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Dr. Fazarmirad
Dr. Ketchersid
M
T
W
R
Notation. Three variants of vector notation are used, all of the following are the same vector: (1,2,3),h1,2,3i,i+ 2j+ 3k.
Problem 1 (10 pts). A polar curve is given by r= 1 + cos(θ)
(a) Find dxdy.
We havex= (1 +cos(θ)) cos(θ) andy = (1 + cos(θ)) sin(θ)
dx
dθ =−sin(θ)−2 cos(θ) sin(θ) =−sin(θ)−sin(2θ) dy
dθ = cos(θ) + cos
2(θ)−sin2(θ) = cos(θ) + cos(2θ)
dy dx =
cos(θ) + cos2(θ)−sin2(θ) −sin(θ)−2 cos(θ) sin(θ) =
cos(θ) + cos(2θ) −sin(θ)−sin(2θ)
(b) At what points(x, y)is the tangent line vertical? We must find where dx/dθ= 0 whiledy/dθ6= 0.
dx
dθ = 0⇐⇒ −sin(θ)(1 + 2 cos(θ)) = 0
⇐⇒θ= 0, π or θ= 2π/3.4π/3
At θ=π,dy/dθ= 0 so this one must be considered carefully. (there is a cusp here not a vertical tangent line). so vertical tangents occur atθ ∈ {0,2π/3,4π/3} the corresponding points are{(0,2),(−1/4,√3/4),(−1/4,−√3/4)}.
(b) What is the arc length of the curve on[0,2π]?
Z 2π
0
s r2+
dr dθ
2 dθ=
Z 2π
0
p
(1 + cos(θ))2+ (−sin(θ))2dθ
=
Z 2π
0
p
2 cos(θ) + 2dθ=√2
Z 2π
0
p
1 + cos(θ)dθ
This received 4 points. The integral could also be set up as Z 2π
0 dx dθ, dy dθ dθ=
Z 2π
0 s dx dθ 2 + dy dθ 2 dθ
I’ll leave it to you to see that this reduces to the same as above. To finish you need Z
p
1 + cos(θ)dθ
This can be done by setting1 + cos(θ) = cos2 θ2+ sin2 θ2+ cos2 θ2−sin2 θ2= cos2 θ2
or by using the double angle formulacos2 θ2= 12(cos(θ) + 1). Either way leads to Z
p
1 + cos(θ)dθ√2 =
Z 2π
0 cos θ 2
dθ= 2√2
Z π 0 cos θ 2
dθ= 4√2 sin
θ 2 π 0
Exercise 2 (20 pts). Consider the curve r(t) =h2 sin(t),2 cos(t),sin(2t)i.
(a) [15 pts] Decompose accelerationddt22r att= π4 into its normal and tangential components. (Make sure that the following are clear,T,N,aT, andaN amd writer00(t) =aTT+aNN.)
dr
dt = (2 cos (t),−2 sin (t),2 cos (2t)) d2r
dt2 = (−2 sin (t),−2 cos (t),−4 sin (2t))
dr
dt(π/4) = (
√
2,−√2,0)
d2r
dt2(π/4) = (− √
2,−√2,−4)
Since ddt22r(π/4)and
dr
dt(π/4)are already orthogonal we have
T(π/4) = 1 2(
√
2,−√2,0)
N(π/4) = 1 2√5(−
√
2,−√2,−4)
aT(π/4) = 0
aN = 2
√ 5
So
d2r
dt2(π/4) = 0T(π/4) + 2 √
5N
(b) [5 pts] Compute the curvature at π4,κ π4.
dr
dt(π/4)× d2r
dt2(π/4) = 4(
p
(2),p(2),1)
dr
dt(π/4)× d2r
dt2(π/4)
= 4√5
κ(π/4) = 4 √
5 23 =
Problem 3 ( 20 pt). Consider
f(x, y, z) = ln(xy2+yz+xz)
(a)[10 pts] 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 whichf is 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)[5 pts] What is the rate of increase of f 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) [5 pts] Find the directional derivative of Du(f)(1,1,1) in the the direction of v = (−2,2,1).
u= kvvk = 1
3(−2,2,1)and
Du(f)(1,1,1) =∇f(1,1,1)u=
1
3(2,3,2) 1
Problem 4 (15 pts). Consider f(x, y) =px2+y. (a) [5 pts] On what subset ofR2 is f continuous.
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 parabolay=−x2.
(b) [10 pts] 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
Exercise 5 (15 pts). The ideal 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) [10 pts] Find the total differentialdp
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) [5 pts] A gaseous substance is in a cubicle container of dimensions2m on each side, the temperature inside is measured at90 K, and the amount of substance measured at40 mol. If the error in measurement of temperature is±3K, and error in moles of substance±0.004
mol. What is the approximate maximal error in computed pressure? Since the container is fixed there is no error in measurement in volume. (TakeR = 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 sodV = 0and we have: The approximate error is bounded by
90R
8 (0.004)−
(40)(90)R
64 (0) +
40R
Exercise 6 (15 pts). Find the equation of the tangent plane to f(x, y) = arctan xy2+x at (1,1).
We have∇f =
1
x−3y,
−3
x−3y
= x−31 y(1,−3)so∇f(7,2) = (1,−3)
f(x, y)≈f(7,2) +∇f(7,2)(x−7, y−2) = ln(1) + (1,−3)(x−7, y−2) = (x−7)−3(y−2)
Exercise 7 (15 pts). Compute the following limits if they exist, otherwise show that they do not exist.
(a) [5 pts] lim (x,y)→(0,0)
sin(x2+y2)
x2+y2
Here the easiest thing to notice is to look at what happens on the circlesr =x2+y2. The limit limr→0ln(r) = limr→0 ln(x−1x) = limr→0−x= 0. So it is clear that the answer is0.
(b) [5 pts] lim (x,y)→(0,0)
y4−x2 x+y2
Here we have x
2−y2
2x−2y = x+y
2 for x6=y and so (x,ylim)→(1,1)
x2−yy
2x−2y = 1. (c) [5 pts] lim
(x,y)→(0,0)
y4−x2 x2+y2
The limits along every path of the formy=mx(straight line) is0, however, the limit along x=ay2 is
lim
y→0
ay4 a2y4+y4 =
1
a2+ 1
