#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
#
#Attendance Question #1:
#Let a = 5th digit of your RUID
#Let b = 2nd digit of your RUID
#
#Let f(x,y) = x^a+bx^2a*y^3
#Find the two partial derivatives of f(x,y):
#
#Answer: RUID - 191003667 -> a = 0 and b = 9
#f(x,y) = x^0 + 9x^(2*0)*y^3 = 1+9x^0 * y^3 = 1+9y^3
#f_x(x,y) = df/dx = (1+9y^3)' = 0; 1+9y^3 is treated as a constant
#f_y(x,y) = df/dy = (1+9y^3)' = 0+27y^2 = 27y^2
#
#Attendance 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 the relationship(equation):
#
#x^a*y^2*z^b + x^2*y^3b*z^3 = e^x*y*z
#
#Answer: a = 0 and b = 9
#Equation = y^2*z^9 + x^2y^27*z^3 = e^xyz
#
#Differentiate With Respect to X:
#Z'(y^29z^8) + z'(3x^2y^27z^2) + 2xy^27z^3) = z'(yxe^y(xz)) + yze^y(xz)
#
#Z'(y^29z^8) + z'(3x^2y^27z^2)-z'(yxe^y(xz)) = yze^y(xz) - (2xy^27z^8)
#
#Dz/dx = Z' = (ze^y(xz) =2xy^26z^8)/(9yz^8+3x^2y^26z^2-xe^y(xz))
#
#Differentiate With Respect To Y:
#
#y^29z^8z' + 2yz^9 + x^2(3y^27z^2z'+27y^26z^3)
#
#Z'(y^2*9z^8+x^23y^27z^2-yxe^xyz) = xze^xyz-2yz^9-(27z^2y^26z^3)
#
#Dz/dy = z' = xze^xyz-2yz^9-(27z^2y^26z^3)/(y^2*9z^8+x^23y^27z^2-yxe^xyz)
#
#Attendance Problem #3:
#Do the Second Part Find f_y(1,1)
#
#(x^2+y^2)' = (xyz+xyz^5)'
#
#Differentiate With Respect To Y:
#
#2y = x(yz)'+x(yz^5)'
#
#2y = x(yz'+z) + x(5yz^4z'+z^5)
#
#Focus on point x=1, y=1 (z=1)
#
#2(1) = (1)(1z' + 1) + (1)(5(1)(1)^4z' + (1)^5)
#
#2 = 2+6z'
#
#0 = 6z'
#Z' = 0
#f_y(1,1) = 0
#
#Attendance Problem #4:
#Is the answer the same as f_y(1,1) also 0?
#
#Yes,f_y(1,1) is also 0.
#
#Could you have predicted it without doing the calculation?
#Yes, the derivative in terms of y is essentially the same as the derivative #in terms of x due to the variables lacking different exponents and #containing the same values.
#
#Attendance Problem #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: z = x^0 + y^9 + 0*9*x*y = 1 + y^9
#
#Plug in to check: 2+0*9 = 2 = 1+(1)^9 = 2
#
#f_x = d/dx(1+y^9) = 0 as 1+y^9 is treated as a constant
#
#f_y = d/dy(1+y^9) = 9y^8
#
#f_x(1,1) = 0
#f_y(1,1) = 9
#
#Equation z - 2 = 0(x-1) + 9(y-1)
#Z-2 = 9(y-1) + 2 
#Z = 9y-7