5 Steps to a 5 AP Calculus BC 2019

More Applications of Derivatives 199

8 7 6 5 4 3 2 1

0123456789

t

v

v(t)

(feet/sec)

(seconds)

Figure 9.7-1

31. Find the Cartesian equation for the curve
    defined byr=4 cosθ.


32. The motion of an object is modeled by
    x=5 sint, y= 1 −cost. Find the
    y-coordinate of the object at the moment
    when itsx-coordinate is 5.


33. Calculate 4u− 3 vifu=

〈

6,− 1

〉

and

v=

〈

−4, 3

〉

.

34. Determine the symmetry, if any, of the
    graph ofr=2 sin(4θ).


35. Find the magnitude of the vector 3i+ 4 j.

9.8 Solutions to Practice Problems

Part A The use of a calculator is not

allowed.

1. Equation of tangent line:
   y=f(a)+f′(a)(x−a)
   f′(x)=

1

4

( 1 +x)−^3 /^4 (1)=

1

4

( 1 +x)−^3 /^4

f′(0)=

1

4

and f(0)=1;

thus,y= 1 +

1

4

(x− 0 )= 1 +

1

4

x.

f(0.1)= 1 +

1

4

(0.1)= 1. 025

2. f(a+Δx)≈ f(a)+ f′(a)Δx

Letf(x)=^3

√

xand f( 28 )=

f( 27 + 1 ).

Thenf′(x)=

1

3

(x)−^2 /^3 ,

f′( 27 )=

1

27

, and f( 27 )=3.

f( 27 + 1 )≈ f( 27 )+f′( 27 )( 1 )≈

3 +

(

1

27

)

( 1 )≈ 3. 037

3. f(a+Δx)≈ f(a)+f′(a)Δx
   Convert to radians:
   46
   180

=

a

π

⇒a=

23 π

90

and 1◦=

π

180

;

45 ◦=

π

4

.

Let f(x)=cosxandf(45◦)=

f

(

π

4

)

=cos

(

π

4

)

=

√

2

2

.

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