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

MA 3972-MA-Book April 11, 2018 12:11

Take a Diagnostic Exam 29

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.

24. See the figure below.
    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
      
      
      
      
    
    
    
    
  
  
  
  

25. v(t)=t

3

3

− 2 t^2 + 5

a(t)=v′(t)=t^2 − 4 t

See the figure below.

The graph indicates that for 0≤t≤6, the

maximum acceleration occurs at the endpoint

t=6.a(t)=t^2 − 4 tanda(6)= 62 −4(6)= 12.

26. y=x^3 ,x≥0;

dy

dx

= 3 x^2

f′(x)= 12 ⇒

dy

dx

= 3 x^2 = 12

⇒x^2 = 4 ⇒x= 2

Slope of normal=negative reciprocal of slope

of tangent=−

1

12

.

Atx=2,y=x^3 = 23 =8; (2, 8)

y− 8 =−

1

12

(x−2).

Equation of normal line:⇒y=−

1

12

(x−2)+ 8

ory=−

1

12

x+

49

6

.

27. f(x)=
    lnx
    x
    ; f′x=
    (1/x)(x)−(1)lnx
    x^2

=

1

x^2

−

lnx

x^2

y=−x^2 ;

dy

dx

=− 2 x

Perpendicular tangents

⇒(f′(x))

(

dy

dx

)

=− 1

⇒

((

1

x^2

)

−

lnx

x^2

)

(− 2 x)=− 1.

Using the [Solve] function on your calculator,

you obtainx≈ 1. 37015 ≈ 1 .370.