Last:
First:
Please circle your instructor and TA. Thank you!
Instructor: Dr. Fazarmirad Dr. Ketchersid
TA: Marcel
Erica
Mylinh
Instructions. All work must be shown and exact answers are expected. You are allowed calculators, but you should only use calculators tocheck your work not to perform your work for you. For example,sin(2π/3) =√3/2 will be accepted,sin(π/4) =.70...will not.
Problem 1 (10pts). Let r(t) =t3i−2t2j+ 16k. (a) (5pts) Find the arc length of the curve on[0,1].
r0(t) =h3t2,−4t,0i
kr0(t)k=p9t4+ 16t2=|t|p9t2+ 16 so the arc length is
Z 1 0
kr0(t)kdt= Z 1
0
|t|p9t2+ 4dt= 1 18
Z 25 16
u1/2du= 1 18 2 3u 3/2 25 u=16= 1 27(5
3−43)
(b) (5pts) Find the curvatureK att= 1.
r00(t) =h6t,−4,0i
r0(1) =h3,−4,0i
r00(t) =h6,−4,0i kr0(1)k= 5
r0(1)×r00(1) = det
i j k
3 −4 0 6 −4 0
=h0,0,12i
kr0(1)×r00(1)k= 12
K(1) = kr
0(1)×r00(1)k
Problem 2 (10 pts). Consider f(x, y) =y3−xy2+x−4y
(a) (5pts) Find the direction and magnitude of the maximal rate of increase of f at
P(1,2).
∇f(x, y) =h1−y2,3y2−2xy−4i
so
∇f(1,2) = (−3,4)andk∇f(1,2)k= 5 and the direction of maximal increase is
∇f(1,2)
k∇f(1,2)k =
1
5(−3,4) the magnitude of increase in that direction is5.
(b) (5pts) Find the directional derivative atP in the direction from P to Q(4,−2). The direction is
u=
−−→
P Q
k−P Q−→k =
h3,−4i
5 and the directional derivative is
∇f(1,2)·h3,−4i
5 = 1
5h−3,4i ·3,−4i=
Problem 3 (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 4 (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 points
n
±√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 5 (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 6 (10 pts). Suppose z =f(x−y), that is, z =f(u) where u =x−y. Show that
∂z ∂x +
∂z ∂y = 0. ∂z
∂x + ∂z ∂y =
df du
∂u ∂x+
df du
∂u ∂y =
df du −
Problem 7 (10pts). Evaluate the following by first converting to polar: Z 2 √ 2 0 Z √
16−y2
y
ex2+y2dx dy
The region being integrated over is the region in the first quadrant bounded byx=y and
x2+y2= 16. This is the polar rectangle0≤r≤4and 0≤θ≤π/4. So Z 2 √ 2 0 Z √
16−y2
y
ex2+y2dx dy= Z π/4
0 Z 4
0
er2r dr dθ= Z π/4
0
dθ
Z 4
0
er2r dr= π 4
1 2
Z 16
0
eudu= π 8(e
Problem 8 (10 pts). Compute Z Z
R
sin(y2)
y dA
whereR={(x, y) : 0≤x≤π and √x≤y≤√π}. You may need to play with the order of integration.
Z π 0
Z
√
π
√
x
sin(y2)
y dy dx=
Z
√
π
0
Z y2 0
sin(y2)
y dx dy=
Z
√
π
0
y sin(y2)dy= 1 2
Z π 0
The final three problems involve sequences and series. For each problem make sure to explain your reasoning: (1) state which tests are being used and (2) verify the hypotheses of the test.
Problem 9 (10pts). Determine if the series
∞
X i=1
(n+ 1)2
n(n+ 2) converges or diverges. If convergent, then find its sum.
(n+ 1)2
Problem 10 (10 pts). Determine whether the seriesP∞i=0anxn converges on0≤x≤1 given thatan>0 and P∞n=0an converges.
Problem 11 (10 pts). Find radius of convergence for the series
∞
X n=2
(−1)nxn
4n ln(n).
lim n→∞
(−1)n+1xn+1
4n+1 ln(n+ 1)
(−1)nxn
4nln(n)
= |x| 4
ln(n+ 1) ln(n) =
|x|
4
