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MATH 2400 Homework 4 - Due Friday, June 29 Summer 2018 In these problems you should show all of your work in complete mathematical “sentences”, writing complete English sentences when you explain your logic. Please always staple your homework and label it with your section number. Late homework will not be accepted. 1. Verify the conclusion of Clairaut’s Theorem (uxy = uyx ) for u = cot(x2 + 2y). 2. Use implicit differentiation to find ∂z ∂x and ∂z ∂y for cos(xyz) = x − y − z. 3. Determine which functions solve Laplace’s equation uxx + uyy = 0: (a) u = x2 + y 2 (b) u = x2 − y 2 (c) u = x3 + 2xy 2 (d) u = ln p x2 + y 2 (e) u = e−x cos y − e−y cos x. 2y 4. Determine for which values of (x, y) the function f (x, y) = √2x2 x +y 2 is continuous. 5. Find an equation of the tangent plane to the surface z = xexy at the point (2, 0, 2). 6. Find an equation of the tangent plane to the parametric surface hu2 , v 2 , z = uvi when u = 1 and v = 2. 7. Suppose you need to know an equation of the tangent plane to a surface S at the point P (2, 1, 3). You don’t have an equation for S but you know the curves r1 (t) = h2 + 3t, 1 − t2 , 3 − 4t + t2 i r2 (u) = h1 + u2 , 2u3 − 1, 2u + 1i both lie on S. Find an equation of the tangent plane at P . 8. List all the ways in which the functions f (x, y) = x2 + y 2 , g(t) = ht − 1, et i, and k(a, b) = hb, ai can be composed. 9. Use the chain rule to find the indicated partial derivatives. u = x2 + yz, x = pr cos θ, ∂u ∂u ∂u y = pr sin θ, z = p + r; , , when p = 2, r = 3, θ = π. ∂p ∂r ∂θ

Accounting_Teacher
School: University of Maryland

Hi, here are the solutions :). if you have questions, let me know :)

MATH 2400

Homework 4 - Due Friday, June 29

Summer 2018

In these problems you should show all of your work in complete mathematical “sentences”, writing
label it with your section number. Late homework will not be accepted.
1. Verify the conclusion of Clairaut’s Theorem (uxy = u yx) for u = cot(x2 + 2y).
uxy = ∂
∂y

∂u
∂x

∂u
∂x

===-

∂x

2(x2+y2)
2(x2+y2)*∂/∂x

(x2 +y2)
2(x2+y2)(2x +∂)

= ∂u
∂x

= -2x

∂ ∂u
∂y ∂x

2

(x2 +y2)

∂y

= -2x(2

-2x

2-1(x2+y2)

= -6x

∂y

2(x2+y2)

∂y

-2x

2(x2+y2)

(x2+y2)*cot(x2+y2)) ∂
∂y
* cot(x2+2y)

cot(x2+y2)

2(x2+y2)

∂ ∂
∂x ∂y

2(x2+y2)

* ∂
∂y

(x2+y2) (-

uxy= -8x

== -2

cot(x2+y2)

- (2)

2(x2+y2)

∂x

= (-2) * (2

-2

2(x2+y2))

2-1 (x2+y2)

* ∂
∂x

(x2+y2))

(x2+y2)

= -6

(x2+y2) (-

(x2+y2) * cot(x2+y2)*

∂x

(x2+y2)

2(x2+y2) *cot(x2+y2)(2x+0)
=-4
=-8x 2(x2+y2...

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Anonymous
Outstanding Job!!!!

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