Day 4 Homework
Use
your calculator on problems 10 and 13c only.
t. rf x=t2-land y=e".find!.
2.
3.
ax
If a particle moves in the xy-plane so that at any time f > 0, its position vector is
L t t + - \ ^ r\ ^
\ln(/- )t ). 3t' ), hnd its velocity vector at time f = 2.
A particle moves in the xy-plane so that at any time f, its coordinates are given by
x=t5
4.
5.
-1
and
y=3ta -2t3. Find,its accelerationvector att= I.
If a particle moves in the xy-plane so that at time f its position vector is
I r
t^r\
(ri" [ :, - i),t,')
.find the velociry vector
A particle moves on the curve
/
at rime t =
= lnx so that its x-component has derivative
*'(t)=r+ I for / > 0.At time f = 0, the particle
the position of the particle at time f =
6.
is at the point (1, 0). Find
1.
A particle moves in the xy-plane in such
(+ t, f).If
ft
L.
a way
that its velocity vector is
the position vector ar t = 0is (S, O), find the position
of
the particle at t = 2.
7.
A particle moves along the curve
what is the value of
!!
dt
8.
xy= 10. If x = 2
x=t3
-;f
4dt
=
Z,
I
Tle position of a particle moving in the x7-plane
equations
and
-r8r+5
and
is given by the parametric
!=t3 -6t2 +9t+4. Forwhatvalue(s)of
/ is the particie at restl
9.
A curve Cis definedbythe parametric equations x=t3 andy=t2 -5t1Z.Write
the equation of the line tangent to the graph of C at the point (8, -4).
10. A particle moves
r(t)
= 5r + 3sin
I
inthe xy-plane
and
so that the position of the particle is given by
y(,)= (t - r)(t -
cos
r). Find the velocity vector
at the time
when the particle's horizontal position is x = 25.
ll.
a
The position of a particle at any time t > 0 is given Uy
x(r) = t2 -3 andy(r) =
(a) Find the magnitude of the velocity vector at time f =
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2010 The College Board
5.
i,t.
Vectors
(b) Find the total distance traveled by the particle from f
= 0 to f =
)-.
(c) Find ? uta function of x.
dx
12. Point
P(*, y)
moves
av
in the xy-plane in such a way that
-1=2t for />0.
dxl
-dt=-t+l
5.
and
dt
(a) Findthe coordinates ofPin termsof f given that,when
t= L, x=ln1 andy=0.
(b) Write an equation expressingT in terms of x.
(c) Find the average rate of
chan ge of y
with respect to x
as f varies
from 0 to 4.
(d) Find the instantaneous rate of chan ge of y with respect
to r when / = l.
13. consider the curve c given by the parametric equations x 2
3 cos / and
=
-
-!<r<f,.
!=3+2sint,fo,
(a) Find
Idx
uru
fun*,on
oi.
(b) Find the equation of the tangent line at the point where
t=! .
4
(c) The curve. C intersects the
twice.
Approximate
the
length
of the curve
7-axis
between the two y-intercepts.
Answers to Day 4 Homework
dy df ,'l
^13
dr=dr=ALt
J _ 3tt e'- 5te
l.
dx dx
A
dr"
=- 2
-l
Vlt'-r)
ay\ f zt+s -\
-l=\--.o/iSo
dtl \r"+5t' f
3. ,(,) = (#,#)
4.
= (sto , rzt3
- at
,,\
,h\=(2"
\/
\14'
)
.
o(t) =
(#, #)
u(,)=
(*,*l=(,*,[,, -;),6,) *
=
(zot,,
36t2
- Dt),
sou(r) =
,(:)=(3cosz, tn)=(-z,zn)
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@
2010 The College
(zo, z+),
Boud
/ \
fl
,
t2
r(r) = J\t +t)dr = -
'(o)=
since
-r t
t= c so'(r)=
x(l)=
*
C.
L*r *t.
I ,nd ,(t)= Irf f l" position =(2.,"f 1)l
or,,".. r(,;=,.Jj(,'. ,rl'='
:and
y(r)=\"i;j
Jiition
=[,; '11;]
\./
x(o)=5 so c=5 and"(,)=
'(r)= J(t+r)at=,*++c.
.44
r
|
,
,.
/'
y(,)=Jf ar=r+n. y(o)=0so D=0and r(,)=T.
**.
\ 2
Position vector =
(,
or, since ,(2) =s
.
Jr'(r
t, *\
4l
At t =2,
Position
+ t)at = e and y(2) =o +
=
J'r'
7v,
t*L*s.
+).
dt = 4,
position =(0, +).
.7
When
x=2,y=5.
dv dx
x:+v-=U
dt dt
(z)(:) +s4=oso
\ ./\ / dt
or find tnut !=
dx6dt 5'
d'=-9
dt
dx -+.
xt
t
o,
Then substituting rnto
"
girr,
!dx= 4L
dx" +4 = Idr
dt
dt
*'(t)=3P -3t- l8 =:(r-:)(r +z)= 0 when / = 3 and t =-2.
v'(t)=3t2 -12t +9 =:(r- t)(r-3) = 0 when t =3
and
t =1.
The particle is at rest wiren v (r) = (0, O) so at rest when / = 3.
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O 2010 The College Board
so thut
9.
8
t=2
t3 =
t2
-5t+2=-4
t2
-5t+6=0
(t -s)(t -2)= o
t=3, t=2
et (t, - 4) when t =2
dyl = u-sl
I
--
=-&1,=z----r
3t' l,=z
Tangent line equatio
12
ni y +4 = -
*(r - 8)
10. 5t+3sin /= 25 when t = 5.445755...
'(,)
=
(#' #)= (s + 3cos /,-r + cos r + (r - r).in r)
,(s.++szss...)= (z.oos , -2.22s)
lI.
(a) Magnitude when / = 5 is
,r)'
1,
(b)
=
Jzaoo
or fiJz6
?('ui
J\
,, dy 2t2
,",;=r=t=tlx+3
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@
2010 the College Board.
-')
)
rz. (a)xO=
"(r)=
.l
Ifiar= n(r+l)+C.
When t =t, x= ln2 so C =0.
rn(r+ r)
/t)= [ztat=f + D. when t=1,!=0
t(t)=P
r+ 1= e'
D=-1.
-t
(*,r)=(rn(r+
(b)
so
r),/ -r)
so t = e'
(c) Average rate
-l
and
y
=(r,
-r)t
-, = er, -2e'
.
rs-(-r)
v(+)- v(o)
-!g
- ='\ul-'14x\n)- xla) r(+)-x(o) ln5- lnt= ln5
or change
(d) Instantaneous rate ofchange
=
dyl
2t
dxl. =- I
-4
tI=l
t+l
13.
(a) o!
dx=',"?tl
3sinr=?cotr
3
(b)
-*ll
2))
y-(r*.,6) =11,-(,
-\
/3(.(.-
(c) x= 0 when / = -0.84106867..., 0.84106867...
rength
= Jl,llj;
(:rin r)'
+
(z"os t)' dt = 3.7 56 or
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2010 The College Board.
3.7 57