Calculus
Homework # 9
Homework # 9
10.1 # 6: (3 pts, p. 700) Sketch the curve with the given vector equation. Indicate with an
arrow the direction in which t increases.
r(t) = ⟨t3 , t2 ⟩
Here is a plot of this vector equation:
8
7
6
y = t2
5
4
3
2
1
0
-4
-3
-2
-1
0
1
2
3
4
x = t3
Note that the value of t increases as the function goes to the right.
10.2 # 12: (3 pts, p. 681) Find the derivative of the vector function:
r(t) = at cos 3t i + b sin3 t j + c cos3 t k
Treak the i, j, and k, as if they were constants and the derivative of this function is:
r′ (t) = a(cos 3t − 3t sin 3t) i + 3b cos t sin2 t j − 3c cos2 t sin t k
10.3 # 22: (4 pts, p. 714) Use Theorem 10 to find the curvature.
r(t) = t i + t2 j + et k
See Theorem 10 on page 710. The formula for curvature needs r′ (t), |r′ (t)|, r′′ (t), and
1
Calculus
Homework # 9
|r′ (t) × r′′ (t)|, these are:
r′ (t) = i + 2t j + et k
√
|r′ (t)| = 1 + 4t2 + e2t
r′′ (t) = 2 j + et k


i j k


r′ (t) × r′′ (t) = 1 2t et 
0 2 et
( t
)
= 2te − 2et i − et j + 2 k
√(
)2
|r′ (t) × r′′ (t)| =
2tet − 2et + e2t + 4
√
= 4t2 e2t − 4te2t + 5e2t + 4
Now the curvature becomes:
|r′ (t) × r′′ (t)|
|r′ (t)|3
√
4t2 e2t − 4te2t + 5e2t + 4
=
(e2t + 4t2 + 1)3/2
κ(t) =
2