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Module MA2E02 (Frolov), Multivariable Calculus, 2017 Homework Sheet 3 Due: by 12:00 on the first Friday after the corresponding tutorial session Name and student number: 1. Consider the function p f (x, y, z) = −2 + 2y + x4 + yz − sin(3y − 2z) , and the point P = (−1, 2, 3) . (a) Calculate f (−1, 2, 3), fx (−1, 2, 3), fy (−1, 2, 3), fz (−1, 2, 3). (b) Find a unit vector in the direction in which f increases most rapidly at the point P . (c) Sketch the projection of the vector onto the xz-plane (d) Find a unit vector in the direction in which f decreases most rapidly at the point P. (e) Sketch the projection of the vector onto the xy-plane (f) Find the rate of change of f at the point P in these directions. Show the details of your work. Solution: (a) We find f (−1, 2, 3) = p −2 + 2y + x4 + yz − sin(3y − 2z) = 3 , P fx (x, y, z) = fy (x, y, z) = −2 cos(2x − 3z) 2f (x, y, z) 2 + z − 3 cos(3y − 2z) 2f (x, y, z) fz (x, y, z) = y + 2 cos(2z − 3x) 2f (x, y, z) 2 fx (−1, 2, 3) = − , 3 ⇒ ⇒ ⇒ fy (−1, 2, 3) = fz (−1, 2, 3) = 1 , 3 2 . 3 (b) f increases most rapidly in the direction of its gradient. The gradient and its magnitude are equal to 2 1 2 ∇f (−1, 2, 3) = − , , , ||∇f (−1, 2, 3)|| = 1 . 3 3 3 Therefore, the unit vector in the direction of the gradient is ∇f (−1, 2, 3) 2 1 2 u= = − , , ≈ (−0.666667, 0.333333, 0.666667) . ||∇f (−1, 2, 3)|| 3 3 3 1 z 0.7 0.6 0.5 0.4 0.3 0.2 0.1 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 x -0.1 (c) The projection of the vector u onto the xz-plane is the vector 2 2 uxz = − , 0 , ≈ (−0.666667, 0, 0.666667) , 3 3 see the picture. (d) f decreases most rapidly in the direction opposite to its gradient, so the unit vector is 2 1 2 ∇f (−1, 2, 3) = ,− ,− v=− ≈ (0.666667, −0.333333, −0.666667) . ||∇f (−1, 2, 3)|| 3 3 3 (e) The projection of the vector v onto the xy-plane is the vector 1 2 , − , 0 ≈ (0.666667, −0.333333, 0) . vxy = 3 3 It is shown below y 0.1 0.2 0.3 0.4 0.5 0.6 0.7 x -0.05 -0.10 -0.15 -0.20 -0.25 -0.30 -0.35 (f) The rate of change of f at P in the direction of u is equal to ||∇f (−1, 2, 3)|| = 1 , and the rate of change of f at P in the direction of v is equal to −||∇f (−1, 2, 3)|| = −1 . 2 2. Let r = p x2 + y 2 . (a) Show that ∇r = r , r where r = x i + y j . Solution: We have ∂r 2x x = p = , ∂x r 2 x2 + y 2 ∂r 2y y = p = , ∂y r 2 x2 + y 2 which proves the formula. (b) Show that ∇f (r) = f 0 (r)∇r = f 0 (r) r. r Solution: We use the chain rule to get ∂f (r) ∂r x = f 0 (r) = f 0 (r) , ∂x ∂x r ∂f (r) ∂r y = f 0 (r) = f 0 (r) , ∂y ∂y r which proves the formula. 3. Consider the surface p 13 3 2 2x − y + 10 z = f (x, y) = ln 2 (a) Calculate z0 ≡ f (1, −2), fx (1, −2), fy (1, −2). (b) Find an equation for the tangent plane to the surface at the point P = (1, −2, z0 ) where z0 = f (1, −2). (c) Find points of intersection of the tangent plane with the x-, y- and z-axes. (d) Sketch the tangent plane. (e) Find parametric equations for the normal line to the surface at the point P (1, −2, z0 ). (f) Sketch the normal line to the surface at the point P (1, −2, z0 ). Show the details of your work. Solution: (a) We first simplify p 13 3 1 2 z = ln 2x − y + 10 = ln(2x3 − y 2 + 10) − ln 2 . 2 3 and compute z0 1 ln 2 − 4 + 10 − ln 2 = 0 . 3 Then, we compute the partial derivatives at P (1, −2, 0) z0 = z|x=1,y=−2 = ∂ 1 6x2 z= ∂x 3 2x3 − y 2 + 10 3 ⇒ ∂ 1 z|x=1,y=−2 = . ∂x 4 ∂ 1 −2y z= 3 ∂y 3 2x − y 2 + 10 ⇒ ∂ 1 z|x=1,y=−2 = . ∂y 6 (b) The tangent plane equation is given by 1 1 1 1 1 z = 0 + (x − 1) + (y + 2) = x + y + . 4 6 4 6 12 1 (c) (− 31 , 0, 0) = (−0.33333, 0, 0), (0, − 21 , 0) , (0, 0, 12 ) ≈ (0, 0, 0.08333) (d) The tangent plane is the one through the points in (c). (e) The normal line to the surface (and the tangent plane) is given by 1 1 r = i − 2j + t − i − j + k . 4 6 (f) The normal line is perpendicular to the plane. 4. Consider the function f (x, y) = x4 − 2x2 y + 2y 2 − 2y − 5 Locate all relative maxima, relative minima, and saddle points, if any. Solution: We first find all critical points fx (x, y) = 4x3 − 4xy = 0 , fy (x, y) = −2x2 + 4y − 2 = 0 . From the second equation we find y in terms of x x2 1 + , y= 2 2 and substituting it to the first equation, we derive the following equation for x 2x − 2x3 = 0 . There are three solutions to this equation x = 0 , x = −1 , x = 1 , and, therefore, three critical points 1 (x = 0 , y = ) , 2 (x = −1 , y = 1) , (x = 1 , y = 1) . Computing the values of f at critical points, we get 1 11 f (0, ) = − , 2 2 f (−1, 1) = −6 , 4 f (1, 1) = −6 . To find out if they are maximum, minimum or saddle points we use the second derivative test. To this end we compute ∂ 2f ∂ 2f ∂ 2f 2 (x, y) = −4x , (x, y) = 12x − 4y , (x, y) = 4 , ∂x2 ∂y 2 ∂x∂y and ∂ 2f ∂ 2f D(x, y) = − ∂x2 ∂y 2 Computing D and ∂2f ∂x2 ∂ 2f ∂x∂y 2 = 32x2 − 16y , for the three critical points, we get 1 D(0, ) = −8 , 2 ∂ 2f 1 (0, ) = −2 , 2 ∂x 2 and therefore (0 , 12 ) is a saddle point. D(−1, 1) = 16 , ∂ 2f (−1, 1) = 8 , ∂x2 and therefore (−1, 1) is a relative minimum. D(1, 1) = 16 , ∂ 2f (1, 2) = 8 , ∂x2 and therefore (1, 1) is a relative minimum too. The graph of the function is shown below 5