#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:

#Let a :=5th digit of your RUID; B:=2nd digit of your RUID

#Let f(x,y):= x^a + b*x^(2*a)*y^3

#Find the two partial derivatives of f(x,y)


#Answer to attendance question 1:

f(x,y):=x^0 + 8x^0y^3
f(x,y):= 1+8y^3

f'x=0
f'y=24y^2

#attendence 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 + x^2*y^(3b)*z^3 = exp(x*y*z)

#Answer to attendance question 2:


x^0y^2 + x^2y^24z^3

y^2 + x^2y^24z^3 = exp(x*y*z)




#Attendence question 3:

# Do the second part, find f_y(1,1)


#Answer to attendance problem 3:

I may have done it wrong on paper but I also got 0


#Attendence problem 4:

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

I think maybe because x and y are the same in the equation and in the point? 
There's x^2 and y^2 and x*y+x*y so turning them into constants is the same as well. 


#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 to attendance problem 5:

z=x^8 + y^0 + 0 at the point (1,1,2)

f = x^8 + 1 -  z

f_x(x,y,z)= 8x^7

f_z(x,y,z)= -1 = 0


8x^7i -k

(1,0,-1)

(x,y,z)(1,0,-1)=(1,1,2)(1,0,-1)
X-z=1-2

x-z=-1