5 Steps to a 5 AP Calculus BC 2019

More Applications of Derivatives 193

the path. The acceleration vector is

〈

d^2 x

dt^2

,

d^2 y

dt^2

〉

and the magnitude of acceleration is

|a|=

√(

d^2 x

dt^2

) 2

+

(

d^2 y

dt^2

) 2

. The vectorTtangent to the path attisT(t)=
∥r′(t)
∥r′(t)∥∥and

the normal vector attisN(t)=
∥T′(t)
∥T′(t)∥∥.

Example 1

The position functionr=

〈

t^3 ,t^2

〉
=t^3 i+t^2 jdescribes the path of an object moving in the
plane. Find the velocity and acceleration of the object at the point (8, 4).

Step 1: The velocityv=

〈

dx

dt

,

dy

dt

〉

=

〈

3 t^2 ,2t

〉

. At the point (8, 4),t=2. Evaluated at
t=2, the velocityv=

〈

12, 4

〉

. The speed|v|=

√

144 + 16 ≈ 12 .649.

Step 2: The acceleration vector

〈

d^2 x

dt^2

,

d^2 y

dt^2

〉

=

〈

6 t,2

〉

. Evaluated att=2, the acceleration
is

〈

12, 2

〉

. The magnitude of the acceleration is|a|=

√

144 + 4 ≈ 12 .166.

Example 2

The left field fence in Boston’s Fenway Park, nicknamed the Green Monster, is 37 feet high
and 310 feet from home plate. If a ball is hit 3 feet above the ground and leaves the bat at
an angle of
π
4
, write a vector-valued function for the path of the ball and use the function

to determine the minimum speed at which the ball must leave the bat to be a home run. At
that speed, what is the maximum height the ball attains?

Step 1: The horizontal component of the ball’s motion, the motion in the “x” direc-

tion, isx=s·cos

π

4

·t=

s

√

2

2

t. The vertical component follows the parabolic

motion model y = 3 +s ·sin

π

4

t −

1

2

gt^2 , whereg is the acceleration due to

gravity. The path of the ball can be represented by the vector-valued function

r=

〈

s·

√

2

2

t,3+

s·

√

2

2

t− 16 t^2

〉

.

Step 2: In order for the ball to clear the fence, its height must be greater than 37 feet when

its distance from the plate is 310 feet.

s·

√

2

2

t=310, solved fort, givest=

620

s·

√

2

seconds. At this time, 3+

s·

√

2

2

t− 16 t^2 = 3 +

s

√

2

2

(

620

s

√

2

)

− 16

(

620

s

√

2

) 2

, and

this value must exceed 37 feet. Setting 3+

s·

√

2

2

(

620

s·

√

2

)

− 16

(

620

s·

√

2

) 2

= 37

and solving givess ≈ 105 .556. The ball must leave the bat at 105.556 feet per

second in order to clear the wall.