{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

part_diff_sol

# part_diff_sol - MIT OpenCourseWare http/ocw.mit.edu 18.02...

This preview shows pages 1–4. Sign up to view the full content.

MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 200 7 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2. Partial Differentiation 2A. Functions and Partial Derivatives 2A-1 In the pictures below, not d l of the level curves are labeled. In (c) and (d), the picture is the same, but the labelings are different. In more detail: b) the origin is the level curve 0; the other two unlabeled level curves are .5 and 1.5; c) on the left, two level curves are labeled; the unlabeled ones are 2 and 3; the origin is the level curve 0; d) on the right, two level curves are labeled; the unlabeled ones are -1 and -2; the origin is the level curve 1; The crude sketches of the graph in the first octant are at the right. 2A-3 a) both sides are mnxm-' yn-' x-Y . -x -(Y b) f, = fx, + = (fx), = ' --- f, - x) + = - + f y x = (x yI2 (x yI3 ' (x yI2 ' (x + Y ) ~ c) f, = -2xsin(x2+y), f,, = (f,), = -2xcos(x2 +y); f, = -sin(x2+y), f,, = -cos(x2+y).2x. d) both sides are fl(x)g'(y). 2A-4 (f,), = ax + 6y, (f,), = 22 + 6y; therefore f,, = f,, H a = 2. By inspection, one sees that if a = 2, f (x, y) = x2y + 3xy2 is a function with the given f, and f,. 2A-5 a) w, = aeax sin ay, w,, = a2eax sin ay; W, = eaxa cos ay, w,, = eaxa2 (- sin ay); therefore wyy = -w,,. 22 2(y2 - x2) b) We have w, = - + WXX = + If we interchange x and y, the function x2 y2 ' (x2 y2)2 ' w = ln(x2 + y2) remains the same, while w,, gets turned into w,,; - - since the interchange just changes the sign of the right hand side, it follows that w,, = -w,,. 2B. Tangent Plane; Linear Approximation 2B-1 a) z, = y2, z, + = 2xy; therefore at (1,1,1), we get z, = 1, z, = 2, so that the tangent plane is z = 1 (x - 1) + 2(y - I), or z = x + 2y - 2.
2. PARTIAL DIFFERENTIATION 1 b) w, = -y2/x2, W, = 2ylx; therefore at (1,2,4), we get w, = -4, w, = 4, so that the tangent plane is w = 4 - 4(x - 1) + 4(y - 2), or w = -4x + 4y. x x Y 2B-2 a) z, = - - - by symmetry (interchanging x and Y), z, = -; then the I / - z z xo Yo xo Yo tangentplaneisz=zo+-(x-xo)+-(y-yo), or z = --x+--Y ,sincexi+~:=z:. zo 20 zo zo b) The line is x = xot, y = yot, z = zot; substituting into the equations of the cone and the tangent plane, both are satisfied for all values of t; this shows the line lies on both the cone and tangent plane (this can also be seen geometrically). 2B-3 Letting x, y, z be respectively the lengths of the two legs and the hypotenuse, we have z = I / - thus the calculation of partial derivatives is the same as in 2B-2, and 3 4 7 we get Az M -Ax + -Ay. Taking Ax = Ay = .01, we get Az M -(.01) = .014. 5 5 5 2B-4 From the formula, we get R = R1R2 . From this we calculate R1+ R2 dR 2 , and by symmetry, 4 4 1 Substituting R1 = 1, R2 = 2 the approximation formula then gives AR = -ARl+ -AR2. 9 9 4 1 5 By hypothesis, lARil 5 .l, for i = 1,2, so that 1 ARl 5 -(.I) + -(.I) = -(.I) M .06; thus n 9 9 9 2B-5 a) Wehave f ( x , y ) = ( ~ + y + 2 ) ~ , fX=2(x+y+2), fy=2(x+y+2). Therefore at (0,0), f,(O, 0) = f, (0,O) = 4, f (0,O) = 4; linearization is 4 + 42 + 4y; at (1,2), fx(l,2) = fy(l,2) = 10, f (l,2) = 25; linearization is 10(x - 1) + 10(y - 2) + 25, or lox + 10y - 5. linearization at (0,O): 1 + x; linearization at ( 0 , ~ / 2 ) :-y dV dV dV 2B-6 We have V = rr2h, - = 2 ~ r h , - " AT+ dr a h = r r 2 ; A V M ( ~ ) , ( = ) , ~ h .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 16

part_diff_sol - MIT OpenCourseWare http/ocw.mit.edu 18.02...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online