Solutions to the Practice Test for Midterm 3 You must show all your ...

Solutions to the Practice Test for Midterm 3You must show all your work. The number of points earned on eachproblem will be determined by how well you have justified your work.1. Find the second partial derivatives of fÝx,yÞ = e x y 2 ? x 2 y + xlny.f x Ýx,y,zÞ = e x y 2 ? 2xy + lnyf y Ýx,y,zÞ = 2e x y ? x 2 + x yf xx Ýx,y,zÞ = e x y 2 ? 2yf yy Ýx,y,zÞ = 2e x ? xy 2f xy Ýx,y,zÞ = 2e x y ? 2x + 1 y2. Find an equation of the tangent plane to the surface r3Ýu,vÞ = 3u i + uv 2 j ? v 3 k at the pointÝ?6,?2,?1Þ.We first look for the direction numbers of the normal vector by finding the tangent vectors:r3 u = 3 i + v 2 jr3 v = 2uv j ? 3v 2 kThe normal is the cross product of these two vectors:i j k3 v 2 0 = ?3v 4 i + 9v 2 j + 6uvk0 2uv ?3v 2The point Ý?6,?2,?1Þ occurs when u = ?2,v = 1. So the normal vector at the point is-3 i + 9 j ? 12kTherefore, the tangent plane has the equation ?3Ýx + 6Þ + 9Ýy + 2Þ ? 12Ýz + 1Þ =? 3x ? 12 + 9y ? 12z = 03. Let xsinÝy + zÞ + ycosÝx + zÞ + ze x = 0. Find /z /z, and ./x /yWe know that /z/x= ? /F//x/F//z, and/z/y= ? /F//y/F//z ./F//x = sinÝy + zÞ ? ysinÝx + zÞ + ze x/F//y = xsinÝy + zÞ + cosÝx + yÞ/F//z = xcosÝy + zÞ ? ysinÝx + zÞ + e x/z= ? sinÝy+zÞ?ysinÝx+zÞ+zex/x xcosÝy+zÞ?ysinÝx+zÞ+e x/z= ? xsinÝy+zÞ+cosÝx+yÞ/y xcosÝy+zÞ?ysinÝx+zÞ+e x4. Find the maximum rate of change of fÝx,yÞ = x 3 siny ? tany at the point Ý1,2Þ. In whichdirection does it occur?The maximum value of the directional derivative is |4fÝx,yÞ| =|Ö3x 2 siny,x 3 cosy ? sec 2 y×|.At the pointÝ1,2Þ,|4fÝ1,2Þ| = |Ö3sin2,cos2 ? sec 2 1×| = 9sin 2 2 + cos 2 2 ? 2cos2sec 2 1 + sec 4 2 = 6.6189

The maximum rate of change occurs in the direction of4fÝx,yÞ = Ö3sin2,cos2 ? sec 2 2× r Ö2.7279,?6.1905×5. Find the absolute maximum and minimum values of fÝx,yÞ = x 2 y + x 2 y 2 ? xy in the closedtriangular region D with vertices Ý?3,0Þ,Ý3,0Þ, and Ý0,3Þ.We first find the critical points by setting the first partials equal to 0.f x Ýx,yÞ = 2xy + 2xy 2 ? y = yÝ2x + 2xy ? 1Þ = 0 ì y = 0 or 2x + 2xy ? 1 = 0f y Ýx,yÞ = x 2 + 2x 2 y ? x = xÝx + 2xy ? 1Þ ì x = 0 or x + 2xy ? 1 = 0We see immedieately that Ý0,0Þ is a critical point.If y = 0 and x + 2xy ? 1 = 0 then x + 2xÝ0Þ ? 1 = x ? 1 = 0 ì x = 1, and Ý1, 0Þ is a criticalpoint.If 2x + 2xy ? 1 = 0 then x ® 0.If 2x + 2xy ? 1 = 0 and x + 2xy ? 1 = 0, then x = 2x ì x = 0 which is not possible.Therefore, there are two critical points:Ý0,0Þ ,Ý1,0Þ.fÝ0,0Þ = 0,fÝ1,0Þ = 0We now find the value of the function along the edges of D.L 1 := áÝx,yÞ| ? 3 ² x ² 0,y = 0â ì fÝx,0Þ = 0.L 2 := áÝx,yÞ|y = ?x + 3,0 ² x ² 3â ì fÝx,?x + 3Þ = Ý?x + 3ÞÝx 2 + x 2 Ý?x + 3Þ ? xÞ =? 7x 3 + x 4 + 13x 2 ? 3xWe check for the extreme values of this function by taking the derivative:f v Ýx,?x + 3Þ = 4x 3 ? 21x 2 + 26x ? 3 = 0, Solution is : áx = .12837â,áx = 1.7151â,áx = 3.4065â ì Ý.12837,2.8716Þ,Ý1.7151,1.2849Þ,Ý3.4065,?.4065Þ will yield any extremevalues:fÝ.12837,2.8716Þ = .12837 2 2.8716 + .12837 2 2.8716 2 ? .12837Ý2.8716Þ = ? .18542fÝ1.7151,1.2849Þ = 1.7151 2 1.2849 + 1.7151 2 Ý1.2849Þ 2 ? 1.7151Ý1.2849Þ = 6.4323Ý3.4065,?.4065Þ is not in the triangular region.L 3 := áÝx,yÞ|y = x + 3,?3 ² x ² 0â ì fÝx,x + 3Þ = Ýx + 3ÞÝx 2 + x 2 Ýx + 3Þ ? xÞ =7x 3 + x 4 + 11x 2 ? 3xWe check for the extreme values of this function by taking the derivative:f v Ýx,x + 3Þ = 4x 3 + 21x 2 + 22x ? 3 = 0, Solution is : áx = ?3.7153â,áx = ?1.6566â,áx = .12186â, Solution is : áx = ?3.7153â,áx = ?1.6566â,áx = .12186â ì Ý?3.7153, ? .7153Þ,Ý?1.6566,1.3434Þ,Ý.12186,3.12186Þ will yield any extreme values:?3.7153, ? .7153 is not in the triangular regionfÝ?1.6566,1.3434Þ = ?1.6566 2 1.3434 + 1.6566 2 Ý1.3434Þ 2 ? ?1.6566Ý1.3434Þ = 3.491 5Ý.12186,3.12186Þ is not in the triangular regionThe absolute minimum is fÝ.12837,2.8716Þ = ? .18542, andthe absolute maximum is fÝ1.7151,1.2849Þ = 6.4323.6. Use Lagrange multipliers to find the maximum and minimum values of f subject to the givenconstraint.fÝx,y,zÞ = x 2 + y 2 + z 2 ; x ? y + z = 12x = V2y = ?V2z = VBy the first and third equations we have x = zBy the first and second equations we have y = ?xUsing the constraint x ? y + z = 1 we have x + x + x = 1 ì x = 1/3, z = 1/3,y = ?1/3

fÝ1/3,1/3,1/3Þ = 1/3 2 + 1/3 2 + 1/3 2 = 1 . 3Choosing any Ýx,y,zÞ such that x ? y + z = 1, such as Ý1,0,0Þ ì fÝ1,0,0Þ = 1, we can deducefÝ1/3,1/3,1/3Þ = 1 is a minimum value.3