(a) Find the equation of the plane that passes through the points P, Q, and R.

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1 Math 040 Miterm Exam 1 Spring 014 S o l u t i o n s 1 For given points P (, 0, 1), Q(, 1, 0), R(3, 1, 0) an S(,, 0) (a) Fin the equation of the plane that passes through the points P, Q, an R P Q = 0, 1, 1, P R = 1, 1, 1 The normal vector of the plane is ī j k n = =, 1, The plane passes through the point P (, 0, 1) an has a normal vector n =, 1, 1 Its equation is (x ) + y (z + 1) = 0, or x + y z 5 = 0 (b) Fin the volume of the parallelepipe with ajacent eges P, Q, R an S V = P S P S = 0,, 1 ( ) P Q P R = P S n = 0,, 1, 1, 1 = 1 = 3 Fin parametric equations an symmetric equations for the line of intersection of the planes x y + z 5 = 0 an y z + = 0 The corresponing normal vectors are n 1 = 1, 1, 1 an n = 0, 1, 1 The irectional vector of the line is ī j k v = = 0, 1, Now we nee to fin a point that lies on the line Its x, y, an z coorinates must satisfy to both equations of planes First equation gives x = y z + 5 = y z = = 3 since y z + = 0 from the secon equation Put y 1, z = 1 that satisfy the secon equation Then the point P (3, 1, 1) lies on the line The vector equation of the line x, y, z = 3, 1, 1 + t 0, 1, 1 gives parametric equations x = 3, y 1 + t, z = 1 + t an symmetric equations x = 3, y + 1 = z 1

2 3 Evaluate the limit, if it exists If it oes not exist explain why lim (x,y) (0,0) y tan x x xy Note, that path x = 0 is not allowe since it is not in the omain of the function If you use this path then you probably lost several points Path 1 y = 0: Path y = x: lim cos x lim lim (x,y) (0,0) lim (x,y) (0,0) y tan x x xy = lim sin x x y tan x x xy = lim 0 x = 0 x tan x x x lim tan x x lim Two paths give two ifferent results for the limit Hence, the limit oes not exist sin x x cos x 4 Using the linear approximation of f(x, y) = cos x y approximate the number cos(00) 098 The linearization of f(x, y) at (0, 1) is L(x, y) = f(0, 1) + f x (0, 1)x + f y (0, 1)(y 1) We have f(0, 1) = 1, f x sin x y, f x (0, 1) = 0, f y cos x y 3/, f y(0, 1) 1 Then L(x, y) = 1 1 (y 1) cos(00) = f(00, 098) L(00, 098) = (098 1) = = 101 Note, that just a result for L(00, 098) is not consiere to be the answer since the question is to approximate the number cos(00) (a) Fin the graient of f(x, y) = 4 ln y cos x 4 ln y sin x f x (x, y) = cos x, f y(x, y) = 4 y cos x 4 ln y sin x f = cos x, 4 y cos x (b) Evaluate the graient foun in (a) at the point P (π, ) f(π, ) = 0, Page

3 (c) Fin the rate of change of f at P in the irection of the vector ū = 5 1, 3 The rate of change is f(1, ) ū = = 5, 1 ū = 5, ū ū = 0, 1, 3 5 = 6 5 ū ū = ū 1 ū = 5 1 5, 3 5 = 1, Fin the curvature of at the point (0, 1, 0) r(t) = e t sin t ī + e t j + t k At the point (0, 1, 0) t = 0 The curvature is r (t) = e t ( sin t + cos t), e t, t, κ = r (0) r (0) r (0) 3 r (t) = e t (4 sin t + cos t + cos t sin t), 4e t, = e t (3 sin t + 4 cos t), 4e t, r (0) = 1,, 0, r (0) = 4, 4,, r (0) r (0) = 4,, 4 = 6, r (0) = 5 κ = Fin the curvature of y = x4 8 + x 4 at the point ( 1, 3 ) 8 The curvature is κ = y (1) 1 + (y (1)) 3/ y = x3 + x, y (1) = 1, y = 3x + 1, y (1) = κ = (1 + 1) 3/ = 3/ = = 1 A right solutin can be obtaine if you use a vector function r(x) = x, x4 8 + x 4, 0 8 Use ifferentials to estimate the amount of paint neee to apply a coat of paint 005 cm thick to four sies of a rectangular builing with the base of 1 18 meters an the hight of 10 meters Note: 1 cm = 001 m, the roof will not be painte, sies only Hint: Use V (x, y, z) = xyz, z = 0, where z is the hight Let x be the with, y be the length, an z be the hight Then the volume is V = xyz, x = 1 m, y = 18 m, z = 10 m, x = y = 005 cm = 10 3 m Page 3

4 V V = V x x + V y y + V z z, where x x, y y, z z = 0 Hence V V x (1, 18, 10) x + V y (1, 18, 10) y V x = yz, V x (1, 18, 10) = 180, V y = xz, V y (1, 18, 10) = 10 V = = 03 m 3 The amount of paint neee to apply a coat is 03 m 3 9 Fin the maximum rate of change of the function f(x, y, z) = ln(xy z 3 ) at the point (, 4, 3) an the irection in which it occurs The maximum rate of change occurs in the irection of the graient vector of f(x, y, z) = ln x + ln y + 3 ln z: f = 1 x, y, 3, f(, 4, 3) = 1 z, 1, 1 Hence the maximum rate of change occurs in the irection of the vector 1, 1, 1 The maximum rate of change is 1, 1, 1 3 = Note ln(xy z 3 ) ln x + ln y + 3 ln z, in general since these two functions have ifferent omains But you can check that both functions have the same erivative an we can use the secon function to evaluate the erivative faster 10 Fin z x an z y if xz + yz = 1 Implicit ifferentiation Differentiate both sies wrt x: (xz + yz ) x = (1) x, z + xz x + yzz x = 0, (x + yz)z x z, z x z x + yz Differentiate both sies wrt y: (xz + yz ) y = (1) y, xz y + z + yzz y = 0, (x + yz)z y z, z y x + yz (= zz x) Alternative solution: Put F (x, y, z) = xz + yz 1 Then F x = z, F y = z, F z = x + yz, z x F x z F z x + yz, z y F y z F z x + yz z 11 Fin an sketch the omain of the function f(x, y) = ln(y + x ) x 1 Page 4

5 Domain is efine by the ineqaulities y + x 0 an x 1 Or y x, x 1 In the xy-plane the omain is the upper region boune by the parabola y x (ashe curve) with the portions of the line x = 1 above the parabola remove from the region (ashe line) 1 Fin all the secon partial erivatives for the function f(x, y) = ln(x y) f x = (x y) 1, f y (x y) 1, f xx = ()( )(x y) 4(x y) 4 (x y), f yx = f xy = ( 1)( 1)(x y) = (x y) = (x y), f yy ( 1)( 1)(x y) (x y) 1 (x y) 13 For the given surface xy = x z + 1 an the point P (3,, ) (a) Fin an equation of the tangent plane to the surface at P The equation of the surface can be written in the form of level surface F (x, y, z) = xy x + z = 1 Then F x = y x, F x (3,, ) 4, F y = x, F y (3,, ) = 3, F z = z, F z (3,, ) = 4 An equation of the tangent plane to the surface at P is 4(x 3) + 3(y ) + 4(z ) = 0 or 4x + 3y + 4z = 0 or 4x 3y 4z + = 0 * Alternative solution: z = (x xy + 1) 1/ (it is positive since at the point P z = > 0) Then z x = x y x y (x = xy + 1) 1/ z, z y = x x (x = xy + 1) 1/ z At the point P (3,, ) we have z x = 6 = 1, z y An equation of the tangent plane to the surface at P is z = 1 (x 3) + ( 3 4) (y ), z = + x y + 3, z = x 3 4 y + 1 If you multiply both sies by 4, then the last equation can be written as 4z = 4x 3y + or 4x 3y 4z + = 0 which is the same as before (b) Fin symmetric equations of the normal line to the surface at P x 3 4 = y = z 3 4 or x 3 4 = y 3 = z 4 or Page 5

6 x 3 = y 3 (z ) 4 bonus problem Calculate (z x ) (z y ) (z xy ) 3 if xyz = 1 x [xyz = 1], yz + xyz x = 0, z x z x, z xy = y [z x] = [ z ] z y y x x = z xy y [xyz = 1], xz + xyz y = 0, z y z y, Then (z x ) (z y ) (z xy ) 3 = z4 x y Note: xyz = 1 z3 x 3 y 3 (xyz) = z4 x y z4 x y = 0 * Alternative solution: z = (xy) 1, z x x y 1, z y x 1 y, z xy = x y Then (z x ) (z y ) (z xy ) 3 = (x 4 y )(x y 4 ) (x 6 y 6 ) = x 6 y 6 x 6 y 6 = 0 bonus problem Calculate zz xy z x z y if = ln(xyz) ln x + ln y + ln z = x [ln x + ln y + ln z = ], 1 x + z x z = 0, z x z x, y [ln x + ln y + ln z = ], 1 y + z y z = 0, z y z y, z xy = y [z x] = y [ z ] z y x x = z xy Then zz xy z x z y = z xy z xy = 0 * Alternative solution: z = e (xy) 1, z x e x y 1, z y e x 1 y, z xy = e x y Then zz xy z x z y = (e x 1 y 1 )(e x y ) ( e x y 1 )( e x 1 y ) = e 4 x 3 y 3 e 4 x 3 y 3 = 0 Page 6