Math 265 (Butler)
Practice Midterm I — B (Solutions)
1. Find the projection of h2s, 1, s − 1i onto the vector h−2t, 5 − t2 , 4ti.
The formula for projection of a vector u onto a vector v is
u·v
v.
v·v
Applying it to our case with u = h2s, 1, s − 1i and v = h−2t, 5 − t2 , 4ti we have
that the projection is
h2s, 1, s − 1i · h−2t, 5 − t2 , 4ti
h−2t, 5 − t2 , 4ti
2
2
h−2t, 5 − t , 4ti · h−2t, 5 − t , 4ti
(2s)(−2t) + (1)(5 − t2 ) + (s − 1)(4t)
h−2t, 5 − t2 , 4ti
=
2
2
2
2
(−2t) + (5 − t ) + (4t)
−4st + 5 − t2 + 4st − 4t
h−2t, 5 − t2 , 4ti
= 2
4t + 25 − 10t2 + t4 + 16t2
5 − 4t − t2
= 4
h−2t, 5 − t2 , 4ti
2
t + 10t + 25
5 − 4t − t2
h−2t, 5 − t2 , 4ti.
=
(t2 + 5)2
(On a side note we see that this projection only depends on t, even though the
vector we were projecting involved s.)
2. A particle travels along the parametric curve he−t cos t, e−t sin ti starts at (1, 0) at
time t = 0 and then spirals into the origin (0, 0) as t → ∞. How far will the particle
have traveled when it reaches the origin? (In other words, what is the arc length of the
parametric curve for 0 ≤ t < ∞.)
−t
−t
We
0 have
r(t) = he cos t, e sin ti. So the next thing for us to do is to compute
r (t), first we have
r0 (t) = − e−t cos t − e−t sin t, −e−t sin t + e−t cos t ,
so that
0 q
2
2
r (t) =
− e−t cos t − e−t sin t + − e−t sin t + e−t cos t
s
e−2t cos2 t + 2e−2t cos t sin t + e−2t sin2 t
=
+e−2t cos2 t − 2e−2t cos t sin t + e−2t sin2 t
q
√
√
= e−2t (cos2 t + sin2 t + cos2 t + sin2 t) = 2e−2t = 2e−t
Therefore the arc length is
Z ∞
Z
0 r (t) dt =
0
=
=
=
=
√ −t
2e dt
0
Z s√
lim
2e−t dt
s→∞ 0
√ −t s
lim − 2e s→∞
√ −s 0 √ lim − 2e + 2
s→∞
√
2.
∞
(Of course technically this is an improper integral, but this one is one of the
nicer improper integrals and so we can get away without having all of the details
filled in as we did here.)
3. Find the curvature of r(t) = ti + 21 t2 j +
√
2 2 3/2
t k
3
at time t = 2.
Using the equation sheet we have
0
r (2) × r00 (2)
κ(2) =
.
r0 (2)3
We first calculate the derivatives
√
r0 (t) = i + tj + 2t1/2 k,
√
2 −1/2
r00 (t) = j +
t
k.
2
So we have
r0 (2) = i + 2j + 2k,
1
r00 (2) = j + k.
2
Next we calculate the inner product, we have
i j k 1
1
r0 (2) × r00 (2) = 1 2 2 = i + 0j + k − 0k − 2i − j = −i − j + k.
2
2
0 1 1 2
And so we have
q
p
1
(−1)2 + (− 12 )2 + (1)2
− i − j + k
9/4
3
2
κ(2) = = .
= √
3 = p 2
3
2
2
3
i + 2j + 2k
54
( 9)
( (1) + (2) + (2) )
4. A particle moving
space has acceleration a(t) = 2, sin(πt),
6t . At time t = 0
in three
the particle is at 3, 0, 1 while at time t = 2 the particle is at 1, −2, 5 . What is the
velocity of the particle at time t = 1?
We have
Z
Z
v(t) =
a(t) =
Z
2 dt,
Z
sin(πt) dt,
6t dt
1
2
= 2t + C, − cos(πt) + D, 3t + E .
π
If we had some initial conditions on velocity we would almost be done. But we don’t
have any information on velocity and so we keep going. We have
Z
Z
r(t) =
v(t) =
Z
1
2
(2t + C) dt,
− cos(πt) + D dt, (3t + E) dt
π
1
2
3
= t + Ct + F, − 2 sin(πt) + Dt + G, t + Et + H .
π
Z Now we have information about position so we can start figuring out the constants.
We have
r(0) = hF, G, Hi = h3, 0, 1i.
So now we have F , G and H. Updating we now have
1
3
2
r(t) = t + Ct + 3, − 2 sin(πt) + Dt, t + Et + 1 .
π
Our other initial condition gives us
r(2) = h7 + 2C, 2D, 9 + 2Ei = h1, −2, 5i.
Using this we can solve for our constants. We have
7 + 2C = 1
2D = −2
9 + 2E = 5
so
so
so
C = −3
D = −1
E = −2
Putting these into v(t) we have
1
2
v(t) = 2t − 3, − cos(πt) − 1, 3t − 2 .
π
Therefore we have that the velocity at t = 1 is
1
v(1) = − 1, − 1, 1 .
π
5. Find the the line in parametric form which is tangent to the curve x = 2t2 + t, y = 2t2 ,
z = t3 − t at time t = 2.
This in vector form is r(t) = h2t2 + t, 2t2 , t3 − ti. To find the tangent line
we will need to have r(2) and r0 (2), so first we calculate the derivative to get
r0 (t) = h4t + 1, 4t, 3t2 − 1i. We now have
r(2) = h10, 8, 6i
and
r0 (2) = h9, 8, 11i.
We now can easily write the desired line in parametric form giving us the following:
x = 10 + 9t
y = 8 + 8t
z = 6 + 11t