5 Steps to a 5 AP Calculus BC 2019

30 STEP 2. Determine Your Test Readiness

16. 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

17. 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.

18. 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

19. 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.

20. 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.

21. 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