# LAST NAME : FIRST NAME :

```LAST NAME :
FIRST NAME :
QUIZ 4, Version A : MATH 251, Section 505
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first name : . . . . . . . .
GRADE : . . . . . . . .
”An Aggie does not lie, cheat or steal, or tolerate those who do”
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1. [25pts] Find an equation of the tangent plane of z = 2x2 + 4y 2 at (1, 1, 6).
2. [25pts] Find the differential of the function z =
3. [25pts] Find
√
yex .
dy
for xy = sin(x + y 2 ).
dx
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) = 2x2 + 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 − 6 = fx (1, 1)(x − 1) + fy (1, 1)(y − 1).
fx (x, y) = 4x so fx (1, 1) = 4 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 − 6 = 4(x − 1) + 8(y − 1)
⇔
4x + 8y − z = 6 .
2.
dz(x, y) = fx (x, y)dx + fy (x, y)dy =
√
1
yex dx + √ ex dy .
2 y
3. We use here the implicit Theorem on f (x, y) = 0 where f (x, y) = xy − sin(x + y 2 ). So, we assume
that fy (x, y) = x − 2y cos(x + y 2 ) 6= 0. Then the Theorem says that y is an implicit function of x in
a neighborhood of these points and
dy
−fx (x, y)
y − cos(x + y 2 )
.
(x) =
= −
dx
fy (x, y)
x − 2y cos(x + y 2 )
4. g, f, x, y are all of them differentiable. By the Chain Rule,
∂x ∂f
∂y ∂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 + 2s(s2 + t2 )
.
2s + t + (s2 + t2 )2
```