Answers and Explanations 321
Answers
401–500
them into the velocity function to see whether the velocity is positive or negative on
those intervals. Using t = 1, you have v(1) = 3(1)
2
– 4 < 0, and using t = 10, you have
v(10) = 3(10)
2
– 4 > 0; therefore, the particle is moving downward on the interval
0 2
3
,






 and upward on the interval 2
3
, ∞





 .
497.
downward on 0 3
4
,

 
) ; upward on 3
4
, ∞
( )
Note that because the velocity is the rate of change in position with respect to time,
you want to find when the velocity is positive and when the velocity is negative.
Assume that when the velocity is positive, the particle is moving upward, and when
the velocity is negative, the particle is moving downward.
Begin by taking the derivative of the position function to find the velocity function:
y
t
t
v y
t
=
− −
= ′ = −
4
6 2
2
8 6
Then find when the velocity is equal to zero:
8 6 0
6
8
3
4
t
t
− =
= =
Now take a point from the interval 0 3
4
,
( ) and a point from 3
4
, ∞
( ) and substitute them
into the velocity function to see whether the velocity is positive or negative on
those intervals. If you use t = 1
2
, you have v 1
2
8 1
2
6 0
( ) ( )
=
− < , and if t = 1, you have
v(1) = 8(1) – 6 > 0; therefore, the particle is moving downward on the interval 0 3
4
,

 
)
and upward on the interval 3
4
, ∞
( ) .
498.
–25, 25
Let x and y be the two numbers so that x – y = 50. The function that you want to
minimize, the product, is given by
P x y
x y
( , ) = ( )( )
Solve for x so you can write the product in terms of one variable:
x y
x
y
− =
= +
50
50
Then substitute the value of x into P:
P
y y
y
y
= ( + )( )
=
+
50
50
2
Next, find the derivative, set it equal to zero, and solve for y:
′
P
y
y
y
=
+
=
+
− =
2
50
0 2
50
25