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QUIZ 2, Version A : MATH 251, Section 505
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Write up your result, detail your calculations if necessary and BOX your final answer
1. [25pts] Find an equation of the tangent plane of z = (x + 1)2 + 4y 2 at (−1, 1, 4).
2. [25pts] Find the differential of the function z =
√
x ln(1 + y).
3. [25pts] Find the gradient of f (x, y) = x2 y + xey at (1, 0).
4. [25pts] Use the chain rule to find gs with g(s, t) = f (x(s, t), y(s, t)) where f (x, y) = ln(x + y 2 ) and
x(s, t) = 2s + t and y(s, t) = s2 + t2 .
1. Let f (x, y) = (x + 1)2 + 4y 2 . f is a polynomial function ( defined on R2 ). The surface S is given as
a graph of function of 2 variables, so we apply the formula seen in class : an equation of the tangent
plane is of the form
z − 4 = fx (−1, 1)(x + 1) + fy (−1, 1)(y − 1).
fx (x, y) = 2(x + 1) so fx (−1, 1) = 0 and fy (x, y) = 8y so fy (−1, 1) = 8. Then an equation of the
tangent plane to the surface S = {(x, y, z) | f (x, y) = z} is
z − 4 = 8(y − 1)
2.
⇔
8y − z = 4 .
√
1
x
dz(x, y) = fx (x, y)dx + fy (x, y)dy = ln(1 + y) √ dx +
dy .
1+y
2 x
3. f est a function defined on R2 . She has also continuous partial derivatives. We have fx (x, y) = 2xy+ey
and fy (x, y) = x2 + xey . Its gradient at (1, 0) is
∇f (1, 0) =< fx (1, 0), fy (1, 0) >=< 1, 2 > .
4. g, f, x, y are all of them differentiable. By the Chain Rule,
∂y ∂f
∂x ∂f
+
,
∂s ∂x ∂s ∂y
1
2y(s, t)
= 2.
+ 2s.
,
2
x(s, t) + y (s, t)
x(s, t) + y 2 (s, t)
gs (s, t) =
=
2 + 4s(s2 + t2 )
.
2s + t + (s2 + t2 )2