Download Find the linearization L(x, y) of f(x, y) = xy 2 + y cos(x − 1)

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Math 234 - Quiz 5
Problem 1 (4 points)
Find the linearization L(x, y) of f (x, y) = xy 2 + y cos(x − 1) at P0 = (1, 2). Then find an upper
bound for the magnitude |E| of the error in the approximation f (x, y) ≈ L(x, y) in the region
R : |x − 1| ≤ 0.1, |y − 2| ≤ 0.1.
Solution:
We have that f (1, 2) = 4 + 2 cos(0) = 6. Also, the partial derivatives of f (x, y) at the given point
are
fx (1, 2) = y 2 − y sin(x − 1)|(1,2) = 4,
fy (1, 2) = 2xy + cos(x − 1)|(1,2) = 4 + 1 = 5.
Thus, the linearization is
L(x, y) = 6 + 4(x − 1) + 5(y − 2).
To find a bound for the erroe we must first find a bound for all second oreder partial derivatives in
the given region.
|fxx | = | − y cos(x − 1)| ≤ |y| ≤ 2.1
|fyy | = |2x| ≤ 2(1.1) = 2.2
|fxy | = |2y − sin(x − 1)| ≤ |2y| + 1 ≤ 2(2.1) + 15.2
An upper bound for the second order partial derivatives in the region is 5.2. Thus,
1
|E| ≤ 5.2(|x − 1| + |y − 2|)2 ≤ 2.6(0.1 + 0.1)2 .
2
Problem 2 (4 points)
Find all the local maxima, local minima, and saddle points of the function
f (x, y) = x2 − 4xy + y 2 + 6y + 2.
Solution:
To find the critical points we must set the first order partial derivatives equal to zero. That is, we
must solve the system of equations
fx = 2x − 4y = 0
fy = −4x + 2y + 6 = 0.
From the first equation we have that x = 2y. Substituting this into the second equation we have that
1
2
−8y + 2y + 6 = 0.
That is, y = 1 and x = 2. We must now determine whetherthere is a local max, min or saddle point
at point (2, 1). We have that fxx = 2, fyy = 2 and fxy = −4. Thus, the determinat is
D(2, 1) = 2 ∗ 2 − (16) = −12 < 0.
Hence, there is a saddle point at (2, 1).