# Attendance Question 1:
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# Let a be the 5th digit of your RUID and let b be the 2nd digit of your RUID
# let f(x, y) = x^a + b*a^(2*a)*y^3
# Find the two partial derivates of f(x, y) 
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# Answer 1: 
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# Because both are 0, a=1st digit and b=8th digit so a = 2 and b = 2
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# f(x, y) = x^2 + 2*2^(2*2)*y^3 = x^2 + 32*y^3
# f_x(x,y) = 2x
# f_y(x,y) = 96*y^2
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# Attendance Question 2:
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# With a and b as above, find f_x (z(x, y)) and f_y (z(x,y)) if x,y,z are realted by the relationship (equation)
# f(x,y) = x^a * y^2 * z^b + x^2 * y^(3*b) * z^3 = exp(x*y*z)
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# Answer 2:
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# x^2 * y^2 * z^2 + x^2 * y^(3*2) * z^3 = exp(x*y*z)
# 
#wrt x:
# 2*x*y^2*z^2 + 2*x^2*y^2*z * dz/dx + 2*x*y^6*z^3 + 3*x^2*y^6*z^2 * dz/dx = y*z*e^x*y*z + x*y*e^x*y*z * dz/dx
# dz/dx * (2*x^2*y^2*z + 3*x^2*y^6*z^2 - x*y*e^x*y*z) = y*z*e^x*y*z - 2*x*y^2*z^2 - 2*x*y^6*z^3
# dz/dx = (y*z*e^x*y*z - 2*x*y^2*z^2 - 2*x*y^6*z^3) / (2*x^2*y^2*z + 3*x^2*y^6*z^2 - x*y*e^x*y*z)
# 
# wrt y:
# 2*x^2*y*z^2 + 2*x^2*y^2*z * dz/dy + 6*x^2*y^5*z^3 + 3*x^2*y^6*z^2 * dz/dy = x*z*e^x*y*z + x*y*e*^x*y*z * dz/dy
# dz/dy * (2*x^2*y^2*z + 3*x^2*y^6*z^2 - x*y*e^x*y*z) = x*z*e^x*y*z - 2*x^2*y*z^2 - 6*x^2*y^5*z^3
# dz/dy = (x*z*e^x*y*z - 2*x^2*y*z^2 - 6*x^2*y^5*z^3) / (2*x^2*y^2*z + 3*x^2*y^6*z^2 - x*y*e^x*y*z)
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# Attendance Question 3:
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# Do the second part, find f_y(1,1) for the function x^2 + y^2 = x*y*z + x*y*z^5
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# Answer 3:
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# wrt y:
# 2*y = x*z + x*y*z' + x*z^5 + 5*x*y*z^4 * z'
# 2*1 = 1*1 + 1*1*z' + 1*1^5 + 5*1*1*1^4 * z'
# 2 = 1 + z' + 1 + 5*z'
# 2 = 2 + 6*z'
# 0 = 6*z'
# z' = 0
# f_y(1,1) = 0
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# Attendance Question 4:
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# Is the answer the same, is f_y(1,1) also 0? Could you have predicted it without doing the calculation?
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# Answer 4:
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# Yes, the answer is the same for f_y(1,1) and f_x(1,1), as they both equal 0. No, you could not have predicted it
# without doing the calculation because the value of the partial derivative of y at f_y(1,1)could be different 
# from the value of the partial derivative of x at f_x(1,1).
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# Attendance Question 5:
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# 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)
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# Answer 5:
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# z = x^2 + y^2 + 4*x*y at (1,1,6)
# f_x(x,y) = 2*x + 4*y --> f_x(1,1) = 2*1 + 4*1 = 6
# f_y(x,y) = 2*y + 4*x --> f_y(1,1) = 6
# z - 6 = 6*(x - 1) + 6*(y - 1)
# z = 6*(x - 1) + 6*(y - 1) - 6
# z = 6*x + 6*y - 6
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