Review 1 1) State the largest possible domain o deinition or the unction (, ) = 3 - ) Determine the largest set o points in the -plane on which (, ) = sin-1( - ) deines a continuous unction 3) Find the equation o the plane tangent to z = -sin( ) at the point P = (1, 1, 0) 4) A stead-state temperature unction u = u(, ) or a thin lat plate satisies Laplace's equation u + u = 0 Show that u = esin() satisies Laplace's equation 5) Find the maimum and minimum values attained b (, ) = - + 4 on R, where R is the triangle region with vertices (0, 0), (, 0), and (0, ) 6) Given (, ) = ( + )sin( ); P(1, 4), Q(105, 399), use the eact value (P) and the dierential d to approimate the value (Q) 7) Find z and z as unctions o,, and z assuming that z = (, ) satisies the equation e z + e + ze = 7 8) Suppose that the temperature w (in degrees Celsius) at the point (, ) is given b w = (, ) = 5 + 000 + 0003 In what direction u should a grasshopper hop rom the point (10, 0) in order to get warmer as quickl as possible? Find the directional derivative in this optimal direction 9) Given (,, z) = - z, (a) what is the direction in which is increasing most rapidl at the point (1,, 3)? (b) What is the directional derivative o in that direction? (c) What is the directional derivative o at the point (1,, 3) in the direction o i + j + k? 10) A rectangular bo has its lower our vertices in the -plane and its upper our on the ellipsoidal surace with equation 6 + 3 + z = 450 Use the method o Lagrange multipliers to ind the maimum possible volume that such a bo can have 11) Find and classi the critical points o (, ) = cos( ) cos( ) 1
SOLUTIONS Problem 1 The cubic root is deined or an real number and thereore the largest possible domain o deinition o the unction entire planer 3 (, ) = is the Problem The unction arcsinis deined and continuous on the interval [ 1,1] The unction unction or equivalentl, arcsin( ) is deined and continuous onr Thereore the is deined and continuous in the region 1 1, 1 + 1, ie the region between two quadratic parabolas including the parabolas themselves Problem 3 An equation o the tangent plane to the surace z (, ) point P ( 0, 0, z0) is (see eg Rule () on page 916) In our case z z = (, )( ) + (, )( ) 0 0 0 0 0 0 0 = at the = =, (, ) π cos( π ) and (, ) π cos( π ) whence (1,1) = π and (1,1) = π An equation o the tangent plane can be written as z= π( 1) + π( 1) or z = π+ π 3π
Problem 4 We compute u = e sin(), u = e cos(), u u + = 0 u = 4e sin(), whence u = 4e sin() Problem 5 We ollow the procedure outlined in Rule (9) on page 95 (see also Eample 7) The partial derivatives are (, ) =, (, ) = 4+ 4 Thereore the onl critical point o is (0, 1) and this point is not inside R but on its boundar We look now what happens on the boundar o R I = 0, then unction is 0 when 0 then (0, ) = + 4,0 The smallest value o this = or = ; the largest value is when = 1 I = 0 (,0) =,0, with the smallest value 0 when = 0 and the = Finall on the third side o the triangle = (, ) = + 4,0, with the smallest value 0 when = 0 = largest value 4 when and and the largest value 4 when Summarizing we see that the smallest value o the unction in R is 0 and it is attained at (0, 0) or (0, ) The largest value is 4 attained at (, 0) Problem 6 We use the ormula (, ) ( a, b) + d = ( a, b) + ( a, b)( a) + ( a, b)( b) Notice that π π π (, ) = sin + ( + ) cos and
π π π (, ) = sin ( + ) cos Thereore (4,1) = (4,1) = 1and (105,399) 5 + 05 01 = 504 Problem 7 Dierentiating both parts o the relation b we get z e + e + ze = 7 whence z z z z e + e + e + e = 0, + + z z e e = z e e Similarl, dierentiating the original relation b we obtain whence z z z e z + + e + ze + e = 0 z e + ze + ze = z e + e z,
Problem 8 B Theorem 15 on page 939 the grasshopper should move in the direction o the gradient vector, (10,0) = (10,0) i+ (10,0) j= 04 i+ 1j which is proportional to the vector i+ 3j The corresponding directional derivative is the norm o the gradient vector (10,0) = 04 + 1 = 016 = 16 5 10 Problem 9 (a) =, = z, = (1,,3) = i 6j 4k z (b) The directional derivative in the direction o is (1,,3) = + 6 + 4 = 14 748 (c) The unit vector in the direction i+ j+ k is 1 u= i+ j+ k The 3 3 3 1 directional derivative is u = 6 4= 4 3 3 3 Problem 10 First notice that such a bo must have lower vertices at the points ),(, ),(, ), and (, ) where and are positive real numbers The volume o such a bo is V = 4z Notice also that 9 6 3 z = V We know that the product o three non-negative 4 numbers with a ied sum takes its largest value when the numbers are equal; hence 6 3 z 150 = = = andv ma = 4 5 50 75= 500 6 1474
π π π Problem 11 First we compute (, ) = sin cos and π π π (, ) = cos sin From these epressions we see that the critical points o are either o the orm ( k, n) or (k+ 1,n+ 1) where k and n are integer numbers To ind the character o these critical points we compute the derivatives o the second order π π π = cos cos, 4 π π π = = sin sin, 4 π π π = cos cos, 4 In the case o a critical point ( k, n) we have = 0 I both k and n are both even or both odd then π = = and b Rule 3 on page 947 we have a 4 local maimum On the other hand i one o these numbers is even and the second is odd then π = = and the point is a local minimum 4 Finall we notice that at a point(k+ 1,n+ 1) = = 0but 0 whence such a point is a saddle point