Math 53 Homework 4 Solutions

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1 Math 5 Homework 4 Solutions Problem 1. (a) z = is a paraboloid with its highest point at (0,0,) and intersecting the -plane at the circle + = of radius. (or: rotate the parabola z = in the z-plane about the z-ais). z (b) z = + : the top half of a right circularcone(theheightz isequaltothe distance + from the z-ais). (or: rotate the half-line z =, 0 in the z-plane about the z-ais). z (c) maima on the -ais; saddle points on the -ais surrounding a local min at (0,0) (not directl visible but implied b the picture). (d) e / = c / = lnc. So the level curve f(,) = c is a straight line with slope lnc through the origin #6: If we start at the origin and move along, sa, the -ais, the z-values of a cone with its verte at the origin increase at a constant rate, so we epect the level curves to be equall spaced. A paraboloid centered on the z-ais, on the other hand, has z-values which change slowl near the origin and more quickl as we move further awa; thus we epect the level curves to be spaced more widel apart near the origin and more closel together further awa. (See pictures above). So contour map I must correspond to the paraboloid, and map II to the cone #61: graph C and contour map II: indeed, the function is the same if is interchanged with, so the graph and contour map are smmetric about the plane = ; moreover the function is constant over each hperbola =constant (since its value onl depends on that of ), and its values oscillate between 1 and #68: the level sets + + 5z = c form a famil of ellipsoids for c > 0 (and the origin for c = 0)

2 14. #9: Let f(,) = ( 4 4 )/( + ). On the -ais, f(,0) = for 0, so f(,) 0 as (,) (0,0) along the -ais. On the other hand, approaching (0,0) along the -ais, f(0,) =, so f(,) along the -ais. Since f has two different limits along two different lines, the limit does not eist. 14. #1: f(,) = We can see that the limit along an line through +. (0,0) is 0, as well as along various other paths approaching (0,0). So it is reasonable to epect that the limit eists and equals 0. One wa to prove it is to notice that, since = +, we have 0. Since 0 as + (,) (0,0), this bound implies that f(,) 0 as (,) (0,0). (Alternative solution: in polar coordinates, f(rcosθ,rsinθ) = (rcosθ)(rsinθ) = r rcosθsinθ, so indeed f(,) r and f(,) 0 as (,) (0,0).) 14. #9: lim (,) (0,0) + + = lim (rcosθ) +(rsinθ) r 0 + r = lim r 0 +(rcos θ +rsin θ) = #10: Starting at (,1), where f(,1) = 10, and moving in the positive - direction, we reach the net contour line (f = 1) after approimatel 0.6 units. This represents an average rate of change of f Or, moving in the negative -direction, we reach the net contour line (f = 8) after about 0.8 units, representing an average rate of change of f 0.8 =.5. Either of these (or even better, their average) would be a reasonable estimate of f (,1). Similarl for f : moving in the positive -direction the value of f decreases from 10 to 8 after approimatel 0.9 units, a rate of change of f Or, moving in the negative -direction, f increases from 10 to 1 after about 1 unit, which corresponds to f 1 =. Either value is a reasonable estimate of f (,1). 14. #4: w(u,v) = ev u+v w u = ev (u+v (b the quotient rule) and ) = (u+v v)e v (u+v ). w v = (u+v )(e v ) (v)(e v ) (u+v ) 14. #47: Differentiating + +z = 1 with respect to gives: ( + +z ) = z (1), so +0+6zz = 0, and = z. (Note: and are independent variables, while z is implicitl a function of,). Similarl, ( + +z ) = z (1), so 4 +6zz = 0, which gives = z. 14. #69: Differentiating w = /( + z) with respect to first then then z we find: w/ = 1/( +z); w/ = /(1/( +z)) = 1/( +z) ; and w/z = /z ( 1/( +z) ) = +4/( +z). Similarl: w/ = /( +z) ; w/ = 1/( +z) ; w/ = 0.

3 14. #77: u(,,z) = ( + + z ) 1/ u = 1 ()( + + z ) / = ( + +z ) /, and u = ( + +z ) / ( )()( + +z ) 5/ = ( + +z )+ ( + +z ) 5/ = z ( + +z ) 5/. B smmetr, u = z ( + +z ) 5/ and u zz = z ( + +z ) 5/; thus u +u +u zz = z + z +z ( + +z ) 5/ = 0. (Note: if ou ve alread taken Phsics 7B, ou might know that the electric potential solves the Laplace equation in regions of space that contain no electric charges. The calculation we just did confirms this for the electric potential of a charged particle at the origin, which is a constant multiple of ( + +z ) 1/.) 14. #79: B the (single variable) chain rule, (f( + at)) = f ( + at), while (+at) t (f(+at)) = t f (+at) = af (+at). Similarl for the partial derivatives of g( at), (g( at)) = g ( at) and t (g( at)) = ag ( at). Given u(,t) = f(+at)+g( at), we have u t = af (+at) ag ( at), and, differentiating again, u tt = a f (+at)+a g ( at); similarl, u = f (+at)+ g ( at) and u = f (+at)+g ( at). So u tt = a u. (Note: phsicall, f(+at) represents a wave whose shape is given b f and which travels towards negative at speed a: indeed, for fied t, the graph of f( + at) is obtained from that of f() b shifting b at to the left. Similarl, g( at) represents a wave with profile g and travelling towards positive at speed a. The given u(, t) is the superposition of a left-moving and a right-moving wave.) 14. #88: Note: the equation PV = mrt can be used to solve for an one of P,V,T as a function of the two others. The notation P/V means that we view P as a function of V and T, and consider the rate of change of P with respect to V (with T fied). Similarl for the other partials in the statement of the problem. With this understood: P = mrt/v, so P/V = mrt/v ; V = mrt/p, so V/T = mr/p; and T = PV/mR, so V/P = V/mR. Multipling, we get P V T V T P = mrt mr V V P mr = mrt PV = 1. This apparentl paradoical equalit is actuall a special case of a more general identit (see 14.5 problem # 58). It illustrates the dangers of reling on notation (no, ou can t simplif products of partial derivatives!). The reason wh there is no parado is that each individual quantit in the product corresponds to a different situation: P/V assumes we var V (and hence P) keeping T fied, while for V/T it is P that is kept fied, and for T/P it is V that remains constant. 14. #97: If f = + 4 then f = 4; however, if f = then f =. Since f and f are continuous everwhere but f (,) f (,), the eistence of such a function f would contradict Clairaut s theorem.

4 Problem. a) At (, 1 ), changing while keeping fied results in lower values of f if increases, and higher values if decreases, so f < 0. Similarl, f > 0. At the point (1,1), f < 0 (same argument); and f = 0 (the line parallel to the -ais through (1,1) is tangent to the level curve f = 0, and the value of f(1,) passes through a maimum for = 1, so at this point we have f = 0). b) The two critical points where f = f = 0 are near P 1 = (0.,0.7) and P = (1.5,1.5). The level curve through P 1 onl consists of P 1 itself (for slightl lower values of f, the level curves are small ovals, shrinking to the point P 1 as one approaches the local maimum). The level curve through P has two branches that intersect each other at P. Starting from P 1, the value of f decreases in ever direction. P 1 is a local maimum. Starting from P, the value of f decreases if we move North or South, but increases if we move East or West. P is a saddle point. Problem. a) f = , and f = + 1, so f (, 1 ) = 4, f (, 1 ) = ; f (1,1) = 1, f (1,1) = 0. b) f = + 1 = 0 gives = 1, and then f = = 0 can be rewritten as: ( 1) 6( 1) = 0, or = 0. The two roots of this quadratic are (± 49)/4, i.e. and 5 4. Remembering that = 1, this gives the two critical points ( 1, ) and (, 5 4 ). Net we calculate f = 6 6, (f = f = 1), and f =. At ( 1, ): f = 4 and f =, both negative: this eplains wh moving East/West or North/South alwas leads to a decrease in f. Once we learn about the second derivative test, we ll observe that, since f f f = ( 4)( ) 1 = 7 > 0 and f = 4 < 0, this is a local maimum. At (, 5 4 ): f = and f =, which eplains wh moving East or West (varing onl) causes f to increase, while North or South (varing onl) causes it to decrease. Once we learn about the second derivative test, we ll observe that, since f f f = ( ) 1 = 7 < 0, this is a saddle point #5: Differentiating z = f(,) = sin( + ), we get f = sin( + ) + cos(+) and f = cos(+). So f ( 1,1) = 1 and f (1,1) = 1, and the tangent plane is z 0 = 1( ( 1)) 1( 1), or z = (or ++z = 0) #18: Let f(,) = 1 +1, so f = ( 1)/( + 1) and f = 1/( + 1). Evaluating at (0,0): f(0,0) = 1, f (0,0) = 1 and f (0,0) = 1. So the linear approimation is f(0,0)+f (0,0) +f (0,0) = #1: f = 1 ()( + +z ) 1/ = ( + +z ) 1/ ; f = ( + + z ) 1/ and f z = z( + +z ) 1/. Evaluating at (,,z) = (,,6), where f = = 7, we get f (,,6) = /7, f (,,6) = /7, and f z (,,6) = 6/7. So the linear approimation is: f(,,z) 7+ 7 ( )+ 7 ( )+ 6 7 (z 7). Evaluatingat(.0,1.97,5.99): 7+ 7 (0.0)+ 7 ( 0.0)+6 7 ( 0.01) = v T T 14.4 #8: T(u,v,w) =, so dt = du+ 1+uvw u v v vw uw v uv du+((1+uvw) v (1+uvw) (1+uvw) dv+ dv + T w dw = (1+uvw) dw = v wdu+dv uv dw (1+uvw). 4

5 14.4 #: Let and be the two sides; then the area is A(,) =. Since da = A A d+ d = d+d, at (,) = (0,4) we have the linear approimation A + = Given 0.1 and 0.1, the maimum error in the area is about A 4(0.1)+0(0.1) = 5.4 cm #8: Here P = 8.1T/V, so dp = (8.1/V)dT (8.1T/V )dv, hence ( T P 8.1 V T V ) ( 5 V = ) The pressure will decrease b about 8.8 kpa. Problem 4: a) r = ( b+ b 4c)/, so r b = 1 ( 1+b/ ) b 4c, and r c = 1/ b 4c. If r 0 = b 0 /+ b 0 4c 0/, r = r r 0, b = b b 0 and c = c c 0, then r 1 ( ) ( ) 1+b 0 / b 0 4c 0 b+ 1/ b 0 4c 0 c. In particular, if b 0 = 5, c 0 = 4, r 0 = 4, r ( 4 ) b+( 1 ) c. Hence, for b =.01 and c =.0, r ( 4/)(.01) + ( 1/)(.0) =.0 and r 4.0 (the actual root is ) b) In the formula r ( 4/) b+( 1/) c, 4/ > 1/. Since the coefficient of b has larger magnitude, r is more sensitive to changes in b than to changes in c. 5