:
Due on
1
Matt Redmond
Math 283 - HW 3
Due 10/19
Problem 1
Validate the following equation
b × (c × d) = (b · d) c − (b · c) d
My grandpa has always said to attack that hard part first. In this problem I will use his approach and attack
the left hand side first.
î
ĵ k̂ ~b × c1 c2 0 = ~b × 0î + 0ĵ − d1 c2 k̂
d
0 0 1
î ĵ
k̂
= b1 b2
b3 = −b2 d1 c2 î + b1 d1 c2 ĵ + 0k̂
0 0 −d c 1 2
Now shifting gears and following my Grandmothers advice and ”doing the easier work second” we compute:




c1 î
d1 î
= (b1 d1 + b2 0 + b3 0)  c2 ĵ  − (b1 c1 + b2 c2 + b3 0)  0ĵ 
0k̂
0k̂

 

(b1 d1 c1 + b2 d2 c1 + b3 c1 0)î
b1 c1 d1 + b2 c2 d1 + b3 d1 0î
=  +(b1 d1 c2 + b2 c2 0 + b3 c2 0)ĵ  −  b1 c1 d2 + b2 c2 0 + b3 00ĵ 
+(b1 d1 0 + b2 00 + b3 00)k̂
b1 c1 0 + b2 c2 0 + b3 00k̂
After cancelations, we get that;


(−b2 c2 d1 )î
 (b1 d1 c2 )ĵ 
(0)k̂
And we see that we obtain the same vector on both sides of the equation, thus it is verified!
Problem 2
A point on a rim of a wheel of radius R=1 that rolls (ω = 1) without slipping along the x-aris. The trajectory
in a x-y coordinate system is called a Cycloid:
~r(t) = (R sin ωt + ωRt) î + (R cos ωt + R) ĵ
1
Find ~v and ~a.
d~r(t)
= (Rω cos ωt + ωR) î + (−Rω sin ωt) ĵ
dt
d~v (t)
~a =
= −Rω 2 sin ωt î − Rω 2 cos ωt ĵ
dt
~v =
2
Find ~at and ~an .
We know that:
at =
~a · ~v
|v|
~v
|v|
Page 2 of 5
Matt Redmond
Math 283 - HW 3
Due 10/19
So, we will first compute |v|
|v| =
=
p
R2 ω 2 cos2 ωt + 2ω 2 R2 cos ωt + ω 2 R2 + R2 ω 2 sin2 ωt
q
R2 ω 2 cos2 ωt + sin2 ωt + 2ω 2 R2 cos ωt + R2 ω 2
p
= R2 ω 2 + ω 2 R2 (2 cos ωt + 1)
√
= Rω 2 cos ωt + 2
We then compute ~a · ~v .
~a · ~v = −Rω 2 sin ωt (Rω cos ωt + ωR) + Rω 2 cos ωt (Rω sin ωt)
Then we can combine these two and obtain
~a · ~v
|v|
2
!
~
a·~
v
|v|2
.
−Rω 2 sin ωt (Rω cos ωt + ωR) + Rω 2 cos ωt (Rω sin ωt)
=
2R2 ω 2 (cos ωt + 1)
−R2 ω 3 sin ωt (cos ωt + 1) + Rω 2 cos ωt (Rω sin ωt)
=
2R2 ω 2 (cos ωt + 1)
− sin ωt ω cos ωt sin ωt
=
+
2
2 (cosωt + 1)
We now multiply this by vector ~v to get:
i
− sin ωt ω cos ωt sin ωt h
(Rω (cos ωt + 1)) î − (Rω sin ωt) ĵ
+
at =
2
2 (cosωt + 1)
=
− sin ωt
Rω 2 cos ωt sin ωt(cos ωt + 1)
(Rω (cos ωt + 1)) +
2
2 (cos ωt + 1)
2
Rω sin ωt −Rω 2 cos ωt sin2 ωt
ĵ
+
2
2 (cos ωt + 1)
î
We are able to re-write sin2 (ωt) = (1 − cos (ωt)) (1 + cos (ωt)). This helps us canclle the denominator for
the ĵ component. We now simplify at to:
2
Rω
ω R
at =
(− sin ωt cos ωt − sin ωt) +
cos ωt sin ωt
î
2
2
+
Rω
−Rω
2
2
sin (ωt) +
cos ωt − cos ωt ĵ
2
2
We are now able to see some simple calculations that can greatly simplify our value att . First by setting
R = ω = 1 we can combine terms that we couldent have otherwise. First, in the î component we can add
the + and - sin ωt cos ωt to cancell them out. In the ĵ component we can use the pythagoriean identity
sin2 t + cos2 t = 1 to leave us with just a cos(t) term. Our final tangent vector becomes:
at =
− sin (t)
cos(t)
î +
ĵ
2
2
Page 3 of 5
Matt Redmond
Math 283 - HW 3
Due 10/19
We can now use; ~an = ~at − ~a to obatin the normal vector,an .
− sin t sint
cos t
~an = ~at − ~a =
î +
+
+ cos t ĵ
2
2
2
~an =
sin t
3 cos t
î +
ĵ
2
2
3
Find the arc length from (x, y) : (0, 2) to (π, 0).
Using: Arc Length =
R t2 √
~r0 · ~r0 dt We calculate ~r0 · ~r0 first.
t1
~r0 · ~r0 = R2 ω 2 cos2 ωt + 2ω 2 R2 cos ωt + ω 2 R2 + R2 ω 2 sin ωt
= R2 ω 2 + 2ω 2 R2 cos ωt + ω 2 R2
= 2ω 2 R2 (1 + cos ωt)
Next, we will use the given information to solve for our bounds, t1 and t2 of our integral.
Initially:
~r(t) = 0 = R sin ωt1 + ωRt1
~r(t) = 2 = R cos ωt1 + R
Where R = ω = 1, thus:
0 = sin(t) + t1 → t1 = 0
Solving for t2 we solve:
0 = cos t + 1
cos t = −1 → t2 = π
So we know, t1 = 0 and t2 = 1. Plugging all this into our Arc Length equation we obtain:
Z πp
2ω 2 R2 (1 + cos ωt) dt
0
√
π
Z
2ωR
√
1 + cos t dt
0
We now do a simple u-subsitution to replace t, with 2u, so we can use a trigometric identity. We let
1
u = 2t → 2u = t. Differentiating we get du = dt
2 . However, we do not have a 2 to take out of our current
1
integral, So we put in a 2 and a 2 at the same time, to have the equivelent form of 1. We also must change
our bounds on the integral because of our u-subsitution. Thus our upper limit becomes π2 and our lower
limit becomes 0. Now our integral is:
Z
√
= 2 2ωR
π
2
√
1 + cos2u du
0
After this u-sub, not only would my parents be proud, be we can also use a trigometric identity to solve the
rest of the integral. We use
Page 4 of 5
Matt Redmond
Math 283 - HW 3
Due 10/19
2 cos2 (u) = 1 + cos (2u)
Our integral now becomes:
Z
√
2 2ωR
π
2
√
2 cos2 u du
0
√ √ Z
2 2 2
π
2
cos u du
0
π
4 [sin u]02 = 4 (1 − 0) = 4
Problem 3
Prove for any vector field ~v , ∇ · (∇ × ~v ) = 0 First taking the Curl, we obtain:
î
ĵ
k̂ ∂
∂
∂
∂
∂
∂
∂ =
(v
)
−
(v
)
î
−
(v
)
−
(v
)
ĵ
∇ × ~v = ∂x
3
2
3
1
∂y
∂z ∂y
∂z
∂x
∂z
v1 v2 v3 ∂
∂
(v2 ) −
(v1 ) k̂
∂x
∂y
∂ ∂v3
∂v2
∂v3
∂v1
∂ ∂v1
∂ ∂v2
∇ · (∇ × ~v ) =
−
−
−
î +
ĵ +
k̂
∂x ∂y
∂z
∂y ∂z
∂x
∂z ∂x
∂y
2
2
2
∂ v3
∂ 2 v3
∂ 2 v1
∂ 2 v2
∂ v1
∂ v2
=
−
−
−
+
+
∂x∂y ∂x∂y
∂y∂z
∂z∂y
∂z∂x ∂x∂z
+
All of which these terms cancell, giving us:
0+0+0=0
Thus validating this statement!
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