Math 53 Homework 5 Solutions

Transcription

1 14. #: dw dt = w = 14. #7: s = t = Math Homework Solutions dx dt + w (t t 1+t t (1 t) ) (1+t) e (1 t)/(1+t). dy dt + w dz dt = tey/z x z ey/z xy z ey/z s + s = (x y)4 (st) (x y) 4 t = (x y) 4 (st t ). t + t = (x y)4 s (x y) 4 (st) = (x y) 4 (s st). 14. #1: g(u,v) = f(x(u,v),y(u,v)) where x(u,v) = e u +sinv and y(u,v) = e u + cosv. So g u = f x x u +f y y u = e u f x +e u f y, and g v = f x x v +f y y v = cosvf x sinvf y. At (u,v) = (0,0), (x,y) = (1,); so g u (0,0) = 1 f x (1,)+1 f y (1,) = + = 7, and g v (0,0) = 1 f x (1,) 0 f y (1,) =. 14. #18: w u = w w v = w u + w u + w u ; v + w v + w v. 14. #4: Call the sides a and b, and the angle between them θ; so the area A = 1 da absinθ is assumed to be constant. So dt = A da a dt + A db b dt + A dθ θ dt = 0, i.e., 1 da bsinθ dt + 1 db asinθ dt + 1 dθ abcosθ = 0. Solving for dθ/dt, we get: dt dθ dt = bsinθda/dt+asinθdb/dt. Using the given values for a,b,θ,da/dt,db/dt, abcosθ we get dθ dt = ( ) = rad/s #4: (a) chain rule: r = cosθ+ sinθ, θ = ( rsinθ)+ rcosθ. ( ) ( ) (b) = cos θ + ( ) r cosθsinθ + sin θ, and ( ) ( ) = r sin θ ( ) θ r cosθsinθ + r cos θ. So ( ) + 1 ( ) [ ( ) ( ) ] ( ) ( ) r r = + (cos θ +sin θ) = +. θ 14. #1: s = s + s = s +r. So z r s = ( s ) + ( r ) = s ( ) +r ( ) + r r r r ( ( ) = s r + ( ) ) ( ( ) +r r r + ( ) r ) + 1

2 = 4rs z +4s z +4r z z +4rs + ( ) z = 4rs + z +4(r +s ) z +. Explanation: to compute z/ r s, we differentiate the expression for / s with respect to r (keeping s constant). The first step involves the product rule. The second one is more subtle. To calculate r ( ), we use the chain rule once more. If this feels confusing, just set g = / (think of this as some new function of x and y), and note that we are trying to calculate g r. The chain rule gives g r = g r + g r. Remembering that g = /, the first partials of g with respect to x and y are actually second partial derivatives of z. The chain rule is used similarly to differentiate / with respect to r #1: We can approximate the directional derivative at K by the average rate of change of pressure between the points where the line through K and S (red on the figure) intersects the contour lines closest to K. In this case we measure that, going about 1/6 of the way towards S, ( s 0 km), the pressure drops from 1000 to 996 mb ( P 4 mb). So DûP P s 4 0 = 0.08 (millibars per km) #9: (a) f = f x,f y,f z = xyz yz,x z xz,x y xyz. (b) at (x,y,z) = (, 1,1), f =,,. (c) û = 0, 4, is a unit vector, so D ûf = f û =,, 0, 4, = #7: (a)givenaunitdirectionvectorû, recallthatdûf = f û = f cosθ (where θ is the angle between f and û. Since the minimum value of cosθ is 1, occurring for θ = π, the minimum value of Dûf is f and occurs when û is in the opposite direction of f. (b) f = 4x y xy,x 4 x y, so at the point (, ) f decreases fastest in the direction of f(, ) = 1, 9 = 1,9 (or the corresponding unit vector) #8: f(4,6) is perpendicular to the level curve of f that passes through (4,6); so we sketch a portion of level curve through (4,6) (using the nearby level curves as guidelines), and draw a line perpendicular to it. The direction of the gradient vector is parallel to this line, and pointing towards increasing values of f. (towards the lower-right, making about a 6 angle with the horizontal direction). Next we estimate the magnitude f, which equals the directional derivative of f at (4,6) in the direction of f. We estimate this by finding the average rate of change along the direction perpendicular to the level curve. The points where the line previously drawn intersects the contour lines f = and f = are 0. units apart, so f = 1 and s 0., giving f f s 1 0. =. Hence we sketch the gradient vector with length. (Diagram omitted). (Note: we could also have tried to estimate f x and f y separately and use those to sketch f; this is much less accurate, especially concerning the direction of f.) 14.6 #41: Let f(x,y,z) = (x ) +(y 1) +(z ) : then we are considering the level surface f = 10. f = 4(x ),(y 1),(z ), so f(,,) = 4,4,4.

3 (a) f(,,) = 4,4,4 is a normal vector for the tangent plane at (,,), so an equation of the tangent plane is 4(x )+4(y )+4(z ) = 0, or 4x+4y+4z = 44 (or x+y +z = 11). (b) The normal line has direction 4,4,4, so parametric equations are x = +4t, y = +4t, z = +4t. (or using 1,1,1, x = +t, y = +t, z = +t) #49: f = y,x, so f(,) =,. So the tangent line has equation, x,y = 0 (x )+(y ) = 0, which simplifies to x+y = 1. y f(,) 4 1 x #6: first note that the point (1,1,) is on both surfaces. Let f(x,y,z) = x + y + z, so the ellipsoid is f = 9: then f = 6x,4y,z, so the tangent planetotheellipsoidhasnormalvector f(1,1,) = 6,4,4, andanequationofthe tangent plane is 6x+4y+4z = 18 or x+y+z = 9. The sphere is the level surface g = 0whereg(x,y,z) = x +y +z 8x 6y 8z+4,and g = x 8,y 6,z 8. So the tangent plane at (1,1,) has normal vector g(1,1,) = 6, 4, 4, giving the equation 6x 4y 4z = 18 or x+y+z = 9. The tangent planes are the same, so the surfaces are tangent to each other at (1,1,). (Note: it would have been enough to show that the normal vectors are parallel to each other, without determining the equations of the tangent planes.) Problem 1. a) (i) We need the vector û = a,b to be tangent to the level curve through (, 1 ) (whose shape we can estimate from the neighboring ones). Indeed, the directional derivative Dûf = f û is zero when û is perpendicular to f, i.e. tangent to the level curve through (, 1 ). Hence, the two directions in which df/dt = 0 are the two unit vectors which are tangent to the level curve at (, 1 ). (One is about 0 counterclockwise from î, the other is in the opposite direction i.e. about 10 clockwise from î). Estimating the shape of the level curve through (, 1 ) and drawing its tangent line L, we find that L lies on the side of the level curve where f(x,y) f(, 1 ). So, moving along L in either direction, the value of f reaches a minimum at (, 1 ). (ii) The directional derivative is largest in the direction of the gradient f (perpendicular to level curve, towards larger values of f), i.e. about 10 counterclockwise from î. It is smallest in the direction of f, i.e. about 60 clockwise from î. The nearest given level curves (f = 0.8 and f = 1) are about 0. cm apart, i.e. 0.1 unit, in that direction, hence we estimate df/dt f/ t 0./0.1 = in the direction of f, and in the opposite direction. b) f x = x 6x+y+1, and f y = x y+1, so f x (, 1 ) = 4, f y(, 1 ) =. Using the chain rule (or the definition of the directional derivative), df/dt = f x dx/dt + f y dy/dt = 4 a+ b when t = 0.

4 (i) We want 4 a+ b = 0, i.e. a = b. So we want unit vectors of the form b,b. We need b,b = b = 1, hence b = ± 1. The two solutions are therefore, 1 and, 1. (ii) The largest directional derivative is in the direction of f = 4,, i.e. the unit vector a,b = f f = 1,. The smallest value is in the direction of f, i.e. the unit vector 1,. The directional derivatives in those two directions are f a,b = ± f = ± 4 ±1.68. x Problem : f(x,y,z) = x +y +z, y x +y +z, z x +y +z. f(,6, ) = 7, 6 7, 7 gives the direction of maximum rate of change, and the maximum rate is D dir( f) f = f = 1. Since f(x,y,z) is the distance from the origin to (x,y,z), the answer makes sense geometrically: distance from the origin increases fastest when moving radially outwards, and the rate of increase is 1. Problem. a) g = g x,g y,g z = x,y, 6 = 8,4, 6 at (4,,). The direction of greatest decrease is that of g, i.e. the unit vector g g = 8, 4,6 = 4,, b) Let x = x 4, y = y, z = z ; then the line in the direction of 4,, can be parametrized by x = 4t, y = t, z = t. (Dividing by 9 is unnecessary and makes calculations more complicated.) At P0 = (4,,), we have g =, and g = 8,4, 6 (from part (a)), so linear approximation gives g(x,y,z) g(p 0 )+ g(p 0 ) x, y, z = +8 x+4 y 6 z = +8( 4t)+4( t) 6(t) = 8t. Therefore, g = 0 when 8t 0, or t 1/9. At t = 1/9, (x,y,z) = (4 4t, t, + t) = (4 4 9, 9, + 9 ). Evaluating g at this point, we find 0.04, fairly close to #7: u t = α k e α k t sinkx, and u xx = k e α k t sinkx, so indeed u t = α u xx. 14. #81: by product rule and chain rule, c(x,t) = (4πD) 1/ t 1/ e x /(4Dt) c t = 1 /(4Dt) + x (4πD) 1/e x 4Dt (4πDt) 1/ e x /(4Dt) = Dt+x /(4Dt) 8π 1/ D /, t /e x t / c = x 4Dt (4πDt) 1/ e x /(4Dt) = x 4π 1/ D / t /e x /(4Dt), and c = 1+x(x/4Dt) /(4Dt) 4π 1/ D / t / e x = Dt+x /(4Dt) 8π 1/ D /. t /e x Comparing these expressions, we find that indeed c/ t = D c/. 4

5 14.7 #: From the contour plot, there appears to be a local minimum near (1,1) (enclosed by oval-shaped level curves indicating that as we move away from the point in any direction the values of f are increasing). Moreover, the shape of the level curves near the origin is characteristic of a saddle point at (0,0). To verify these guesses, we have f(x,y) = 4+x +y xy f x = x y and f y = y x. We have critical points when f x = f y = 0. The first equation x y = 0 gives y = x, and substituting into the second equation gives (x ) x = 0, hence x(x 1) = 0, hence x = 0 or x = 1. So we have two critical points (0,0) and (1,1). The second partial derivatives are f xx = 6x, f yy = 6y, f xy =, so D(x,y) = f xx f yy f xy = 6xy 9. Then D(0,0) = 0 9 < 0 so f has a saddle point at (0,0); and D(1,1) = 6 9 > 0, with f xx (1,1) = 6 > 0, so f has a local minimum at (1,1) #11: f(x,y) = x x+xy f x = x +y and f y = 6xy. Then f x = 0 x +y = 1, f y = 0 xy = 0. Thus, at a critical point, either x or y is 0, and then the other one is ±1. There are four critical points: (±1,0) and (0,±1). Next, f xx = 6x, f xy = 6y, f yy = 6x, so D = f xx f yy f xy = 6x 6y. D(1,0) = 6 > 0 and f xx (1,0) = 6 > 0, so (1,0) is a local minimum. D( 1,0) = 6 > 0 and f xx ( 1,0) = 6 < 0, so ( 1,0) is a local maximum. D(0,±1) = 6 < 0, so (0,1) and (0, 1) are saddle points #4: We want to minimize the distance from (4,,0) to (x,y,z), d = (x 4) +(y ) +z, where z = x + y. Instead, it is easier to minimize d = f(x,y) = (x 4) +(y ) +(x +y ). Since f x = (x 4)+x = 4x 8 and f y = (y )+y = 4y 4, the only critical point is (x,y) = (,1). This point must correspond to the minimum distance (f(x,y) when x and/or y become large). For x = and y = 1, the equation of the cone gives z = or z = ±. Hence the points on the cone closest to (4,,0) are (,1,± ) #49: Let (x,y,z) be the corner opposite the origin. Since z = 1 (6 x y) and the volume is xyz, we want to maximize f(x,y) = xyz = 1 xy(6 x y). f x = 1 y(6 x y), and f y = 1 x(6 x 4y). Setting f x = f y = 0 gives x+y = 6 and x+4y = 6, hence the only critical point is (x,y) = (,1), which geometrically must yield a maximum. Thus the volume of the largest box is V = f(,1) = 4.