Homework 11: Functions of Several Variables I Name Due on Tuesday, in class. Submit your solutions (work and answers) on this page only! Let z = f (x, y) = 4 − x2 − y 2 .

(1) Sketch the graph of the function. (Hint: first square both sides, like in class)

(2) Find and sketch the domain of f .

(3) Find and sketch the contours f (x, y) = c for c = −1, 0, 2, 4, 5, if they exist.

(4) Find and sketch the domain of g(x, y) = ln(4 − x2 − y 2 ).

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Homework 12: Multivariable Functions II: Limits and Continuity Name Due on Tuesday, in class. Submit your solutions (work and answers) on this page only! (1) Find lim(x,y)→(1,3) xy . x2 +y 2

(2) Find lim (x,y)→(1,1) x=y x2 −y 2 x−y

(hint: factor)

(3) Find lim (x,y)→(2,0)
2x−y=4

√

2x−y−2 2x−y−4

(hint: conjugate)

(4) Show that lim(x,y)→(0,0) and C3 {y = x2 }.

2x4 −3y 2 x4 +y 2

does not exist by finding the limit along the three paths: C1 {x = 0}, C2 {y = 0}

(5) Show that lim(x,y)→(0,0) cos

2x4 y x4 +y 4

=1

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Homework 13: Multivariable Functions III: Partial Derivatives Name Due at the beginning of our next class period. Submit your solutions (work and answers) on this page only! (1) Compute all first and second order partial derivatives of f (x, y) = x3 y 4 + ln( x ). y

(2) Find the equation of the tangent plane to the graph of the function z = f (x, y) = exp(1 − x2 + y 2 ) at (x, y) = (0, 0). Convert to normal form.

(3) Find the equation of the tangent plane to the surface r(u, v) = u3 −v 3 , u+v +1, u2 at (u, v) = (2, 1). Convert to normal form.

(4) Suppose that fx (x, y) = 6xy + y 2 and fy (x, y) = 3x2 + 2xy. Compute fxy and fyx to determine if there is a function f (x, y) with these first derivatives. If so, integrate to find such a function.

(5) Show that the function u(x, y) = ln(

x2 + y 2 ) is Harmonic (i.e., it satisfies Laplaces equation uxx + uyy = 0).

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Homework 14: Multivariable Functions IV: Chain Rule Name Due on Tuesday. Submit your solutions (work and answers) on this page only! (1) Show that u = r2 cos(2θ) satisfies Laplace’s equation in polar coordinates: urr + 1 ur + r
1 u r 2 θθ

= 0.

(2) Sand is falling onto a canonical pile of volume V = πr2 h/3. Suppose that when the height is 5m and the base radius is r=2m, the height is increasing at .4 m/s and the base radius is increasing at .7 m/s. At what rate is the volume increasing?

(3) A particle is moving on the surface r(u, v) = u2 − v 2 , 2uv, u2 + v 2 along the curve with with u(t) = t2 and v(t) = t3 . Use the chain rule to find the velocity of the particle at t = −2.

√ (4) Suppose that w = ln(x2 + y 2 + z 2 ) with x = s − t, y = s + t and z = 2 st. Find

∂w ∂s

and

∂w ∂t .

(5) (Implicit Differentiation) Find zx and zy when x2 + 2y 2 = 3z 2 + 4xyz + 8.

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Homework 15: Multivariable Functions V: Directional Derivatives and the Gradient Submit your solutions (work and answers) on this page only!

Name

(1) Find the gradient of f (x, y, z) = ex sin(y) + ey sin(z) + ez sin(x) at P (0, 0, 0). Then, find the directional derivative of f at P in the direction of 1, −1, 2 .

(2) Suppose that the temperature in a solid is given by f (x, y, z) = xy 2 z 3 . Suppose that a bug is at the point (2, 2, 2) in the solid. In what direction should the bug move to cool down as quick as possible? In what direction might the bug move to remain the same temperature?

(3) Find the equation of the tangent plane to xyz + x2 + z 3 = 14 + 2y 2 at the point P (5, −2, 3).

(4) Show that the surfaces z = xy and z = 3 x2 − y 2 intersect perpendicularly at the point (2, 1). 4

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Homework 16: Multivariable Functions VI: Extrema Name Submit your solutions (work and answers) on this page only! For the following problems, let f (x, y) = 8 x y − 2 x2 − y 4 . (1) Find the linear and quadratic approximations L(x, y) and Q(x, y) to f (x, y) at the point (2, 1). Do you know what type of surface Q(x, y) is?

(2) Use the linear and quadratic approximations from problem 1 to estimate the value of f at (2.1, 0.9), and compare these to the exact value. That is, compute L(2.1, 0.9) and Q(2.1, 0.9), then compare these to f (2.1, 0.9) (using your calculator).

(3) The critical points of f (x, y) are (0, 0), (4, 2), (−4, −2), as you can verify. Use the max/min/saddle tests to classify these critical points as maximum, minimum or saddle.

(4) Find the quadratic approximation Q(x, y) to f (x, y) at the critical point (4, 2), and determine if its graph is an elliptic paraboloid opening up, an elliptic paraboloid opening down, or a hyperbolic paraboloid.

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Answers
(1.1) (1.2) (1.3) (1.4) (1.5) (1.6) (2.1) (2.2) (2.3) (2.4) (2.5) (3.1) (3.2) (3.3) (3.4) Circle x2 + z 2 = 3 in the xz plane. Unit box in the first quadrant. 0√ x, y, z ≤ 2 ≤ 5 2 (x2 + (y + 7)2 + z 2 = 49 √ Center (0, −1/3, 1/3), Radius 29/3 −4, −2, 5 = −4i − 2j + 5k 9 2 6 11 11 , − 11 , 11 3 Direction: 7 , − 6 , 2 ; Midpoint: (2.5, 1, 6) 7 7 −→ − − → 1 1− M N = N − M = A+C − A+B = 2 (C − B) = 2 BC 2 2 o o Horiz: 20 cos(23 ), Vert: 20 sin(23 ) 0, 3, 4 , 10, 8, −6 48.19o or 0.84107 radians 56.31o , 90o , 123.69o √ |a + b| = (a + b) · (a + b) = a · a + 2a · b + b · b = √ √a · a + b · b since a · b = 0. Likewise |a − b| = a · a + b · b. 3.1429 −1 cα = √2 , cβ = √14 , cγ = √3 14 14 −17, −1, 11 √
3 2

(3.5) (4.1) (4.2) (4.3) (4.4) (4.5) (5.1) (5.2) (5.3) (5.4) (5.5) (5.6) (6.1) (6.2) (6.3) (6.4) (6.5) (6.6) (7.1) (7.2) (7.3) (7.4)

√ (7.5) 3 3 (8.1) 3/2 √ √ √ 8s+π 2 8s+π 2 8s+π 2 sin + , (8.2) cos √ 2 √ 2 √ 2 8s+π 2 8s+π 2 8s+π 2 sin cos − 2 2 2 (8.3) 1/t √ t + (9.1) κ(t) = 2/(e√ e−t )2 √ (9.2) 2/3 3, −2/3 3 (10.1) 1, t, t2 /2 (10.2) 0, 1, t (10.3) 0, 0, 1 (10.4) t2 /2, −t, 1 (10.5) (t2 + 2)/2 (10.6) 1 + t2 /2 (10.7) t + t3 /2 (10.8) 4/(t2 + 2)2 (10.9) 4/(t2 + 2)2 (10.10) t (10.11) 1 2 (10.12) t22 , t22t , t2t+2 +2 +2 (10.13) (10.14) (10.15) (12.1) (12.2) (12.3) (12.4) (12.5) (13.1) (13.2) (13.3) (13.4) (13.5) (14.1) (14.2) (14.3) (14.4) (14.5) t2 , − t22t , t22 t2 +2 +2 +2 2 2t − t2 +2 , 2−t , t22t t2 +2 +2 2 t t22 , t22t , t2t+2 +2 +2

3 49 37 23 − 4299 , 4299 , − 4299 1 − 2t, 2 + 2t, 3 + 3t 7/5, −14/5, 11/5 sketch 0.93027 rad or 53.3o 3x + y − 3z = 1. √ 16/√195 34/ 212 √ 14 2/ √ 1/2 3 −t, −16 + 3t, −13 + 2t (starting point may be different) √ √ √ √ (1 + 2, 2), (1 − 2, − 2) r(t) = 2, −15t, t + π/6 r(t) = t3 + 3, − 5 t2 + 4t + 4, t4 − 5t 2 π/4, ln(2), ln(2)/2

265/29 ≈ 3.0229

3/10 2 1/4 −3, 2, −1/2 2x4 y ≤ 2|y| .. x4 +y 4

+ − t22t , +2

2−t2 , 2t t2 +2 t2 +2

= 0, 1, t

z=e −4x − 12y + 15z = −61

5.2π ≈ 16.336 160, 160, −224

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