Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
This is the Solved Exam of Multivariable which includes Parametrized, Figure, Parallelogram, Drawing, Open Sets, Definition, Lines Intersect, Parameterized etc. Key important points are: Function, Whether, Continuous, Different Directions, Approaching, Vector Field, Jacobian Matrix, Direction, Point, Equation
Typology: Exams
1 / 7
This page cannot be seen from the preview
Don't miss anything!
EXAM II - NOVEMBER 6, 2007
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 6, 2007
1.(7 pts) (i) Consider the function
f (x, y) =
3 xy x^4 +y^2 ,^ if (x, y)^6 = (0,^ 0); 0 , otherwise.
Determine whether f (x, y) is a continuous at (0, 0). Justify your answer. [Hint: try approaching (0, 0) from different directions] Approach (0, 0) along the line y = mx for some m 6 = 0. For any (x, y) on this line that is different from (0, 0), we have
f (x, y) = (^) x 43 x+( mxm 2 )x 2 = 3 mx
2 x^2 (x^2 + m^2 ) =^
3 m x^2 + m^2 As (x, y) โ (0, 0) along y = mx,
f (x, y) โ (^3) mm 2 = m^3.
Thus for different values of m, f (x, y) would approach to different limits as (x, y) โ (0, 0). Hence, the limit
(x,y^ lim)โ(0,0) f^ (x, y) does not exist and therefore f is not continuous at (0, 0).
(8 pts.)(ii) Consider the vector field F (x, y, z) = (2xz, โxy, โz). Find div F and curl F. Here, F = (F 1 , F 2 , F 3 ) where F 1 = 2xz, F 2 = โxy, F 3 = โz. It follows that
div F = โ ยท F = โF โx^1 + โF โy^2 + โF โz^3 = 2z โ x โ 1.
curl F = โ ร F =
i j k โx^ โ โyโ โzโ F 1 F 2 F 2
= (0 โ 0)i โ (0 โ 2 x)j + (โy โ 0)k = 2xj โ yk.
4 EXAM II - NOVEMBER 6, 2007
The gradient of f is โf = (2xeโ^2 y, โ 2 x^2 eโ^2 y) so that โf (1, 0) = (2, โ2). It follows that the directional derivative of f in the direc- tion of u is
Duf (1, 0) = โf (1, 0) ยท (^) โuuโ = (2, โ2) ยท (1 โ,^ 1) 2
(8 pts) (ii) Find an equation for the line tangent to the level curve f (x, y) = 1 at the point (1, 0).
The gradient is always perpendicular to the level set so โf (1, 0) is orthogonal to the level curve f (x, y) = 1 at the point (1, 0). Thus, the tangent is given by
((x, y) โ (1, 0)) ยท โf (1, 0) = 0
or ((x โ 1), y) ยท (2, โ2) = 0 or y = x โ 1.
MATH206A MULTIVARIABLE CALCULUS - PROF. P. WONG 5
Note that (^) โf โx = 3x
(^2) + 6x = 3x(x + 2), โf โy = 3y
(^2) โ 6 y = 3y(y โ 2).
Setting ( โfโx , โfโy ) = (0, 0) yields (0, 0), (0, 2), (โ 2 , 0), (โ 2 , 2) as critical points. (8 pts) (ii) For each of the critical point(s) a found in part (i), find the corresponding Hessian matrix Hf (a).
The Hessian matrix is
Hf =
[ (^) โ (^2) f โx^2
โ^2 f โ^2 f โyโx โxโy
โ^2 f โy^2
6 x + 6 0 0 6 y โ 6
Thus, we have
Hf (0, 0) =
Hf (0, 2) =
and
Hf (โ 2 , 0) =
Hf (โ 2 , 2) =
(8 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.
The points (0, 0) and (โ 2 , 2) are both saddle points since the determinant det Hf < 0 for each of these points. The point (0, 2) is a local minimum since det Hf (0, 2) = 36 > 0 and โโx^2 f 2 (0, 2) = 6 > 0. The point (โ 2 , 0) is a local maximum since det Hf (โ 2 , 0) = 36 > 0 and โโx^2 f 2 (โ 2 , 0) = โ 6 < 0.
MATH206A MULTIVARIABLE CALCULUS - PROF. P. WONG 7
Consider the parametrization x(t) = (cos t, sin t)
where 0 โค t โค ฯ 2.
(9 pts.) Find the work done by the force F over the curve C. That is, find โซ C F (x) ยท dx.
Note that F (x(t)) = (cos t, cos^2 t + sin^2 t) and dx = (โ sin t, cos t) dt.
It follows that โซ C
F (x) ยท dx =
โซ ฯ 2 0
(cos t, cos^2 t + sin^2 t) ยท (โ sin t, cos t) dt
=
โซ ฯ 2 0
(cos t, 1) ยท (โ sin t, cos t) dt
=
โซ ฯ 2 0
cos t sin t + cos t dt
=
โซ ฯ 2 0
cos t(1 โ sin t) dt (use substitution w = 1 โ sin t)
= โ (1^ โ^ sin^ t)
2 2
ฯ 2
0
