Math 241 E1: Calculus III
Midterm II Solution
July 15, 2014
1. Find all critical points of the given function f and identify their nature as local extrema.
f (x, y) = ex
2
+5y 2
Solution. To find critical points, we need fx = fy = 0. We have
(
2
2
fx = 2x · ex +5y = 0
2
2
fy = 10y · ex +5y = 0
Since the exponential function can never be zero, we must have the only critical point (0, 0). To identify this
critical point, we consider the Hessian matrix
!
2
2
2
2
fxx fxy
(4x2 + 2)ex +5y
20xy · ex +5y
H(x, y) =
=
2
2
2
2
fyx fyy
20xy · ex +5y
(10 + 100y 2 )ex +5y
At the origin,
H(0, 0) =
2
0
0
10
and
det H(0, 0) = 20 > 0, fxx (0, 0) = 2 > 0
So (0, 0) is the local minimum of the function f (x, y) = ex
1
2
+5y 2
.
2. Heron’s formula for the area of a triangle whose sides have length x, y, and z is
p
Area = s(s − x)(s − y)(s − z),
where s = 21 (x + y + z) is the so-called semi-perimeter of the triangle.
(a) Use Heron’s formula to find the largest area when s = 12 (x + y + z) = 4. (Hint: Since the area is positive,
you can optimize Area2 to simplify computation.)
Solution. Let f (x, y, z) = Area2 = 4(4 − x)(4 − y)(4 − z) and g(x, y, z) = x + y + z. We need to optimize the
function f (x, y, z) subject to the constraint g(x, y, z) = x + y + z = 8. By Lagrange Multiplier, this happens
when
→
−
→
−
∇f = λ · ∇g for some λ ∈ R
We have
→
−
∇f = − 4(4 − y)(4 − z), −4(4 − x)(4 − z), −4(4 − x)(4 − y)
→
−
∇g = (1, 1, 1)
Hence we have a system of equations

−4(4 − y)(4 − z) = λ



−4(4 − x)(4 − z) = λ

−4(4 − x)(4 − y) = λ



x+y+z =8
We have
−
(1)
(2)
(3)
(4)
λ
= (4 − y)(4 − z) = (4 − x)(4 − z) = (4 − x)(4 − y).
4
If we assume z 6= 4 and x 6= 4, then these equations give us x = y = z, and hence x = y = z = 38 . In this case
s 8
8
8
16
Area = 4 4 −
4−
4−
= √ .
3
3
3
3 3
If x = 4 or z = 4, without lost of generality, assume x = 4. Then we must have λ = 0 in equation (3). But
then, we must have (4 − y)(4 − z) = 0 in equation (1). Again, without lost of generality, assume y = 4, then
z = 0 from equation (4) which is impossible for a triangle.
√
So the minimal area is 316
3
(b) When the area is maximal, the triangle is an: (Circle the correct answer.)
A. Equilateral triangle B. Right triangle C. Isosceles triangle
Solution. We have x = y = z when solving the system.
3. (a) Consider the following path:
→
−
r (t) = ln(cos t), cos t, sin t ,
t ∈ [0,
π
]
3
Compute
of this path.
R the arc-length
(Hint: sec x dx = ln sec x + tan x + C)
Solution.
sin t
, − sin t, cos t = (− tan t, − sin t, cos t)
−
cos t
p
p
√
−
k→
r 0 (t)k = tan2 t + sin2 t + cos2 t = tan2 t + 1 = sec2 t = sec t
→
−
r 0 (t) =
So the arc-length is
Z
L=
0
π
3
π3
√
√
sec t dt = ln sec x + tan x = ln(2 + 3) − ln(1 + 0) = ln(2 + 3).
0
→
−
(b) Consider the line segment l (t) = (2 + t, 5 − 3t, 2t + 1), t ∈ [0, 5]. Compute the arc-length parameter s(t),
then reparametrize the line by arc-length.
√
→
−
→
−
Solution. We have the velocity l 0 (t) = (1, −3, 2), and the speed k l 0 (t)k = 14. So the arc-length parameter is
Z t
Z t√
√
→
−
s(t) =
k l 0 (τ )k dτ =
14 dτ = 14t
0
Then we have t =
√s ,
14
0
so the reparametrization by arc-length is
→
−
l (s) =
s
3s
2s
2+ √ , 5− √ , √ +1
14
14
14
4. Consider the following curve:
→
−
r (t) = 0, t, ln(cos t) ,
π π
t∈ − ,
2 2
−
−
(a) Compute →
r (0) and lim →
r (t).
t→1
−
Solution. →
r (0) = (0, 0, 0) and
−
lim →
r (t) = lim 0, lim t, lim ln(cos t)
t→1
t→1
t→1
t→1
= 0, 1, ln(cos 1)
(b) Compute the moving frame
n→
− →
− →
−o
T , N , B for this curve.
Solution.
sin t
0, 1, −
= (0, 1, − tan t)
cos t
p
−
k→
r 0 (t)k = 1 + tan2 t = sec t
→
−
r 0 (t) =
→
−
→
−
r 0 (t)
1
tan t
T (t) = →
= 0,
,−
= (0, cos t, − sin t)
sec t
sec t
k−
r 0 (t)k
→
−0
T (t) = (0, − sin t, − cos t)
→
−0
k T (t)k = 1
→
−0
→
−
T (t)
N (t) = →
= (0, − sin t, − cos t)
−0
k T (t)k
→
→
− 
−
→
−
i
j
k
→
−
→
− →
−
B (t) = T × N = det  0
cos t − sin t 
0 − sin t − cos t
= (−1, 0, 0)
(c) Compute the curvature κ(t).
Solution.
→
−
1
k T 0 (t)k
=
κ(t) = →
= cos t
sec t
k−
r 0 (t)k