5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 15:14

222 STEP 4. Review the Knowledge You Need to Score High

Thenf′(x)=−sinxand

f′( 45 ◦)= f′

(

π

4

)

=−

√

2

2

f( 46 ◦) =f

(

23 π

90

)

=f

(

π

4

+

π

180

)

f

(

π

4

+

π

180

)

≈ f

(

π

4

)

+

f′

(

π

4

)(

π

180

)

≈

√

2

2

−

(√

2

2

)(

π

180

)

≈

√

2

2

−

π

√

2

360

4. 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 10.7-1.)

[–3, 4] by [–5, 15]

Figure 10.7-1

Atx=2,y=x^3 = 23 =8.

Thus, the point is (2, 8).

5. 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 the normal line.

Atx=ln 2,y=ex=eln 2=2. Thus,

the point of tangency is (ln 2, 2).

The equation of the normal line:

y− 2 =−

1

2

(x−ln 2) or

y =−

1

2

(x−ln 2)+ 2.

6. 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/sec.
   On the interval (1, 3),v(t)=0 and on (3, 5),
   v(t)=−1. (See Figure 10.7-2.)