#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 be the 0(5th), be be the 0(2rd),
Let f(x, y):=x^a+b*x^(2*a)*y^3
Find the two partial derivatives of f(x, y) 

Answer:
#f(x, y):= x^0+0*x^(2*0)*y^3= 1+0=1
#f-x(x, y)=0
#f-y(x, y)=0



Attendance question 2:
With a and b as above, fix 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^(3*b) * z^3

Answer:
#=y^2+x^2*z^3
#fx(z(x, y))=2*y+2*x*z^3
#fy(z(x, y))=2*y+x^2*(3*z^2)



Attendance question 3:
Do the second part, find f-y(1,1)

Answer:
#x^2+y^2=x*y*z+x*y*z^5
#(x^2+y^2)'=(x*y*z+x*y*z^5)'
# 2*y=x*(y*z)'+x*(y*z^4)'
# 2*y=x*(y'*z+y*z')+x*(y'*z^4+y*(z^4)')
# 2*y=x*(z+y*z')+x*(z^4+y*4*(z^3)*z')
When f-y(1,1)
So, 2*1=1*(1+1z')+1*(1+4*1*z')
# 2=2+5z'
# 0=5z'
# z'=0
# Ans: f-y(1,1)=0



Attendance 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. Because derivative equation of dx/dz or dy/dz is same, and both are (1,1)
So the answer is same and we can predict it which is 0.


Attendance question 5:
Let a and b be as above,
Find the equation of 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^0+0*0*x*y
#z=1+1
#f-x(z(x, y))=0 f-y(z(x, y))=0
Ans is z-2=0(x-1)+0(y-1)