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Fall 2006 Group G1 University of Illinois at Urbana-Champaign Math 380 Graded Homework III Due Friday, September 29. 2 1. Find √ the directional derivative of the mapping f dened by f (x, y) = xy + ln(x + 1) in the direction of 3 1 u=( , ). 2 2 2x Correction. One has ∇f (x, y) = (y + , x) ; since u is a unit vector, the derivative of f in the direction 2 x +1 of u is simply ∇f.u, in other words √ √ x 3 3x y+ 2 + . ∇u f (x, y) = 2 x +1 2 2. Given unit vectors u = (ux , uy ) and v = (vx , vy ), and a function z = z(x, y) with continuous second-order partial derivatives, nd a formula for the mixed second order directional derivative ∇u ∇v z . ∂z ∂z By denition, we have ∇v z = ∇z.v = vx + vy . ∂x ∂y Applying the same denition, we obtain ∇u ∇v z = ∇( ∇u ∇ v z = ( ∂z ∂z vx + vy ).u. Therefore, ∂x ∂y ∂2z ∂2z ∂2z ∂2z ∂2z ∂2z ∂2z ∂2z vx + vy , vx + 2 vy ).(ux , uy ) = vx ux + vy ux + vx uy + 2 vy uy . 2 2 ∂x ∂x∂y ∂y∂x ∂y ∂x ∂x∂y ∂y∂x ∂y 3. Show that y is dened implicitly as a function of x in the neighborhood of the point P in the following equations : π • x cos(xy) = 0, P = (1, ) ; 2 • xy + log(xy) = 1, P = (1, 1) (log stands for the natural logarithm). Use implicit dierentiation to compute y 0 (1) and y 00 (1) in the rst case, and y 0 (1) in the second case. 2 Correction. The implicit function theorem tells us that, if F : R → R and λ ∈ R, then the equation ∂F F (x, y) = λ denes implicitly y as a function of x in a neighborhood of a solution P as soon as is ∂y continuous and dierent from 0 at P . ∂F ∂F π • In the rst case we have F (x, y) = x cos(xy), so that = −x sin(xy), which implies that (1, ) = −1. ∂y ∂y 2 Therefore the hypothesis of the theorem is satised, and y is dened implicitly as a function of x near P . Implicit dierentiation yields (cos(xy) − y sin(xy))dx − x sin(xy)dy =0. Since we are looking for y 0 (1), we may π notice that close to (1, ) the equation x cos(xy) = 0 means that cos(xy) = 0, so that implicit dierentiation 2 simply gives −y sin(xy)dx − x sin(xy)dy = 0, or π 2 dy y y0 y y = y 0 (x) = − . Thus y 00 (x) = − + 2 = 2 2 close to P , dx x x x x and this gives y 0 (1) = − , and y 00 (1) = π . (Notice that, in that case, one can actually solve the equation : in π a neighborhood of P , one simply has y = ; dierentiating this equality gives for y 0 (1) and y 00 (1) the values 2x we obtained earlier). ∂F 1 ∂F • In this case F (x, y) = xy + log(xy), so = x + . Therefore, (1, 1) = 2 6= 0, so the equation denes ∂y y ∂y implicitly y as a function of x near P . Implicit dierentiation gives this time y 0 (x) = − y+ x+ 1 x 1 y , or y 0 (x) = y 2 (x)x + y(x) . This shows that y 0 (1) = 1. y(x)x2 + x (Bonus, just to show how atrocious iterated implicit dierentiations can get) If we were asked to compute also y 00 (1), then we would have to take a deep breath, then write that y 00 = (2yy 0 x + y 2 + y 0 )(yx2 + x) − (y 2 x + y)(x2 y 0 + 2xy + 1) (2 + 1 + 1)(1 + 1) − (1 + 1)(1 + 2 + 1) , so y 00 (1) = =0. (yx2 + x)2 (1 + 1)2 4. Show that 2xy − z + 2xz 3 = 5 can be solved implicitly for z as a function of x near (1, 2, 1). ∂2z Compute the rst-order partial derivatives of z at (1, 2), as well as 2 (1, 2). ∂y Correction. We have this time an equation of the type F (x, y, z) = λ ; for this to dene z implicitly near ∂F be continuous and dierent from 0 at that point. A direct computation yields (1, 2, 1) , it is enough that ∂z ∂F ∂F that = −1 + 6xz 2 , so that (1, 2, 1) = 5 6= 0. Therefore the equation denes z implicitly as a function ∂z ∂z of (x, y) near (1, 2, 1). Implicit dierentiation leads to 2ydx + 2xdy − dz + 2z 3 dx + 6xz 2 dz = 0, so that (2y + 2z 3 )dx + 2xdy . This enables us to obtain (2y + 2z 3 )dx + 2xdy = (1 − 6xz 2 )dz , or dz = 1 − 6xz 2 ∂z 2y + 2z 3 = , ∂x 1 − 6xz 2 From (*) we obtain that ∂z 2x = (∗) ∂y 1 − 6xz 2 6 ∂z 2 ∂z (1, 2) = − and (1, 2) = − . We also get ∂x 5 ∂y 5 ∂z 12xz ∂y ∂2z = 2x . ∂y 2 (1 − 6xz 2 )2 Given the values computed above, this yields 12(− 25 ) 48 ∂2z (1, 2) = 2 =− . ∂y 2 25 125 5. Determine whether the function f : R3 → R3 dened by f (x, y, z) = (ex cos(y) + z, x sin(y) sin(z), xz cos(y)), π admits a dierentiable inverse g near (1, , π). If so, give the value of the Jacobian determinant of g at the 2 π point f (1, , π) = (π, 0, 0). 2 The consequence of the implicit function theorem seen in class (called the local inversion theorem) π implies that we only need to check whether the Jacobian determinant of f vanishes or not at (1, , π). The 2 Jacobian matrix of f at (x, y, z) is Correction. ex cos(y) −ex sin(y) 1 Jf(x,y,z) = sin(y) sin(z) x cos(y) sin(z) −x sin(y) cos(z) z cos(y) −xz sin(y) x cos(y) 0 −e 1 This implies that Jf(1, π2 ,π) = 0 0 1. The determinant of this matrix is 0, which proves that f does not 0 0 0 π admit a dierentiable inverse g near f (1, , π) = (π, 0, 0). 2