30 STEP 2. Determine Your Test Readiness
- Solvex= 1 +e−tfort.x− 1 =e−t⇒
−ln (x−1)=t.Substitute iny= 1 +et.
y= 1 +e−ln (x−1)⇒y= 1 +
1
x− 1
⇒y=
x
x− 1
Chapter 8
- Letzbe the diagonal of a square. Area of a
squareA=
z^2
2
dA
dt
=
2 z
2
dz
dt
=z
dz
dt
.
Since
dA
dt
= 4
dz
dt
;4
dz
dt
=z
dz
dt
⇒z= 4.
Letsbe a side of the square. Since the
diagonalz=4,s^2 +s^2 =z^2 or 2s^2 =16. Thus,
s^2 =8ors= 2
√
2.
- See Figure DS-7.
The graph ofgindicates that a relative
maximum occurs atx=2;gis not
differentiable atx=6, since there is acuspat
x=6; andgdoes not have a point of
inflection atx=−2, since there is no tangent
line atx=−2. Thus, only statement I is true.
Figure DS-7
Chapter 9
- y=
√
x− 1 =(x−1)^1 /^2 ;
dy
dx
=
1
2
(x−1)−^1 /^2
=
1
2(x−1)^1 /^2
dy
dx
∣∣
∣∣
x= 5
=
1
2(5−1)^1 /^2
=
1
2(4)^1 /^2
=
1
4
Atx=5, y=
√
x− 1 =
√
5 − 1
=2; (5, 2).
Slope of normal line=negative reciprocal of
(
1
4
)
=−4.
Equation of normal line:
y− 2 =−4(x−5)⇒y=−4(x−5)+2or
y=− 4 x+ 22.
- y=cos(xy);
dy
dx
=[−sin(xy)]
(
1 y+x
dy
dx
)
dy
dx
=−ysin(xy)−xsin(xy)
dy
dx
dy
dx
+xsin(xy)
dy
dx
=−ysin(xy)
dy
dx
[1+xsin(xy)]=−ysin(xy)
dy
dx
=
−ysin(xy)
1 +xsin(xy)
Atx=0, y=cos(xy)=cos(0)
=1; (0, 1)
dy
dx
∣∣
∣∣
x=0,y= 1
=
−(1) sin(0)
1 +0 sin(0)
=
0
1
= 0.
Thus, the slope of the tangent atx=0is0.
- See Figure DS-8.
v(t)=t^2 −t
Setv(t)= 0 ⇒t(t−1)= 0
⇒t=0ort= 1
a(t)=v′(t)= 2 t− 1.
Seta(t)= 0 ⇒ 2 t− 1 =0ort=
1
2
.
Sincev(t)<0 anda(t)>0on
(
1
2
,1
)
, the
speed of the particle is decreasing on
(
1
2
,1
)
.
[
(^012)
0
1
V(t) 0
a(t)
t
- – – – – – – – – – – – – – – – – – – – + + + + +
- – – – – – – – + + + + + + + + + + + + + + + + + +
0
Figure DS-8