3.5. RELATED RATES
203
account the increasing distance between the camera and the rocket. We assume that
the rocket rises vertically. (A similar problem is discussed and pictured dynamically
at http://gvsu.edu/s/9t. Exploring the applet at the link will be helpful to you in
answering the questions that follow.)
(a) Draw a figure that summarizes the given situation. What parts of the picture
are changing? What parts are constant? Introduce appropriate variables to
represent the quantities that are changing.
(b) Find an equation that relates the camera’s angle of elevation to the height of
the rocket, and then find an equation that relates the instantaneous rate of
change of the camera’s elevation angle to the instantaneous rate of change of
the rocket’s height (where all rates of change are with respect to time).
(c) Find an equation that relates the distance from the camera to the rocket to
the rocket’s height, as well as an equation that relates the instantaneous rate
of change of distance from the camera to the rocket to the instantaneous rate
of change of the rocket’s height (where all rates of change are with respect to
time).
(d) Suppose that the rocket’s speed is 600 ft/sec at the instant it has risen 3000
feet. How fast is the distance from the television camera to the rocket changing
at that moment? If the camera is following the rocket, how fast is the camera’s
angle of elevation changing at that same moment?
(e) If from an elevation of 3000 feet onward the rocket continues to rise at 600
feet/sec, will the rate of change of distance with respect to time be greater when
the elevation is 4000 feet than it was at 3000 feet, or less? Why?
⊳
In addition to being able to find instantaneous rates of change at particular points
in time, we are often able to make more general observations about how particular rates
themselves will change over time. For instance, when a conical tank (point down) is filling
with water at a constant rate, we naturally intuit that the depth of the water should increase
more slowly over time. Note how carefully we need to speak: we mean to say that while the
depth, h, of the water is increasing, its rate of change
dh
dt is decreasing (both as a function
of t and as a function of h). These observations may often be made by taking the general
equation that relates the various rates and solving for one of them, and doing this without
substituting any particular values for known variables or rates. For instance, in the conical
tank problem in Activity 3.14, we established that
dV
dt
=
1
16
πh
2 dh
dt
,
and hence
dh
dt
=
16
πh 2
dV
dt
.
203
account the increasing distance between the camera and the rocket. We assume that
the rocket rises vertically. (A similar problem is discussed and pictured dynamically
at http://gvsu.edu/s/9t. Exploring the applet at the link will be helpful to you in
answering the questions that follow.)
(a) Draw a figure that summarizes the given situation. What parts of the picture
are changing? What parts are constant? Introduce appropriate variables to
represent the quantities that are changing.
(b) Find an equation that relates the camera’s angle of elevation to the height of
the rocket, and then find an equation that relates the instantaneous rate of
change of the camera’s elevation angle to the instantaneous rate of change of
the rocket’s height (where all rates of change are with respect to time).
(c) Find an equation that relates the distance from the camera to the rocket to
the rocket’s height, as well as an equation that relates the instantaneous rate
of change of distance from the camera to the rocket to the instantaneous rate
of change of the rocket’s height (where all rates of change are with respect to
time).
(d) Suppose that the rocket’s speed is 600 ft/sec at the instant it has risen 3000
feet. How fast is the distance from the television camera to the rocket changing
at that moment? If the camera is following the rocket, how fast is the camera’s
angle of elevation changing at that same moment?
(e) If from an elevation of 3000 feet onward the rocket continues to rise at 600
feet/sec, will the rate of change of distance with respect to time be greater when
the elevation is 4000 feet than it was at 3000 feet, or less? Why?
⊳
In addition to being able to find instantaneous rates of change at particular points
in time, we are often able to make more general observations about how particular rates
themselves will change over time. For instance, when a conical tank (point down) is filling
with water at a constant rate, we naturally intuit that the depth of the water should increase
more slowly over time. Note how carefully we need to speak: we mean to say that while the
depth, h, of the water is increasing, its rate of change
dh
dt is decreasing (both as a function
of t and as a function of h). These observations may often be made by taking the general
equation that relates the various rates and solving for one of them, and doing this without
substituting any particular values for known variables or rates. For instance, in the conical
tank problem in Activity 3.14, we established that
dV
dt
=
1
16
πh
2 dh
dt
,
and hence
dh
dt
=
16
πh 2
dV
dt
.