Ams 261 - Applied Calculus Iii Cheat Sheet Page 3
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1
2
3i + j + k
i
j + k
Answer:
,
(1)(1) (1)(1)+(1)(1)
1
u v = u
v cos θ
cos θ =
=
3
1+1+1 1+1+1
i + j + k
i j + k
Answer:
u
v = 2i + 0j
2k =
4 + 4 = 2 2
i + j + k
i
j + k
Answer:
calculated in part (ii),
is a point in the plane so the plane
n = 2i + 0j
2k
P (1, 1, 1)
equation is
,
2(x
1)
2(z
1) = 0
2x
2z = 0
i + j + k i
j + k
i + j + k
Answer:
V = (u
v) w =(2i + 0j
2k) ( i + j + k) =
2
2 = 4
r(t) = cos ti + sin tj
T (t)
t = π/3
Answer:
T (t) = r
,
(t)
2
2
r (t) =
sin ti + cos tj, r (t) =
sin
(t) + cos
(t) = 1
r
(t)
,
π
π
π
3
1
T (t) =
sin ti + cos tj
T (
) =
sin(
)i + cos(
)j =
i +
j
3
3
3
2
2
N (t)
t = π/3
Answer:
N (t) = T
,
,
(t)
2
2
T (t) =
cos ti
sin tj
T (t) =
cos
(t) + sin
(t) = 1
T
(t)
,
π
π
π
1
3
N (t) =
cos ti
sin tj
N (
) =
cos(
)i
sin(
)j =
i
j
3
3
3
2
2
t
[0, 2π]
Answer:
2π
2π
2π
2
2π
2
r (t) dt =
sin
(t) + cos
(t)dt =
dt = t
= 2π
0
0
0
0
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