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Exam 2 Solutions
APPM 2350, Calculus 3, Fall 2008
October 20, 2008
- Answers in table form: a.B b.C c.C^ d.D e.B^ f.A^ g.D^ h.B i.D^ j.B
Figure 1 Figure 2 (Problem 1a) (Problem 1e)
(a) The full dependency graph is given in Figure 1. Using all paths from f to v and summing the terms from each path gives B as the answer. (b) The equation fy = 0 must hold at a critical point by the First Derivative Test. For this function fy = ex^ = 0, which does not hold for any โโ < x < โ. Thus, the answer is C. (c) Let g(x, y, z) = f (x, y) โ z, then โg is normal to the g = 0 level surface at any point on the surface. In terms of f , โg = fxi + fyj โ k, so the answer is C. (d) Let y = โx for values of x 6 = 0. This function does not exist along this path, thus, according to the definition of a multivariable limit, the limit as (x, y) โ (0, 0) does not exist, either. A Two Path Test is more difficult: first, consider y = x. The limit of 2x^2 /x as x โ 0 is 0. Then we must create a path that exposes the problem in the denominator. The curve y = โx + x^2 is such a path.
xlimโ 0
x^2 + (โx + x^2 )^2 x + (โx + x^2 ) = lim xโ 0
2 x^2 โ 2 x^3 + x^4 x^2 = lim xโ 0 (2^ โ^2 x^ +^ x
Theferore, D is the answer. (e) The integral is computed on the region in Figure 2. To integrate in y first, consider a fixed x, and then note that a vertical line enters the region at y = 0 and exits at y = x^2. The smallest x-value in the region is 0 and the largest is 1. To switch the order of integration properly, we must choose B. (f) See the box on page 960; the answer is A. (g) The linearization formula is: L(x, y) = f (1, 1) + fx(1, 1)(x โ 1) + fy(1, 1)(y โ 1). We calculate,
f (1, 1) = (1)^2 + (1)^2 + 1 = 3, fx(1, 1) = 2 x
โฃx=1 = 2,
fy(1, 1) = 2 y
โฃy=1 = 2.
Then, L(x, y) = 3 + 2(x โ 1) + 2(y โ 1), which simplifies to D.
(h) Set f (x, y) = xy, then fx = y and fy = x. The differential formula gives
df = fx dx + fy dy = y dx + x dy. (2)
Dividing by f = xy predicts the relative uncertainty: df f =^
y dx xy +^
x dy xy =^
dx x +^
dy y.^ (3) This shows that 1% uncertainty in x and y gives approximately 2% uncertainty in f ; the answer is B. (i) Set f (x, y) = x + y, then fx = 1 and fy = 1. The differential formula gives
df = fx dx + fy dy = dx + dy. (4)
Dividing by f = x + y predicts the relative uncertainty: df f =^
dx x + y +^
dy x + y =^
dx x
x x + y
dy y
y x + y
This shows that x + y very small compared to either x or y yields arbitrar- ily large uncertainty. For actual examples consider x = 101 and y = โ 100 with dx = 1, dy = โ1 (approximately 1% for each). Then the relative un- certainty is predicted to be df /f = .01(101) + (โ.01)(โ100) = 2.01 or 201%. The same calculation for x = 100.1, y = โ100, and dx = 1, dy = โ1 gives df /f = .01(1001) + (โ.01)(โ1000) = 20.01 or 2001%. This can be generalized to arbitrary uncertainty in f , so the answer is D.
(b) From equation (iii) we see that y = โz. Using this in equation (ii) shows ฮป = 0, which plugged into equation (i) gives x = 0. Lastly, equation (iv) gives y = 1/2, so z = โ 1 /2. The unique solution to equations (i)-(iv) is { (x, y, z) =
2 ,^ โ^
, ฮป = 0
(c) f (0, 1 / 2 , โ 1 /2) = 0^2 + (1/ 2 โ 1 /2)^2 = 0 is the minimum.
- (a) The sketch is given below:
(b) Compute the volume with an iterated integral: โซ โซ
R
f (x, y) dA =
โซ (^) x=
x=โ 1
(โซ (^) y=2โ 1 โx 2
y=โ 1 โx^2
y dy
dx =
โซ (^) x=
x=โ 1
y^2 2
)y=2โ 1 โx 2
y=โ 1 โx^2
dx
โซ (^) x=
x=โ 1
4(1 โ x^2 ) โ (1 โ x^2 )
dx
โซ (^) x=
x=โ 1
1 โ x^2 dx =^3 2
x โ x
3 3
)x=
x=โ 1
