200 STEP 4. Review the Knowledge You Need to Score High
- Step 1: Findmtangent.
y=
∣∣
x^3
∣∣
=
{
x^3 ifx≥ 0
−x^3 ifx< 0
dy
dx
=
{
3 x^2 ifx> 0
− 3 x^2 ifx< 0
Step 2: Setmtangent=slope of liney
− 12 x=3.
Sincey− 12 x= 3 ⇒y=
12 x+3, thenm=12.
Set 3x^2 = 12 ⇒x=±2 since
x≥0,x=2.
Set− 3 x^2 = 12 ⇒x^2 =−4. Thus
∅.
Step 3: Find the point on the curve. (See
Figure 9.8-1.)
[−3, 4] by [−5, 15]
Figure 9.8-1
Atx=2,y=x^3 = 23 =8.
Thus, the point is (2, 8).
- Step 1: Findmtangent.
y=ex;
dy
dx
=ex
dy
dx
∣∣
∣∣
x=ln 2
=eln 2= 2
Step 2: Findmnormal.
Atx=ln 2, mnormal=
− 1
mtangent
=−
1
2
.
Step 3: Write equation of normal.
Atx=ln 2,y=ex=eln 2=2. Thus
the point of tangency is (ln 2, 2).
The equation of normal:
y− 2 =−
1
2
(x−ln 2) or
y=−
1
2
(x−ln 2)+ 2.
- Step 1: Findmtangent.
y=−x^2 +4;
dy
dx
=− 2 x.
Step 2: Find the slope of liney− 2 x=b
y− 2 x=b⇒y= 2 x+borm= 2.
Step 3: Find point of tangency.
Setmtangent=slope of line
y− 2 x=b⇒− 2 x=
2 ⇒x=−1.
Atx=−1,y=−x^2 + 4 =
−(−1)^2 + 4 =3; (−1, 3).
Step 4: Findb.
Since the liney− 2 x=bpasses
through the point (−1, 3), thus
3 −2(−1)=borb=5.
7.v(t)=s′(t)=t^2 − 6 t;
a(t)=v′(t)=s′′(t)= 2 t− 6
Seta(t)= 0 ⇒ 2 t− 6 =0ort=3.
v(3)=(3)^2 −6(3)=−9;
s(3)=
(3)^3
3
−3(3)^2 + 4 =− 14.
8.On the interval (0, 1), the slope of the line
segment is 2. Thus the velocityv(t)=2 ft/s.
On (1, 3),v(t)=0 and on (3, 5),
v(t)=−1. (See Figure 9.8-2.)
1
–1
–2
1
v(t)
(^0) 23 45
2
v
t
Figure 9.8-2