Multivariable Calculus Practice Midterm 2
Prof. Fedorchuk
1.
(10 points) Compute the arc length of the curve
4 3 2
~r(t) = t, t 2 , t ,
3
on the interval 1 ≤ t ≤ 3.
D
E
√
1
Solution: We have ~r 0 (t) = 1, 2t 2 , 2t and k~r 0 (t)k = 1 + 4t + 4t2 = 1 + 2t. Hence the
arc length is
Z 3
Z 3
3
0
k~r (t)kdt =
(1 + 2t)dt = (t2 + t) = (12) − 2 = 10.
1
1
1
2.
(10 points) Reparametrize the curve ~r(t) = het , et sin t, et cos ti with respect to arc
length measured from the point (1, 0, 1) in the increasing direction of t.
Solution: We compute that ~r 0 (u) = heu , eu sin u + eu cos u, eu cos u − eu sin ui and that
p
k~r 0 (u)k = e2u + (eu sin u + eu cos u)2 + (eu cos u − eu sin u)2
q
√
= e2u + 2e2u (sin2 (u) + cos2 (u)) = 3eu .
The arc length function is
√
3eu du = 3(et − 1).
0
0
√
Solving for t in terms of s we obtain t = ln(1 + s/ 3). Hence the arc length parameterization of the curve is
D
√
√
√
√
√ √ E
p~(s) = ~r(ln(1+s/ 3)) = eln(1+s/ 3) , eln(1+s/ 3) sin ln(1 + s/ 3) , eln(1+s/ 3) cos ln(1 + s/ 3)
D
√
√
√ √
√ E
= 1 + s/ 3, (1 + s/ 3) sin ln(1 + s/ 3) , (1 + s/ 3) cos ln(1 + s/ 3)
Z
s = s(t) =
t
0
k~r (u)kdu =
Z t√
(We plugged in the expression for t in terms of s into the original equation of the vector
function ~r(t).
1
3.
(10 points)
√
√
Find the tangent lines to the curves r1 (t) = ht, t2 , t3 i and r2 (u) = h1, 2 sin(u), 2 cos(u)i
at the point (1, 1, 1).
Solution: Solving for t and u, we find that t = 1 and √
u = π/4 give√the point (1, 1, 1). We
0
2
0
also compute that r1 (t) = h1, 2t, 3t i and r2 (u) = h0, 2 cos(u), − 2 sin(u)i.
The tangent line to r1 (t) at t = 1 has direction vector r01 (1) = h1, 2, 3i. Hence this tangent
line has parametric equation


x = 1 + s
y = 1 + 2s

z = 1 + 3s
The tangent line to r2 (u) at u = π/4 has direction vector r02 (π/4) = h0, 1, −1i. Hence this
tangent line has parametric equation


x = 1
y =1+s

z = 1 − s
4.
p
x2 − y
(10 points) Let f (x, y) =
.
1+x
a. (5 pts)
Find the partial derivatives ∂f /∂x and ∂f /∂y.
Solution:
p
2x
p
(1 + x) − x2 − y
2 x2 − y
∂f
x(1 + x) − (x2 − y)
x−y
p
p
=
=
=
∂x
(1 + x)2
(1 + x)2 x2 − y
(1 + x)2 x2 − y
∂f
1
=− p
∂y
2 x2 − y(1 + x)
b. (5 pts)
Describe (using inequalities) and sketch the domain of f (x, y).
Solution: The domain is defined by x 6= −1 and x2 − y ≥ 0, or
(
x 6= −1
y ≤ x2
2
The domain is the region on or under the parabola y = x2 with the exception of the
vertical line x = −1.
5.
(10 points)
a. (5 pts) Find the limit
x2 + y 2
(x,y)→(1,2)
xy
lim
Solution:
x2 + y 2
is defined at (1, 2), it is continuous at (1, 2). So the
xy
12 + 22
5
limit is simply the value of the function at (1, 2), which is
= .
2
2
Since the rational function
b. (5 pts)
Prove that the following limit does not exist:
3x2 − y 2
lim
(x,y)→(0,0)
xyex
Solution: Since lim(x,y)→(0,0) ex = e0 = 1, the limit in question exists only if the limit
3x2 − y 2
lim
(x,y)→(0,0)
xy
exists. We proceed to show that the latter limit does not exist.
Take a curve through (0, 0) given by x = y = t. Then along this curve, the function takes
3t2 − t2
values
= 2. Take now a curve through (0, 0) given by x = t and y = −t. Along
t2
3t2 − t2
= −2. Since these two values are distinct,
this curve, the function takes values
−t2
3x2 − y 2
we conclude that the limit of
at (0, 0) does not exist.
xy
3
6.
(10 points)
a. (5 pts)
are zero.
Find all (x, y) such that the first partial derivatives of f (x, y) = xyey
Solution: We compute that fx = yey and fy = x(1 + y)ey . Setting these to zero we obtain
a system
(
yey = 0
x(1 + y)ey = 0
Since ey > 0, this gives
(
y=0
x(1 + y) = 0
The only solution of this system is x = y = 0. Hence the answer is (0, 0).
b. (5 pts)
Find all the second partial derivatives of f (x, y) = xyey .
Solution: We computed that fx = yey and fy = x(1 + y)ey . Therefore, differentiating
once more gives:
fxx = 0
fxy = (1 + y)ey
fyy = x(2 + y)ey
4
7.
(5 points)
a. (5 pts)
Let z = ln(u3 + v 4 ), where u = t4 and v = t2 . Compute dz/dt.
Solution: By the Chain Rule,
dz
∂z du ∂z dv
3u2
4v 3
12t11 + 8t7
3
=
+
= 3
(4t
)
+
(2t)
=
.
dt
∂u dt
∂v dt
u + v4
u3 + v 4
t12 + t8
√
(15 points) A curve in space is given by the vector function ~r(t) = h2 sin(t), sin(t), 5 cos(t)i.
Compute unit tangent vectors to the curve.
8.
√
0
5 sin(t)i. It follows that k~r 0 (t)k =
Solution:
We
compute
that
~
r
(t)
=
h2
cos(t),
cos(t),
−
q
√
√
(2 cos(t))2 + (cos(t))2 + ( 5 sin(t))2 = 5. Therefore, the unit tangent vector at time
t is
~r 0 (t)
2
1
T(t) =
= √ cos(t), √ cos(t), − sin(t)
k~r 0 (t)k
5
5
5
9.
(10 points) Answer the following questions. Give a brief explnation. Each question
is worth 2 point.
Question 1: Can two different level curves of a function intersect?
Answer: No! Different level curves correspond to different values of the function. Since a function cannot take different values at a point, different level curves
never intersect.
Question 2: What are the level surfaces of the function f (x, y, z) = 5x + 4y + 3z.
Answer: These are planes 5x + 4y + 3z = c.
∂ 2f
∂ 2f
and
always equal?
∂x∂y
∂y∂x
Answer: No! Mixed partials are equal for twice continuously differentiable
functions (Clairaut’s theorem), but there are examples of functions whose mixed
partials are not equal.
Question 3: Are the mixed partials
Question 4: Define what it means for a function f (x, y) to be continuous at a point (a, b).
Answer:
lim f (x, y) = f (a, b)
(x,y)→(a,b)
Question 5: A space curve has a unique parameterization.
Answer: No! Space curves have many different parameterizations. For example, a curve ~r(t), t ∈ [a, b], is also parameterized by ~r(2s), s ∈ [a/2, b/2].
6