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.