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# Calculus

#### Solution Preview

1.)
Let s(x,y) = x-y; t(y,z) = y-z; u(z,x) = z-x
hence,
w = f(s, t, u)

del(w)/del(x)
= (del(w)/del(s))*(del(s)/del(x)) + (del(w)/del(u))*(del(u)/del(x))
=(del(w)/del(s))*(1) + (del(w)/del(u))*(-1)
=(del(w)/del(s)) - (del(w)/del(u)) .......(1)

del(w)/del(y)
= (del(w)/del(s))*(del(s)/del(y)) + (del(w)/del(t))*(del(t)/del(y)) =(del(w)/del(s))*(-1) + (del(w)/del(t))*(1)
=-(del(w)/del(s)) + (del(w)/del(t)) .......(2)

del(w)/del(z)
= (del(w)/del(t))*(del(t)/del(z)) + (del(w)/del(u))*(del(u)/del(z)) =(del(w)/del(t))*(-1) + (del(w)/del(u))*(1)
=-(del(w)/del(t)) + (del(w)/del(u)) .......(3)

Add equations 1, 2 and 3:
del(w)/del(x)+del(w)/del(y)+del(w)/del(z)
=(del(w)/del(s))-(del(w)/del(u))-(del(w)/del(s))+(del(w)/del(t))-(del(w)/del(t))+(del(w)/del(u))
= 0
--Hence, porved.

2.)
surface s: z^3 + 3yz - x^3 - 1 = 0
tangent point P(1,-1,2)

because, equation of tangent plane of surface f(x,y,z) = 0 at (x1, y1, z1) is given as:
(del(f)/del(x))*(x-x1)+del(f)/del(y))*(y-y1)+del(f)/del(z))*(z-z1)=0

del(s)/del(x)|(at 1,-1,2) = -3*x^2 = -3*1^2 = -3
del(s)/del(y)|(at 1,-1,2) = 3*z = 3*2 = 6
del(s)/del(z)|(at 1,-1,2) = 3z^2 + 3y = 3*2^2 + 3*(-1) = 9
hence, equation of tangent plane of s=0 at (1,-1,2):
-3*(x-1) + 6*(y+1) + 9*(z-2) = 0
=> -3x+6y+9z +3+6-18 =0
=> 3x-6y-9z+9 = 0
=> x - 2y - 3z ...

#### Solution Summary

Topics included in this problem set include rectangular functions, partial derivatives, tangent planes, extrema, area, and rate of change

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