# TEXAS A&amp;M UNIVERSITY DEPARTMENT OF MATHEMATICS MATH 251-507 / 509 ```TEXAS A&amp;M UNIVERSITY
DEPARTMENT OF MATHEMATICS
MATH 251-507 / 509
Exam 2 version B, 22 Oct 2008
On my honor, as an Aggie, I have neither given nor received unauthorized aid on this work.
Name (print):
In all questions no analytical work =⇒ no points!
Circle your section above (509 is 9:35am, 507 is 11:10am).
1.
Use the differentials to approximate the value of the solution z of the equation
1
cos x − z + exz + y 2 = 0
2
at the point x = 0.02, y = 0.99.
(Hint: The formula involving the differentials works normally here, you just need a way to
find the derivatives of the implicit function z(x, y).)
2.
For the function
f (x, y) = x2 y + xy 2 + y 2 − y
find all ciritical points and determine their character (min / max / saddle point).
3.
Find the absolute maximum and minimum values of the function from the previous question,
f (x, y) = x2 y + xy 2 + y 2 − y,
in the triangular region with the vertices (0, 0), (4, 0) and (0, 4).
4.
Solve only one of the below problems. If both problems are solved you must indicate
clearly which one you wish to be graded. If no such indication is made, I will use the
minimum of the two marks.
(a) For 5 points: Use Lagrange multipliers to derive the formula for the distance from
the point (x0 , y0 , z0 ) to the plane ax + by + cz = d.
(b) For 3 points only: Use Lagrange multipliers to find the distance from the point
(1, 2, 3) to the plane 2x + y − 2z = 7.
5.
(Bonus question +1 point) Suppose p = p(x, y), where x = x(u, v) and y = y(u, v).
Further, you know the following information:
x (1, 2) = 3
y (1, 2) = 4
p (1, 2) = 5
p (3, 4) = 6
∂p
(1, 2) = 7
∂x
∂x
(1, 2) = 11
∂u
∂y
(1, 2) = 15
∂u
∂p
(1, 2) = 8
∂y
∂x
(1, 2) = 12
∂v
∂y
(1, 2) = 16
∂v
∂p
(3, 4) = 9
∂x
∂x
(3, 4) = 13
∂u
∂y
(3, 4) = 17
∂u
∂p
(3, 4) = 10
∂y
∂x
(3, 4) = 14
∂v
∂y
(3, 4) = 18
∂v
Use the chain rule and the above information to compute
Points:
/20
∂p
at the point u = 1, v = 2.
∂v
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