2419Su2011Exam2.pdf

July 28, 2011

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Notation. Any of the three variants of vector notations may be used, all of the following are the same vector: (1,2,3),h1,2,3i,ı+ 2+ 3k.

Instructions.

• All work must be shown andexactanswers are expected. You are allowed calculators, but you should only use these to check your work not to perform your work. For example,sin(2π/3) =√3/2 will be accepted,sin(π/4) =.70...will not.

• You are allowed a single standard81₂ ×11piece of notebook paper.

• Write your final answer in the answer box provided.

Problem 1 (10 pts). Consider f(x, y) =px2+y.

(a) On what subset of_R2 is f continuous? Sketch the region, in thexy-plane.

This function is continuous on its domain (as is all elementary functions). So the answer is just the domain

dom(f) ={(x, y) :x2+y≥0} This is the exterior of as well as the parabola

y=−x2_.

(b) Find all1st and 2nd order partial derivatives of f(x, y)

fx= (2x)(1/2)(x2+y)−1/2 =

x p

x2_+y

fy = (1/2)(x2+y)−1/2 =

1

2px2_+y

fxx= (x2+y)−1/2+ (x)(−1/2)(2x)(x2+y)−3/2= (x2+y)−3/2((x2+y)−x2) =y(x2+y)−3/2

fyy= (−1/4)(x2+y)−3/2

fyx=x(−1/2)(x2+y)−3/2

fxy = (1/2)(−1/2)(2x)(x2+y)−3/2

Problem 2 (10 pts). Compute the following limits if they exist, otherwise show that they do not exist.

(a) lim

(x,y)→(0,0)

y4_−x2

x+y2

Here we have y

4_−x2

y2_+x =y

2_{−x forx6=−y}2 _{and so lim} (x,y)→(0,0)y

2_{−x= 0.}

(b) lim

(x,y)→(0,0)

y4−x2 x2_+y2

Problem 3 (10 pts). Find the equation of the tangent plane to f(x, y) = arctan xy2+x

at (1,−1). Use this to approximatef(.98,−1.1).

We have∇f = 1

1+(xy+x

2 )

2 2(y+ 1),2x

so∇f(1,−1) = 0,1₂

f(x, y)≈f(1,−1) +∇f(1,1)(x−1, y+ 1) = arctan(0) +

0,1

2

(x−1, y+ 1) = 1 2(y+ 1)

The linear approximation (tangent plane) isz= 1₂(y+ 1)andf(.98,−1.1) = 1₂(−1.1 + 1) =

Problem 4 (10 pts). Theideal gas law is

pV =nRT

where p = pressure, V = Volume, T = absolute temperature, R = ideal gas constant, n=number of moles of substance. Assumen, V, andT are variables (Ris constant approx 8.314472 J/K mol) sop=f(n, V, T).

(a) Find the total differential dp

dp=∇p(dn, dV, dT) =

∂p ∂n, ∂p ∂V , ∂p ∂T

(dn, dV, dT)

=

RT

V ,− nRT

V2 ,

nR V

(dn, dV, dT) = RT

V dn− nRT

V2 dV +

nR V dT

(b) A gaseous substance is in a cubicle container of dimensions 2 m on each side, the temperature inside is measured at 90 K, and the amount of substance measured at40

mol. If the error in measurement of temperature is±3K, and error in moles of substance ±0.004mol. What is the approximate maximal error in computed pressure? Since the container is fixed there is no error in measurement in volume. (Take R = 8.314472. You should give a numeric answer, but please show your work for full/partial credit.)

The box is fixed and has no error in measurement so dV = 0 and we have: The approximate error is bounded by

90R

8 (0.004)−

(40)(90)R

64 (0) +

40R

Problem 5 (10 pts). Consider

f(x, y, z) = ln(xy2+yz+xz)

(a) In what direction isf increasing most rapidly at (1,1,1)? (Remember a direction is a unit vector!)

The gradient is

∇f =

y2+z xy2_+yz+xz,

2xy+z xy2_+yz+xz,

x+y xy2_+yz+xz

= 1

xy2_+yz+xz(y

2_{+z,2xy+z, x+y)}

and hence

∇f(1,1,1) = 1

3(2,3,2)

So the direction in whichfis increasing most rapidly is ∇f(1,1,1) k∇f(1,1,1)k =

1

3(2,3,2)

₁ 3 √ 17 = 1 √

17(2,3,2).

(b) What is the rate of increase off at (1,1,1)in the direction you found in (a)? (This is the maximal rate of increase off in any direction.)

This is justk∇f(1,1,1)k= √

17

3 which was calculated in (a).

(c) Find the directional derivativeDu(f)(1,1,1)in the the direction of v= (−2,2,1).

u= _kv_vk = 1

3(−2,2,1)and

Du(f)(1,1,1) =∇f(1,1,1)u=

1

3(2,3,2) 1

Problem 6 (10 pts). Let h(x, y) = xy2−x2−y2. Find all critical points and classify them as local max/local min/saddle point by using the2nd-partials test.

First we must find where∇h(x, y) =hy2_{−2x,2xy−2yi=0. Ify 6= 0, then we getx = 1}

from 2xy+ 2y= 0 and then from y2 = 2x we get y =±√2. If y = 0, then from 2x =y2 we getx= 0. So the critical points are (0,0),(1,√2), and (1,−√2).

LetH(x, y) =hxxhyy−(hxy)2 =−4(x−1)−4y2, then

H(0,0) = 4andfxx =−2<0 so(0,0)is a local maximum

Problem 7 (10 pts). Use the method of Lagrange multipliers to find the minimum and maximum value ofh(x, y) =xy2−x2−y2 on the unit circle x2+y2= 1.

Findx, y such that for some λ

y2−2x= 2λx

2xy−2y = 2λy

Ifx6= 0 andy 6= 0, then

λ= 2

y2

x

−1

λ=x−1

so x= y2 2x and hence

2x2 =y2

which together with y2 = 1−x2 gives 2x2= 1−x2 and sox2= 1₃ and this givesy2= 2/3

so we have the four points pointsn±_√1 3,±

q

2 3

o .

So the maximum value h takes on the unit circle is 2

3√3 −1 at

n

1

√

3,±

q

2 3

o

and the

minimum value is− 2

3√3 −1at

n −_√1

3,±

q

2 3

Problem 8 (10 pts). Use the Chain Rule to find ∂F ∂u,

∂F

∂v whenu= 3, v=−1.:

F =xe(y−z2), x= 2uv, y=u−v, z =u+v

We have

Fx =e(y−z

2₎

Fy =xe(y−z

2₎

Fz =−2zxe(y−z

2₎

xu = 2v yu= 1 zu = 1

xv = 2u yv =−1 zv = 1

∂F

∂u =Fxxu+Fyyu+Fzzu=e

(y−z2₎

(2v+x−2zx)

∂F

∂v =Fxxv+Fyyv+Fzzv =e

(y−z2₎

(2u−x−2zx)

Sincex(3,−1) =−6,y(3,−1) = 4, and z(3,−1) = 2we have

∂F ∂u

(u,v)=(3,−1)

=e0(−2−6−2(2)(−6)) = 16

∂F ∂v

(u,v)=(3,−1)

Problem 9 (10pts). Supposew=f(x, y) is differentiable andx=u−v andy =v−u.

Use the chain rule to show ∂w ∂u +

∂w ∂v = 0.

∂f ∂u =fx

∂x ∂u +fy

∂y

∂u =fx−fy ∂f

∂v =fx ∂x ∂v +fy

∂y

∂v =−fx+fy

So

∂f ∂u+

∂f

Problem 10 (10 pts). Find the direction for the tangent line to the curve of intersection of the two level surfaces f(x, y, z) =x2+xy+xz= 3and g(x, y, z) = ln(x2+y−z2) = 0

at (1,1,1).

Compute the cross product of the two gradients:

∇f = (2x+y+z, x, x) ∇f(1,1,1) = (4,1,1)

∇g= 1

x2_+y−z2(2x,1,−2z) ∇g(1,1,1) = (2,1,−2)

The direction of the tangent line is determined by

(4,1,1)×(2,1,−2) = (−3,10,2)

Problem 11 (10 pts). Find ∂z

∂x(1,1)when x

2_{y+yz+xz= 6}

At(x, y) = (1,1)we have1 + 2z= 6 soz= 5/2

∂z ∂x =−

Fx

Fz

=−2xy+z y+x

so

∂z

∂x(1,1) =−

2 + 5/2

2 =−