER: The equation for the tangent plane at point (2,1) is z = 4*r - 5*s + 2

#Exercise 13:
#The formula for linearization L(a,b) is:
#L(x,y) = f(a,b) + df/dx(a,b)*(x−a) + df/dy(a,b)*(y−b)
#So, first, we need to find partial derivatives of f(x,y):
#df/dx = 2*x*y^3
#df/dy = 3*x^2*y^2
#Calculate their numerical values at (2,1):
#df/dx(2,1) = 2*2*1^3 = 4
#df/dy(2,1) = 3*2^2*1^2 = 3*4 = 12
#Then, find the value of f(2,1):
#f(2,1) = 2^2*1^3 = 4
#Plug all of the values into the linearization formula:
#L(x,y) = 4 + 4*(x-2) + 12*(y-1)
#So, the L(x,y) for (2.01,1.02) and (1.97,1.01) are:
#L(2.01,1.02) = 4 + 4*(2.01-2) + 12*(1.02-1) = 4 + 4*0.01 + 12*0.02 = 4+0.04+0.24 =
# = 4.28
#L(1.97,1.01) = 4 + 4*(1.97-2) + 12*(1.01-1) = 4 + 4*(-0.03) + 12*(0.01) = 
# = 4 - 0.12 + 0.12 = 4
#Calculator values are:
f := (x, y) -> x^2*y^3;
ans1 := f(2.01,1.02)
ans2 := f(1.97,1.01)
#Results: 4.287386441, 3.998495151 ≈ 4.28, 4
#ANSWER: The estimated values are 4.28 and 4, and they match the calculator values

#Exercise 15:
#To estimate the change, we can use the formula:
#df = df/dx(a,b)*(x−a) + df/dy(a,b)*(y−b)
#So, first, we need to find partial derivatives of f(x,y):
#df/dx = 3*x^2*y^(-4)
#df/dy = (-4)*x^3*y^(-5)
#Calculate their numerical values at (2,1):
#df/dx(2,1) = 3*2^2*1^(-4) = 3*4 = 12
#df/dy(2,1) = (-4)*2^3*1^(-5) = (-4)*8 = -32
#Plug all of the values into the formula
#df = 12*(x−2) - 32*(y−1)
#Plug in (2.03,0.9):
#df = 12*(2.03-2) - 32*(0.9-1) = 12*(0.03) - 32*(-0.1) = 0.36 + 3.2 = 3.56
#ANSWER: The estimated change is 3.56

#Exercise 17:
#The formula for linearization L(a,b) is:
#L(x,y) = f(a,b) + df/dx(a,b)*(x−a) + df/dy(a,b)*(y−b)
#So, first, we need to find partial derivatives of f(x,y):
#df/dx = exp(x^2 + y) * 2*x = 2*x*exp(x^2 + y)
#df/dy = exp(x^2 + y) * 1 = exp(x^2 + y)
#Calculate their numerical values at (0,0):
#df/dx(0,0) = 2*0*exp(0^2 + 0) = 0
#df/dy(0,0) = exp(0^2 + 0) = exp(0) = 1
#Then, find the value of f(0,0):
#f(0,0) = exp(0^2 + 0) = exp(0) = 1
#Plug all of the values into the linearization formula:
#L(x,y) = 1 + 0*(x-0) + 1*(y-0) = 1 + y
#So, the L(x,y) for (0.01,-0.02) is:
#L(0.01,-0.02) = 1 - 0.02 = 0.98
#The calculator value is:
f := (x, y) -> exp(x^2 + y);
ans := f(0.01,-0.02);
#Result: 0.9802966981 ≈ 0.98
#ANSWER: The estimated value is 0.98, and it matches the calculator value

#Exercise 23:
#The function to be estimated could be written as:
f := (x,y) -> x^3*y^2;
#The formula for linearization L(a,b) is:
#L(x,y) = f(a,b) + df/dx(a,b)*(x−a) + df/dy(a,b)*(y−b)
#So, first, we need to find partial derivatives of f(x,y):
#df/dx = 3*x^2*y^2
#df/dy = 2*x^3*y
#Calculate their numerical values at (2,1), as this point is very close to the one we need:
#df/dx(2,1) = 3*2^2*1^2 = 3*4 = 12
#df/dy(2,1) = 2*2^3*1 = 2*8 = 16
#Then, find the value of f(2,1):
#f(2,1) = 2^3*1^2 = 8
#Plug all of the values into the linearization formula:
#L(x,y) = 8 + 12*(x-2) + 16*(y-1)
#So, the L(x,y) for (2.01,1.02) is:
#L(2.01,1.02) = 8 + 12*(2.01-2) + 16*(1.02-1) = 8 + 12*(0.01) + 16*(0.02) = 
# = 8 + 0.12 + 0.32 = 8.44
#The calculator value is:
ans := f(2.01,1.02);
#Result: 8.448673280 ≈ 8.44
#ANSWER: The estimated value is 8.44, and it matches the calculator value

#Exercise 25:
#The function to be estimated could be written as:
f := (x,y) -> sqrt(x^2 + y^2);
#The formula for linearization L(a,b) is:
#L(x,y) = f(a,b) + df/dx(a,b)*(x−a) + df/dy(a,b)*(y−b)
#So, first, we need to find partial derivatives of f(x,y):
#df/dx = 1/(2*sqrt(x^2 + y^2)) * 2*x = x/sqrt(x^2 + y^2)
#df/dy = 1/(2*sqrt(x^2 + y^2)) * 2*y = y/sqrt(x^2 + y^2)
#Calculate their numerical values at (3,4), as this point is very close to the one we need:
#df/dx(3,4) = 3/sqrt(3^2 + 4^2) = 3/5
#df/dy(3,4) = 4/sqrt(3^2 + 4^2) = 4/5
#Then, find the value of f(3,4):
#f(3,4) = sqrt(3^2 + 4^2) = 5
#Plug all of the values into the linearization formula:
#L(x,y) = 5 + (3/5)*(x-3) + (4/5)*(y-4)
#So, the L(x,y) for (3.01,3.99) is:
#L(3.01,3.99) = 5 + (3/5)*(3.01-3) + (4/5)*(3.99-4) = 5 + (0.6)*(0.01) - (0.8)*(0.01) = 
# = 5 + 0.006 - 0.008 = 4.998
#The calculator value is:
ans := f(3.01,3.99);
#Result: 4.998019608 ≈ 4.998
#ANSWER: The estimated value is 4.998, and it matches the calculator value

#Exercise 27:
#The function to be estimated could be written as:
f := (x,y,z) -> sqrt(x*y*z);
#The formula for linearization L(a,b,c) is:
#L(x,y,z) = f(a,b,c) + df/dx(a,b,c)*(x−a) + df/dy(a,b,c)*(y−b) + df/dz(a,b,c)*(z-c)
#So, first, we need to find partial derivatives of f(x,y):
#df/dx = 1/(2*sqrt(x*y*z)) * y*z = y*z/(2*sqrt(x*y*z))
#df/dy = 1/(2*sqrt(x*y*z)) * x*z = x*z/(2*sqrt(x*y*z))
#df/dz = 1/(2*sqrt(x*y*z)) * x*y = x*y/(2*sqrt(x*y*z))
#Calculate their numerical values at (2,2,4), as this point is very close to the one we need:
#df/dx(2,2,4) = 2*4/(2*sqrt(2*2*4)) = 4/sqrt(16) = 4/4 = 1
#df/dy(2,2,4) = 2*4/(2*sqrt(2*2*4)) = 4/sqrt(16) = 4/4 = 1
#df/dz(2,2,4) = 2*2/(2*sqrt(2*2*4)) = 2/sqrt(16) = 2/4 = 1/2
#Then, find the value of f(2,2,4):
#f(2,2,4) = sqrt(2*2*4) = sqrt(16) = 4
#Plug all of the values into the linearization formula:
#L(x,y,z) = 4 + 1*(x−2) + 1*(y−2) + 0.5*(z-4)
#So, the L(x,y,z) for (1.9,2.02,4.05) is:
#L(1.9,2.02,4.05) = 4 + (1.9-2) + (2.02-2) + 0.5*(4.05-4) = 4 - 0.1 + 0.02 + 0.025 = 
# = 3.9 + 0.045 = 3.945
#The calculator value is:
ans := f(1.9,2.02,4.05);
#Result: 3.942575300 ≈ 3.945
#ANSWER: The estimated value is 3.945, and it matches the calculator value


#14.5 Homework
#Exercise 7:
#The formula for the gradient is:
#∇h = <dh/dx, dh/dy, dh/dz>
#So first, find all of the first partial derivatives of h(x,y,z):
#dh/dx = y*z^(-3)
#dh/dy = x*z^(-3)
#dh/dz = (-3)*x*y*z^(-4)
#ANSWER: The gradient of h is <y*z^(-3), x*z^(-3), (-3)*x*y*z^(-4)>

#Exercise 11:
#Using the vector r(t) = <cos(t), sin(t)>, we can parametrize it as:
#x = cos(t), y = sin(t)
#The formula for the chain rule is:
#df/dt = (df/dx)*(dx/dt) + (df/dy)*(dy/dt)
#So, first, we find all of the needed partial derivatives:
#df/dx = 2*x - 3*y
#df/dy = (-3)*x
#dx/dt = (-1)*sin(t)
#dy/dt = cos(t)
#Plug all of the results in the formula:
#df/dt = (2*x-3*y)*((-1)*sin(t)) + ((-3)*x)*(cos(t))
#Plug in x = cos(t) and y = sin(t):
#df/dt = (2*cos(t)-3*sin(t))*((-1)*sin(t)) + ((-3)*cos(t))*(cos(t))
#Plug in t = 0:
#df/dt(0) = (2*cos(0)-3*sin(0))*((-1)*sin(0)) + ((-3)*cos(0))*(cos(0)) = 
# = (2*1-3*0)*((-1)*0) + ((-3)*1)*(1) = 0 - 3 = -3
#ANSWER: The numerical value of df/dt at t=0 is -3

#Exercise 13:
#Using the vector r(t) = <exp(2*t), exp(3*t)>, we can parametrize it as:
#x = exp(2*t), y = exp(3*t)
#The formula for the chain rule is:
#df/dt = (df/dx)*(dx/dt) + (df/dy)*(dy/dt)
#So, first, we find all of the needed partial derivatives:
#df/dx = cos(x*y) * y = y*cos(x*y)
#df/dy = cos(x*y) * x = x*cos(x*y)
#dx/dt = exp(2*t) * 2 = 2*exp(2*t)
#dy/dt = exp(3*t) * 3 = 3*exp(2*t)
#Plug all of the results in the formula:
#df/dt = (y*cos(x*y))*(2*exp(2*t)) + (x*cos(x*y))*(3*exp(2*t))
#Plug in x = exp(2*t) and y = exp(3*t):
#df/dt = (exp(3*t)*cos(exp(2*t)*exp(3*t)))*(2*exp(2*t)) + (exp(2*t)*cos(exp(2*t)*y))*(3*exp(2*t)) = 
# = (exp(3*t)*cos(exp(5*t)))*(2*exp(2*t)) + (exp(2*t)*cos(exp(5*t))*(3*exp(2*t))
#Plug in t = 0:
#df/dt(0) = (exp(3*0)*cos(exp(5*0)))*(2*exp(2*0)) + (exp(2*0)*cos(exp(5*0))*(3*exp(2*0)) = 
# = (1*cos(1))*(2*1) + (1*cos(1)*(3*1) = 2*cos(1) + 3*cos(1) = 5*cos(1) ≈ 2.702
#ANSWER: The numerical value of df/dt at t=0 is 5*cos(1), or about 2.702

#Exercise 19:
#Using the vector r(t) = <exp(t), t, t^2>, we can parametrize it as:
#x = exp(t), y = t, z = t^2
#The formula for the chain rule is:
#df/dt = (df/dx)*(dx/dt) + (df/dy)*(dy/dt) = (df/dz)*(dz/dt)
#So, first, we find all of the needed partial derivatives:
#df/dx = y*z^(-1) = t*(t^2)^-1 = t*t^(-2) = t^(-1)
#df/dy = x*z^(-1) = exp(t)*(t^2)^(-1) = exp(t)*t^(-2)
#df/dz = (-1)*x*y*z^(-2) = (-1)*exp(t)*t*(t^2)^(-2) = (-1)*exp(t)*t*t^(-4) = (-1)*exp(t)*t^(-3)
#dx/dt = exp(t)
#dy/dt = 1
#dz/dt = 2*t
#Evaluate all at t=1:
#df/dx(1) = 1^(-1) = 1
#df/dy(1) = exp(1)*1^(-2) = exp(1)
#df/dz(1) = (-1)*exp(1)*1^(-3) = (-1)*exp(1)
#dx/dt(1) = exp(1)
#dy/dt(1) = 1
#dz/dt(1) = 2*1 = 2
#Plug the values into the formula:
#df/dt = 1*exp(1) + exp(1)*1 + (-1)*exp(1)*2 = exp(1) + exp(1) - 2*exp(1) = 0
#ANSWER: The numerical value of df/dt at t=1 is 0

#Exercise 27:
#The formula for the directional derivative of f is:
#∇f . u
#Where u is the unit vector of the direction vector v
#First, we find the gradient, the formula for which is:
#∇f = <df/dx, df/dy>
#So, we find the first partial derivatives of f:
#df/dx = (ln(x^2 + y^2))' = 1/(x^2 + y^2) * 2*x = (2*x)/(x^2 + y^2)
#df/dy = (ln(x^2 + y^2))' = 1/(x^2 + y^2) * 2*y = (2*y)/(x^2 + y^2)
#We find the value of each derivative at (1,0):
#df/dx(1,0) = (2*1)/(1^2 + 0^2) = 2/1 = 2
#df/dy(1,0) = (2*0)/(1^2 + 0^2) = 0
#So, the gradient at (1,0) is <2,0>
#Now, we find the unit vector:
#u = <3,-2>/sqrt(3^2 + (-2)^2) = <3,-2>/sqrt(9+4) = <3,-2>/sqrt(13)
#Finally, we calculate the directional derivative:
#∇f . u = (1/sqrt(13))*(2*3 + 0*(-2)) = (1/sqrt(13))*6 = 6/sqrt(13)
#ANSWER: The requested directional derivative is 6/sqrt(13)

#Exercise 31:
#The direction vector in this exercise would be:
#v = <0-3,0-2> = <-3,-2>
#The formula for the directional derivative of f is:
#∇f . u
#Where u is the unit vector of the direction vector v
#First, we find the gradient, the formula for which is:
#∇f = <df/dx, df/dy>
#So, we find the first partial derivatives of f:
#df/dx = 2*x
#df/dy = 8*y
#We find the value of each derivative at (3,2):
#df/dx(3,2) = 2*3 = 6
#df/dy(3,2) = 8*2 = 16
#So, the gradient at (3,2) is <6,16>
#Now, we find the unit vector:
#u = <-3,-2>/sqrt(((-3)^2 + (-2)^2) = <-3,-2>/sqrt(9+4) = <-3,-2>/sqrt(13)
#Finally, we calculate the directional derivative:
#∇f . u = (1/sqrt(13))*(6*(-3) + 16*(-2)) = (1/sqrt(13))*(-18 - 32) = -50/sqrt(13)
#ANSWER: The requested directional derivative is -50/sqrt(13)

#Exercise 33:
#The direction in which we need to find the rate of change (or directional derivative)
# is:
#v = <5-3,7-9,3-4> = <2,-2,-1>
#The formula for the directional derivative of f is:
#∇f . u
#Where u is the unit vector of the direction vector v
#First, we find the gradient, the formula for which is:
#∇f = <df/dx, df/dy, df/dz>
#So, we find the first partial derivatives of f:
#df/dx = (x*exp(y-z))' = exp(y-z)
#df/dy = (x*exp(y-z))' = x*exp(y-z) * 1 = x*exp(y-z)
#df/dz = (x*exp(y-z))' = x*exp(y-z) * (-1) = (-1)*x*exp(y-z)
#We find the value of each derivative at (3,9,4):
#df/dx(3,9,4) = exp(9-4) = exp(5)
#df/dy(3,9,4) = 3*exp(9-4) = 3*exp(5)
#df/dz(3,9,4) = (-1)*3*exp(9-4) = (-3)*exp(5)
#So, the gradient at (3,9,4) is <exp(5), 3*exp(5), (-3)*exp(5)>
#Now, we find the unit vector:
#u = <2,-2,-1>/sqrt(2^2 + (-2)^2 + (-1)^2) = <2,-2,-1>/sqrt(9) = <2,-2,-1>/3
#Finally, we calculate the directional derivative:
#∇f . u = (1/3)*(2*exp(5) + (-2)*3*exp(5) + (-1)*(-3)*exp(5)) = 
# = (1/3)*((2 - 6 + 3)*exp(5)) = (1/3)*(-1)*exp(5) = (-1/3)*exp(5) ≈ -49.47
#ANSWER: The requested temperature change of the bug is -49.47 degrees Celsius per meter

#Exercise 37:
#We just need to see if the directional value is negative or not.
#So, we do not need to find the directional unit vector.
#We can just use this formula:
#∇f . v = 2*2 + (-4)*1 + 4*3 = 4 - 4 + 12 = 12
#ANSWER: Because the directional dericative is positive,
#f is increasing

#Exercise 39:
#First, we find the gradient of the function:
#df/dx = (sin(x*y + z))' = cos(x*y + z) * y = y*cos(x*y + z)
#df/dy = (sin(x*y + z))' = cos(x*y + z) * x = x*cos(x*y + z)
#df/dz = (sin(x*y + z))' = cos(x*y + z) * 1 = cos(x*y + z)
#Calculate each derivative at (0,-1,pi):
#df/dx(0,-1,pi) = (-1)*cos(0*(-1) + pi) = (-1)*cos(pi) = 1
#df/dy(0,-1,pi) = 0*cos(0*(-1) + pi) = 0
#df/dz(0,-1,pi) = cos(0*(-1) + pi) = cos(pi) = -1
#So, the gradient is <1,0,-1>
#To find the directional derivative, we can use the formula:
#theta = arccos((∇f . u)/(|∇f|*|u|)) = arccos((D_u_f(P))/|∇f|)
#where theta is the angle between the two vectors
#|∇f| = sqrt(1^2 + 0^2 + (-1)^2) = sqrt(2)
#30 = arccos((D_u_f(0,-1,pi)/sqrt(2))
#cos(30) = (D_u_f(0,-1,pi)/sqrt(2)
#sqrt(3)/2 = (D_u_f(0,-1,pi)/sqrt(2)
#sqrt(6)/2 = D_u_f(0,-1,pi)
#ANSWER: The requested directional derivative is sqrt(6)/2

#Exercise 41:
#First, we need to make sure that the point is on the surface:
#f(3,1,2) = 3^2 + 1^2 - 2^2 = 9 + 1 - 4 = 6 - GOOD
#The vector normal to the surface is its gradient vector.
#We need to find the partial derivatives of f(x,y,z):
#df/dx = 2*x
#df/dy = 2*y
#df/dz = (-2)*z
#Find all of the values at (3,1,2):
#df/dx(3,1,2) = 2*3 = 6
#df/dy(3,1,2) = 2*1 = 2
#df/dz(3,1,2) = (-2)*2 = -4
#ANSWER: The vector normal to the surface at (3,1,2) is <6,2,-4>

#Exercise 43:
#We need to find two points where the normal vector (or the gradient)
# is equal to <1,1,-2>
#So, we need to find points where the partial derivatives df/dx, df/dy,
#and df/dz are equal to 1,1,-2, respectively.
#First, we need to find those derivatives:
#df/dx = 2*x/4 = (1/2)*x
#df/dy = 2*y/9 = (2/9)*y
#df/dz = 2*z
#The point that satisfies all of those equations is (2, 9/2, -1)
#We can imagine that point as a multiple of some constant - like c
#(2*c, 9*c/2, (-1)*c)
#Plug this point into the ellipsoid:
#(2*c)^2/4 + (9*c/2)^2/9 + ((-1)*c)^2 = 1
#4*c^2/4 + 81*c^2/4/9 + c^2 = 1
#c^2 + 9*c^2/4 + c^2 = 1
#(1 + 9/4 + 1)*c^2 = 1
#(17/4)*c^2 = 1
#c^2 = 4/17
#c = +- 2/sqrt(17)
#Plugging both 2/sqrt(17) and (-2)/sqrt(17) into the point:
#(2*c, 9*c/2, (-1)*c) = (2*(2/sqrt(17)), 9*(2/sqrt(17))/2, (-1)*(2/sqrt(17)) = 
# = (4/sqrt(17), 9/sqrt(17), (-2)/sqrt(17))
#(2*c, 9*c/2, (-1)*c) = (2*((-2)/sqrt(17)), 9*((-2)/sqrt(17))/2, (-1)*((-2)/sqrt(17)) = 
# = ((-4)/sqrt(17), (-9)/sqrt(17), 2/sqrt(17))
#ANSWER: The two points are (4/sqrt(17), 9/sqrt(17), (-2)/sqrt(17)) and 
#((-4)/sqrt(17), (-9)/sqrt(17), 2/sqrt(17))