#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 5th digit of your RUID. Let b be the second 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)

#MY ANSWER TO ATTENDANCE QUESTION 1: f(x,y) with respect to x = 1 + 9 * y^3 = 0. f(x,y) with respect to y = 1 + 9 * y^3 = 27y^2.

# a = 0 , b = 9

# f(x,y) = x^0 + 9*x^(0) * y^3 = 1 + 9 * y^3

# The derivative of a constant is 0, therefore f(x,y) with respect to x = 1 + 9 * y^3 = 0 and f(x,y) with respect to y = 1 + 9 * y^3 = 27y^2 (This problem is simplified due to 5th digit in RUID being 0.)

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

#MY ANSWER TO ATTENDANCE QUESTION 2: f_x (z(x,y)) = y^2*z^9 + 2x*y^27*z^3 = exp(xyz)*yz and f_y(z(x,y)) = 2y*z^9 + x^2*27y^26*z^3 = exp(xyz)*xz

# a = 0 , b = 9

# x^0 * y^2 * z^9 + x^2 * y^(27) * z^3 = exp(x*y*z)

# f_x (z(x,y)) = y^2*z^9 + 2x*y^27*z^3 = exp(xyz)*yz

# f_y(z(x,y)) = 2y*z^9 + x^2*27y^26*z^3 = exp(xyz)*xz

#ATTENDANCE QUESTION 3: Do the second part, find f_y(1,1) for x^2 + y^2 = x*y*z + (x*y*z^5)'

#MY ANSWER TO ATTENDANCE QUESTION 3: f_y(1,1) for x^2 + y^2 = x*y*z + (x*y*z^5)' = 1.

#z' was established as equal to 0 and z = 1 in this problem. f_y(1,1) = 2y - x - 1.

# f_y(1,1) = 2(1) - 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.

#MY ANSWER TO ATTENDANCE QUESTION 4: Yes the answer is the same, f_y(1,1) is also 0. It could have been predicted without doing the calculation.

#ATTENDANCE 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)

#MY ANSWER TO ATTENDANCE QUESTION 5: 

# a = 0 , b = 9.

# z = 1 + y^9 at the point (1,1,2)

# Fx = 0. Fy = 9y^8 and plug in 1 which gives us Fy(1,1,2) = 9. Fz = 1.

# 0(x-1) + 9(y-1) + 1(z-2)

# 0 + 9y - 9 + z - 2.

# z = -9y + 11.