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The solutions to exam ii of the multivariable calculus course taught by prof. P. Wong, held on november 3, 2006. Problems related to finding the divergence, curl, and jacobian matrix of vector fields, as well as problems involving directional derivatives, tangent planes, and critical points.
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EXAM II - NOVEMBER 3, 2006
Instruction: Read each question carefully. Explain ALL your work and give reasons to support your answers. Advice: DONโT spend too much time on a single problem.
Problems Maximum Score Your Score
1
2 EXAM II - NOVEMBER 3, 2006
div F = โ ยท F = โF โx + โF โy + โF โz ,
it follows that div F = 0.
(5 pts) (ii) Find curl F
curl F = โ ร F = (x โ x, โ(y โ y), z โ z) = (0, 0 , 0).
(5 pts) (iii) What is the Jacobian matrix DF (1, 1 , 1) of F at (1, 1 , 1)?
0 z y z 0 x y x 0
(5 pts) (iv) Find an approximation of F (0. 9 , 1. 1 , 1 .1). The linear approximation asserts that F (x) = F (a) + DF (a) ยท (x โ a)
where a = (1, 1 , 1). It follows that
4 EXAM II - NOVEMBER 3, 2006
Duf (1, 1 , 0) = โf (1, 1 , 0) ยท (^) โuuโ
= (3,^1 ,^ 0) โ^ ยท^ (1,^ โ^1 ,^ 3) 11 = โ^2 11
(12 pts) (ii) Find an equation for the plane tangent to the level surface f (x, y, z) = 9 at the point (3, โ 1 , 2). Since f (x, y, z) = 9 is a level surface, we know that โf (3, โ 1 , 2) = (โ 27 , 23 , 84) is normal to the plane tangent to the level surface at the point (3, โ 1 , 2). Thus, this tangent plane is given by the equation (x โ 3 , y + 1, z โ 2) ยท (โ 27 , 23 , 84) = 0.
In other words, an equation for this tangent plane is
27 x โ 23 y โ 84 z + 64 = 0.
MATH206A MULTIVARIABLE CALCULUS - PROF. P. WONG 5
(6 pts) (ii) For each of the critical point(s) a found in part (i), find the corresponding Hessian matrix Hf (a). First, the Hessian matrix is given by
Hf =
0 โ 6 y
It follows that
Hf ((0, โ^2 3
and Hf ((0, โ โ^2 3
(7 pts) (iii) Use the second derivative test to classify each of the critical point(s) in part (i), i.e., determine whether the critical point is a local max, local min, or saddle point. At the critical point (0, โ^23 ),
det Hf ((0, โ^23 )) =^ โ^243 > 0 and โ
(^2) f โx^2 =^ โ^2.
This implies that (0, โ^23 ) is a local maximum for f. Similarly, at the critical point (0, โ โ^23 ),
det Hf ((0, โ โ^23 )) = โ โ^243 < 0.
It follows that (0, โ โ^23 ) is a saddle point for f.
