 # ENGG1410-E: Short Test 2

```ENGG1410-E: Short Test 2
Name:
Student ID:
Write all your answers on this sheet, and use the back if necessary.
Problem 1 (30%). Let C be the curve r(t) = [5t, t2 , 3] from t = 0 to t = 1. Calculate
R
C
t ds.
Z
Z
1
s
t
t ds =
0
C
Z
dx
dt
2
+
dy
dt
2
+
dz
dt
2
dt
1q
52 + (2t)2 + 02 dt
=
0
Z
=
1
p
t 4t2 + 25 dt.
0
=
=
=
Z
1 1p 2
4t + 25 d(4t2 + 25).
8 0
1
1
(4t2 + 25)3/2 .
12
0
1
(293/2 − 253/2 ).
12
Problem 2 (30%). Let f (x, y, z) = ex y + 5z. Compute the directional derivative of f (x, y, z) in
the direction of [1, 1, 1] at point (1, 2, 3).
√
√
√
Answer: The unit vector in the direction of [1, 1, 1] is u = [1/ 3, 1/ 3, 1/ 3]. Thus, the directional derivative equals
∂f ∂f ∂f
,
,
·u
∇f (x, y, z) · u =
∂x ∂y ∂z
∇f (x, y, z) · u = [ex y, ex , 5] · u
√
√
√
= [2e, e, 5] · [1/ 3, 1/ 3, 1/ 3]
√
= (3e + 5)/ 3.
Problem 3 (40%). Consider the curve C that is the intersection of the following two faces:
x2 + y + z 2 = 5
x + y = 1.
Give a tangent vector of C at point (0, 1, 2).
Answer: From the two equations, we have
x2 − x + z 2 = 4
⇒
⇒ x2 − x + 1/4 + z 2 = 17/4
2
z
x − 1/2 2
√
+ √
= 1
17/2
17/2
Hence we can write C in the parametric form [x(t), y(t), z(t)] where
√
1
17
x(t) =
+
cos t
2
2
√
1
17
y(t) = 1 − x = −
cos t
2
2
√
17
sin t
z(t) =
2
√
√
√
Therefore, a tangent vector is [x0 (t), y 0 (t), z 0 (t)] = [− 217 sin t, 217 sin t, 217 cos t]. The point (0, 1, 2)
√
√
is given by t satisfying ( 17/2) cos t = −1/2 and ( 17/2) sin t = 2. Therefore, a tangent vector at
this point is [−2, 2, −1/2].
2
``` # Rectangular Components in Space Direction Cosines 3-d Rectangular components # Worksheet 6 Chain rule and implicit differentiation 1 Chain rule 2 # ASSIGNMENT 11 SOLUTION 1. Stewart 16.4.21 [5 pts] (1) If C is the 