Name: ________________
Test #3
Applications of Derivatives
Math 160
Calculators are permitted, but show all work.
1.
Solve either of the problems below.
A. A boat is tied to a dock as shown in the picture below. A winch on the dock is connected to the boat,
and when the winch is turned, the rope is drawn in, and the boat moves closer to the dock. The top of
the winch is 5 feet above the point where the rope attaches to the boat. When the winch is turned, the
rope is pulled in at a constant rate of 2 ft per second.
How fast is the boat approaching the dock when it is 12 feet away?
y
doesn't change, so, its rate of change is 0.
r
is decreasing by 2 ft/s, so its rate of change is 2
So, the boat is getting closer to the dock at a rate of 13/6 ft/sec.
5 ft.
r
y
x
x
2
+
y
2
=
r
2
When
x
= 12,
r
= 13
2
x
dx
dt
2
y
dy
dt
=
2
r
dr
dt
x
dx
dt
y
dy
dt
=
r
dr
dt
12
dx
dt
0
=
13
−
2
dx
dt
=
−
13
6
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
B. For the graph of
y
2
x
3
10
y
9
=
0
find the slope,
dy
dx
, of the tangent line.
At what point(s) does this graph have a horizontal tangent?
At what point(s) does this graph have a vertical tangent?
Solve for
y
':
The graph has a horizontal tangent when the numerator is 0, and a vertical tangent when the
denominator is 0.
Horizontal:
so
x
= 0. We need the point (
x
,
y
), so we plug 0 in for
x
in the original equation and find
y
.
y
2
10
y
9
=
0
This is a quadratic equation, giving
y
= 9 and
y
= 1. Thus, the graph has a
horizontal tangent at (0, 1) and (0, 9).
Vertical:
2
y
10
=
0
so
y
= 5. Plugging this in for
y
, we get
25
x
3
−
50
9
=
0
and
x
=
3
16
. The graph
has a vertical tangent at
3
16
,
−
5
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 STAFF
 Derivative, lim, horizontal tangent, maximum, absolute maximum, vertical tangent

Click to edit the document details