#ATTENDANCE QUIZ for Lecture 7 of Math251(Dr. Z.)
#EMAIL  RIGHT AFTER YOU WATCHED THE VIDEO
#BUT NO LATER THAN  Sept. 28, 2020, 8:00PM (Rutgers time)
#THIS .txt FILE (EDITED WITH YOUR ANSWERS)
#TO:
#DrZcalc3@gmail.com
#Subject: aq7
#with an ATTACHMENT CALLED:
#aq7FirstLast.txt
#(e.g. aq7DoronZeilberger.txt)




#LIST ALL THE ATTENDANCE QUESTIONS FOLLOWED BY THEIR ANSWERS

# Question 1:

# Let a be the 5th digit of your RUID, let b be the 2nd digit of your RUID
# Let f(x,y) = x^a + b*x^(2a)*y^3
# Find the two partial derivatives of f(x,y)

# Answer:

# For me a=0 and b=9, so
# f(x,y) = 1 + 9y^3

# f_x(x,y) = 0
# f_y(x,y) = 27y^2




# Question 2:

# With a and b as above, find f_x(z(x,y)) and f_y(z(x,y)) if x,y,z are related by
# x^a * y^2 * z^b + x^2 * y^(3b) *z^3 = exp(x*y*z)

# Answer:

# For me a=0 and b=9, so the equation is
# y^2*z^9 + x^2*y^27*z^3 = exp(x*y*z)

# f_x(z(x,y)):

# y^2*9z^8*z' + 2x*y^27*z^3 + x^2*y^27*3z^2*z' = exp(x*y*z) * (y*z + x*y*z')
# y^2*9z^8*z' + x^2*y^27*3z^2*z' - exp(x*y*z)*x*y*z' = exp(x*y*z)*y*z - 2x*y^27*z^3
# z' * (y^2*9z^8 + x^2*y^27*3z^2 - exp(x*y*z)*x*y) = exp(x*y*z)*y*z - 2x*y^27*z^3
# z' = (exp(x*y*z)*y*z - 2x*y^27*z^3) / (y^2*9z^8 + x^2*y^27*3z^2 - exp(x*y*z)*x*y)


# f_y(z(x,y)):

# 2y*z^9 + y^2*9z^8*z' + x^2*27y^26*z^3 + x^2*y^27*3z^2*z' = exp(x*y*z) * (x*z + x*y*z')
# y^2*9z^8*z' + x^2*y^27*3z^2*z' - exp(x*y*z)*x*y*z' = exp(x*y*z)*x*z - 2y*z^9 - x^2*27y^26*z^3
# z' * (y^2*9z^8 + x^2*y^27*3z^2 - exp(x*y*z)*x*y) = exp(x*y*z)*x*z - 2y*z^9 - x^2*27y^26*z^3
# z' = (exp(x*y*z)*x*z - 2y*z^9 - x^2*27y^26*z^3) / (y^2*9z^8 + x^2*y^27*3z^2 - exp(x*y*z)*x*y)




# Question 3:

# Do the second part, find f_y(1,1)
# x^2 + y^2 = x*y*z + x*y*z^5

# Answer:

# Differentiate with respect to y
# (x^2)' + (y^2)' = (x*y*z)' + (x*y*z^5)'
# 2y = x*z + x*y*z' + x*z^5 + x*y*5z^4

# plug in (x,y,z) = (1,1,1)
# 2 = 1 + 1z' + 1 + 5z'
# 0 = 6z'
# z' = 0




# Question 4:

# Is the answer the same, is f_y(1,1) also 0?
# Could you have predicted it without doing the calculation?

# Answer:

# Yes the answer is the same.
# We could predict this because x and y have the same coefficients and powers on both sides.
# Another way to explain this is that if you switched x and y in the relationship equation, 
# it would still be the same equation.




# Question 5:

# Let a and b be as above
# Find the equation of the tangent plane to the surface
# z = x^a + y^b + a*b*x*y
# at the point (1,1,2+a*b)

# Answer:

# surface equation for me:
# z = 1 + y^9
# find the tangent plane at (1,1,2)

# 2 = 1+1^9 = 2
# this point is on the surface

# z_x = 0
# z_y = 9y^8

# equation of tangent plane:
# z-2 = 0(x-1) + 9(y-1)
# z = 9y - 7