MATH 265 Midterm: MATH 265 Iowa State 265F11PT1BSolutions
15 Feb 2019
School
Department
Course
Professor
Math 265 (Butler)
Practice Midterm I — B (Solutions)
1. Find the projection of h2s, 1, s −1ionto the vector h−2t, 5−t2,4ti.
The formula for projection of a vector uonto a vector vis
u·v
v·vv.
Applying it to our case with u=h2s, 1, s −1iand v=h−2t, 5−t2,4tiwe have
that the projection is
h2s, 1, s −1i · h−2t, 5−t2,4ti
h−2t, 5−t2,4ti · h−2t, 5−t2,4tih−2t, 5−t2,4ti
=(2s)(−2t) + (1)(5 −t2) + (s−1)(4t)
(−2t)2+ (5 −t2)2+ (4t)2h−2t, 5−t2,4ti
=−4st + 5 −t2+ 4st −4t
4t2+ 25 −10t2+t4+ 16t2h−2t, 5−t2,4ti
=5−4t−t2
t4+ 10t2+ 25h−2t, 5−t2,4ti
=5−4t−t2
(t2+ 5)2h−2t, 5−t2,4ti.
(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−tcos t, e−tsin tistarts 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 < ∞.)
We have r(t) = he−tcos t, e−tsin ti. So the next thing for us to do is to compute
r′(t)
, first we have
r′(t) = −e−tcos t−e−tsin t, −e−tsin t+e−tcos t,
so that
r′(t)
=q−e−tcos t−e−tsin t2+−e−tsin t+e−tcos t2
=se−2tcos2t+ 2e−2tcos tsin t+e−2tsin2t
+e−2tcos2t−2e−2tcos tsin t+e−2tsin2t
=qe−2t(cos2t+ sin2t+ cos2t+ sin2t) = √2e−2t=√2e−t
Therefore the arc length is
Z∞
0
r′(t)
dt =Z∞
0
√2e−tdt
= lim
s→∞ Zs
0
√2e−tdt
= lim
s→∞ −√2e−t
s
0
= lim
s→∞ −√2e−s+√2
=√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.)
Document Summary
Practice midterm i b (solutions: 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 v v(cid:19)v. (cid:18) u v. = 4st + 5 t2 + 4st 4t. 4t2 + 25 10t2 + t4 + 16t2h 2t, 5 t2, 4ti h 2t, 5 t2, 4ti. 5 4t t2 t4 + 10t2 + 25h 2t, 5 t2, 4ti (on a side note we see that this projection only depends on t, even though the vector we were projecting involved s. ) 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 < . ) We have r(t) = he t cos t, e t sin ti. =s e 2t cos2 t + 2e 2t cos t sin t + e 2t sin2 t.
