366
6.4. PHYSICS APPLICATIONS: WORK, FORCE, AND PRESSURE
slice tend to zero, we find that the exact area of the region is
A =
b
a
f (x) dx.
In a similar way, if we know that the velocity of a moving object is given by the function
y = v(t), and we wish to know the distance the object travels on an interval [a, b] where
v(t) is nonnegative, we can use a definite integral to generalize the fact that d = r · t
when the rate, r, is constant. More specifically, on a short time interval △t, v(t) is roughly
constant, and hence for a small slice of time, d slice = v(t)△t, and so as the width of the
time interval △t tends to zero, the exact distance traveled is given by the definite integral
d =
b
a
v(t) dt.
Finally, when we recently learned about the mass of an object of non-constant density,
we saw that since M = D · V (mass equals density times volume, provided that density is
constant), if we can consider a small slice of an object on which the density is approximately
constant, a definite integral may be used to determine the exact mass of the object. For
instance, if we have a thin rod whose cross sections have constant density, but whose
density is distributed along the x axis according to the function y = ρ(x), it follows that
for a small slice of the rod that is △x thick, M slice = ρ(x)△x. In the limit as △x → 0, we
then find that the total mass is given by
M =
b
a
ρ(x) dx.
Note that all three of these situations are similar in that we have a basic rule (A = l · w,
d = r · t, M = D · V ) where one of the two quantities being multiplied is no longer constant;
in each, we consider a small interval for the other variable in the formula, calculate the
approximate value of the desired quantity (area, distance, or mass) over the small interval,
and then use a definite integral to sum the results as the length of the small intervals is
allowed to approach zero. It should be apparent that this approach will work effectively
for other situations where we have a quantity of interest that varies.
We next turn to the notion of work: from physics, a basic principal is that work is the
product of force and distance. For example, if a person exerts a force of 20 pounds to lift
a 20-pound weight 4 feet off the ground, the total work accomplished is
W = F · d = 20 · 4 = 80 foot-pounds.
If force and distance are measured in English units (pounds and feet), then the units on
work are foot-pounds. If instead we work in metric units, where forces are measured in
Newtons and distances in meters, the units on work are Newton-meters.
6.4. PHYSICS APPLICATIONS: WORK, FORCE, AND PRESSURE
slice tend to zero, we find that the exact area of the region is
A =
b
a
f (x) dx.
In a similar way, if we know that the velocity of a moving object is given by the function
y = v(t), and we wish to know the distance the object travels on an interval [a, b] where
v(t) is nonnegative, we can use a definite integral to generalize the fact that d = r · t
when the rate, r, is constant. More specifically, on a short time interval △t, v(t) is roughly
constant, and hence for a small slice of time, d slice = v(t)△t, and so as the width of the
time interval △t tends to zero, the exact distance traveled is given by the definite integral
d =
b
a
v(t) dt.
Finally, when we recently learned about the mass of an object of non-constant density,
we saw that since M = D · V (mass equals density times volume, provided that density is
constant), if we can consider a small slice of an object on which the density is approximately
constant, a definite integral may be used to determine the exact mass of the object. For
instance, if we have a thin rod whose cross sections have constant density, but whose
density is distributed along the x axis according to the function y = ρ(x), it follows that
for a small slice of the rod that is △x thick, M slice = ρ(x)△x. In the limit as △x → 0, we
then find that the total mass is given by
M =
b
a
ρ(x) dx.
Note that all three of these situations are similar in that we have a basic rule (A = l · w,
d = r · t, M = D · V ) where one of the two quantities being multiplied is no longer constant;
in each, we consider a small interval for the other variable in the formula, calculate the
approximate value of the desired quantity (area, distance, or mass) over the small interval,
and then use a definite integral to sum the results as the length of the small intervals is
allowed to approach zero. It should be apparent that this approach will work effectively
for other situations where we have a quantity of interest that varies.
We next turn to the notion of work: from physics, a basic principal is that work is the
product of force and distance. For example, if a person exerts a force of 20 pounds to lift
a 20-pound weight 4 feet off the ground, the total work accomplished is
W = F · d = 20 · 4 = 80 foot-pounds.
If force and distance are measured in English units (pounds and feet), then the units on
work are foot-pounds. If instead we work in metric units, where forces are measured in
Newtons and distances in meters, the units on work are Newton-meters.