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DEPARTMENT OF MATHEMATICS AND STATISTICS
UNIVERSITY OF MASSACHUSETTS
MATH 233 EXAM 2 Spring 2007
NAME:
Section Number: Instructor’s Name:
In problems that require reasoning or algebraic calculation, it is not sufficient just to write the answers. You must explain how you arrived at your answers, and show your algebraic calculations.
You can leave answers in terms of fractions and square roots, but if approximate numerical answers are used, they should be rounded off to 4 significant figures.
Total
Perfect Paper −→ 100 Points.
There are seven pages, including this one, in this exam and six problems. Make sure you have them all before you begin!
- For the function f (x, y) = 3x^2 + xy^2 , calculate
(a) (2 points) fx(x, y) = 6x + y^2
(b) (2 points) fy(x, y) = 2xy
(c) (2 points) fxy(x, y) = 2y
(d) (2 points) fxx(x, y) = 6
(e) (4 points) What is the linearization L(x, y) of f (x, y) at (1, 2)? By definition, L(x, y) of f (x, y) at (x 0 , y 0 ) is:
L(x, y) = f (x 0 , y 0 ) + fx(x 0 , y 0 )(x − x 0 ) + fy(x 0 , y 0 )(y − y 0 )
Therefore L(x, y) = f (1, 2)+fx(1, 2)(x−1)+fy(1, 2)(y−2) = 7+10(x−1)+4(y−2). (f) (4 points) Use the linearization L(x, y) in the previous part to estimate f (. 9 , 2 .1). To estimate we evaluate L(. 9 , 2 .1) = 7 + 10(−.1) + 4(.1) = 6.4. (g) (4 points) Calculate the directional derivative of f (x, y) at the point (1, 2) in the direction v = 〈 3 , 4 〉? The directional derivative in the direction of the unit vector u = 〈a, b〉 at a point (x 0 , y 0 ) is Duf (x 0 , y 0 ) = fx(x 0 , y 0 )a + fy(x 0 , y 0 )b
The corresponding unit vector in the direction of 〈 3 , 4 〉 is u = 〈^35 , 45 〉. We plug in to obtain
Duf (1, 2) = fx(1, 2)
Our tangent line will have slope −^34 with respect to the variables x and y (i.e. in the direction of 〈− 4 , 3 , 0 〉).
Suitable parametric equations for such a line are:
x = 3 − 4 t
y = 4 + 3t
z = 5
- Using the hint we know the tangent line lies on both tangent planes (i.e. it will be perpendicular to both planes’ normal vectors).
To find the line’s direction vector we compute the cross product of the normal vectors.
The normal vectors are obtained by viewing our surfaces as level surfaces of the func- tions:
F (x, y, z) = z^2 − x^2 − y^2
G(x, y, z) = x^2 + y^2 + z^2
The normals are ∇F (3, 4 , 5) = 〈− 6 , − 8 , 10 〉 and ∇G(3, 4 , 5) = 〈 6 , 8 , 10 〉.
The direction of the tangent line is 〈− 6 , − 8 , 10 〉 × 〈 6 , 8 , 10 〉 = 〈− 160 , 120 , 0 〉
The parametric equations of a line through (3, 4 , 5) and in the direction of 〈− 160 , 120 , 0 〉 are:
x = 3 − 160 t
y = 4 + 120t
z = 5
This is the same line described in the first solution.
- (18 points) Let f (x, y) = x^3 + xy^2 + 2x^2 + y^2. Find and classify (as local maxima, local minima or saddle points) all critical points of f. (Hint: Apply the second derivative test to check if a critical point is local maximum, minimum or saddle point.) Critical points are points (x 0 , y 0 ) where fx(x 0 , y 0 ) = fy(x 0 , y 0 ) = 0. The partial derivative fy(x, y) = 2xy + 2y = 2y(x + 1) is zero when y = 0 or x = −1. In the case y = 0 fx(x, y) = 3x^2 + y^2 + 4x will vanish only if x = 0, −^43. In the case x = − 1 fx(x, y) = 3x^2 + y^2 + 4x will vanish only if y = ± 1 The four critical points are (0, 0), (−^43 , 0), (− 1 , 1), and (− 1 , −1). To apply the second derivative test we calculate the following second order partial derivatives: fxx(x, y) = 6x + 4 fyy(x, y) = 2x + 2 fxy(x, y) = 2y The test requires we examine
D(x 0 , y 0 ) = fxx(x 0 , y 0 )fyy(x 0 , y 0 ) − fxy(x 0 , y 0 )^2
for each critical point (x 0 , y 0 ). D(0, 0) = 8 D(−^43 , 0) = (^83) D(− 1 , 1) = D(− 1 , −1) = − 4 When y = 0 the critical points will be a min or max (since D is positive). Since fxx(0, 0) = 4 > 0 the function will have a local min at (0, 0). Since fxx(−^43 , 0) = − 4 < 0 the function will have a local max at (−^43 , 0). When x = −1 the critical points will be saddle points (since D is negative).
- (16 points) Find the volume above the rectangle − 1 ≤ x ≤ 1, 0 ≤ y ≤ 5 and below the surface z = 5 + x^2 + 4y^3. By Fubini’s Theorem we can solve this double integral in two ways. We may perform partial integration with respect to x first: ∫ (^5)
0
∫ (^1)
− 1
(5+x^2 +4y^3 )dxdy =
∫ (^5)
0
[5x+
x^3 3
+4xy^3 ]^1 − 1 dy =
∫ (^5)
0
+8y^3 )dy = 50+
Or we may perform partial integration with respect to y first: ∫ (^1)
− 1
∫ (^5)
0
(5+x^2 +4y^3 )dydx =
∫ (^1)
− 1
[5y+x^2 y+y^4 ]^50 dx =
∫ (^1)
− 1
(25+5x^2 +625)dx = 50+
Either way the volume is 1303^13.
- (16 points) Find the integral of the function f (x, y) = x^2 + y on the domain bounded by the two parabolas y = 2x^2 and y = x^2 + 1. This bounded domain is most readily thought of as Type 1. The intersection points of the two defining parabolas are (± 1 , 2). The domain can be written as
D = {(x, y) : − 1 ≤ x ≤ 1 , 2 x^2 ≤ y ≤ x^2 + 1}
We can evaluate the double integral as follows:
∫ ∫
D
f (x, y)dA =
∫ (^1)
− 1
∫ (^) x (^2) +
2 x^2
(x^2 +y)dydx =
∫ (^1)
− 1
[x^2 y+
y^2 2
]x (^2) + 2 x^2 dx^ =
∫ (^1)
− 1
(2x^2 −
x^4 +
)dx =
