5 Steps to a 5 AP Calculus BC 2019

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

The particle is initially at 1 (t=0). It

moves to the left speeding up untilt=1,

when it reaches−1. Then it continues

moving to the left, but slowing down until

t=2at−3. The particle reverses

direction, moving to the right and

speeding up indefinitely.

14. Linear approximation:
    y=f(a)+f′(a)(x−a)a=π
    f(x)=sinxand f(π)=sinπ= 0
    f′(x)=cosxand f′(π)=cosπ=−1.
    Thus,y= 0 +(−1)(x−π)or
    y=−x+π.
    f

(

181 π

180

)

is approximately:

y=−

⎛

⎝^181 π

180

⎞

⎠+π= −π

180

or≈

− 0. 0175.

15. y=f(a)+f′(a)(x−a)
    f(x)=ln (1+x) and
    f(2)=ln (1+2)=ln 3

f′(x)=

1

1 +x

and f′(2)=

1

1 + 2

=

1

3

.

Thus,y=ln 3+

1

3

(x−2).

16. Step 1: Find
    dy
    dx

.

y^2 = 4 − 4 x^2

2 y

dy

dx

=− 8 x⇒

dy

dx

=

− 4 x

y

Step 2: Find

dx

dy

.

dx

dy

=

1

dy/dx

=

1

− 4 x/y

=

−y

4 x

Set

dx

dy

= 0 ⇒

−y

4 x

=0ory= 0.

Step 3: Find points of tangency.

Aty=0,y^2 = 4 − 4 x^2 becomes

0 = 4 − 4 x^2

⇒x=±1.

Thus, points of tangency are

(1, 0) and (−1, 0).

Step 4: Write equations of vertical

tangentsx=1 andx=−1.

17. Step 1: Find
    dy
    dx
    fory=lnxand
    y=x^2 +3.
    y=lnx;
    dy
    dx

=

1

x

y=x^2 +3;

dy

dx

= 2 x

Step 2: Find thex-coordinate of point(s)

of tangency.

Parallel tangents⇒slopes are

equal. Set

1

x

= 2 x.

Using the [Solve] function of your

calculator, enter [( Solve]

1

x

= 2 x,x

)

and obtain

x=

√

2

2

orx=

−

√

2

2

. Since for

y=lnx,x>0,x=

√

2

2

.

18. s 1 (t)=lntands 1 ′(t)=

1

t

;1≤t≤8.

s 2 (t)=sin(t) and

s 2 ′(t)=cos(t); 1≤t≤8.

Entery 1 =

1

x

andy 2 =cos(x). Use the

[Intersection] function of the calculator and

obtaint= 4 .917 andt= 7 .724.

19. Step 1: s(t)=sint
    v(t)=cost
    a(t)=−sint
    Step 2: Setv(t)= 0 ⇒cost=0;
    t=
    π
    2
    and
    3 π
    2

.

Seta(t)= 0 ⇒−sint=0;

t=πand 2π.

Step 3: Determine the directions of

motion. (See Figure 9.8-6.)