MATH 151
Answers to
Test 2
Autumn 2009
1.
A baseball diamond is a square with sides 90 feet long. A baseball player (
P
in the
figure below) is advancing from the second base (
S
) to the third base (
T
) and the umpire
at the home plate (
H
) is watching him.
Let
θ
be the angle between the third baseline
(
TH
) and the line of sight from the umpire to the player (
HP
). How fast is
θ
changing
when the runner is 30 feet from the third base and running at 24 feet per second?
Let
x
be the distance
PT
from the player to the third
base in feet.
In the triangle
HTP
,
TH
= 90
ft
is
constant. The angle
PTH
=
θ
and
x
are changing.
tan
θ
=
x/
90; so 90 tan
θ
=
x
.
Di
ff
erentiate both
sides with respect to time we get 90 sec
2
θ
d
θ
dt
=
dx
dt
.
When
x
= 30, tan
θ
= 30
/
90 = 1
/
3, and so
sec
2
θ
= 1 + tan
2
θ
= 1 + (1
/
3)
2
= 10
/
9.
As
dx
dt
= 24, 90
×
10
9
d
θ
dt
= 24.
Hence
d
θ
dt
=
24
100
.
Answer
:
0
.
24 radians per second
H
F
S
T
P
x
90
θ
2. Find the equation of the tangent line to the graph of the equation at the specified point.
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 Winter '08
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 Calculus, dx

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