4.1. DETERMINING DISTANCE TRAVELED FROM VELOCITY
213
at 3, we know that if 3 s(t) = 3t, then s ′ (t) = 3, so s(t) = 3t is a function whose derivative
is v(t). Furthermore, we now observe that s(1.5) = 4.5 and s(0.25) = 0.75, which are the
respective locations of the person at times t = 0.25 and t = 1.5, and therefore
s(1.5) − s(0.25) = 4.5 − 0.75 = 3.75 miles.
This is not only the change in position on [0.25, 1.5], but also precisely the distance
traveled on [0.25, 1.5], which can also be computed by finding the area under the velocity
curve over the same interval. There are profound ideas and connections present in this
example that we will spend much of the remainder of Chapter 4 studying and exploring.
For now, it is most important to observe that if we are given a formula for a velocity
function v, it can be very helpful to find a function s that satisfies s ′ = v. In this context, we
say that s is an antiderivative of v. More generally, just as we say that f ′ is the derivative of
f for a given function f , if we are given a function g and G is a function such that G ′ = g,
we say that G is an antiderivative of g. For example, if g(x) = 3x 2 + 2x, an antiderivative
of g is G(x) = x 3 + x 2 , since G ′ (x) = g(x). Note that we say “an” antiderivative of g
rather than “the” antiderivative of g because H(x) = x 3 + x 2 + 5 is also a function whose
derivative is g, and thus H is another antiderivative of g.
Activity 4.2.
A ball is tossed vertically in such a way that its velocity function is given by v(t) =
32 − 32t, where t is measured in seconds and v in feet per second. Assume that this
function is valid for 0 ≤ t ≤ 2.
(a) For what values of t is the velocity of the ball positive? What does this tell you
about the motion of the ball on this interval of time values?
(b) Find an antiderivative, s, of v that satisfies s(0) = 0.
(c) Compute the value of s(1) − s(
1
2 ). What is the meaning of the value you find?
(d) Using the graph of y = v(t) provided in Figure 4.6, find the exact area of the
region under the velocity curve between t =
1
2 and t = 1. What is the meaning
of the value you find?
(e) Answer the same questions as in (c) and (d) but instead using the interval [0, 1].
(f) What is the value of s(2) − s(0)? What does this result tell you about the flight
of the ball? How is this value connected to the provided graph of y = v(t)?
Explain.
⊳
3 Here we are making the implicit assumption that s(0) = 0; we will further discuss the different possibilities
for values of s(0) in subsequent study.
213
at 3, we know that if 3 s(t) = 3t, then s ′ (t) = 3, so s(t) = 3t is a function whose derivative
is v(t). Furthermore, we now observe that s(1.5) = 4.5 and s(0.25) = 0.75, which are the
respective locations of the person at times t = 0.25 and t = 1.5, and therefore
s(1.5) − s(0.25) = 4.5 − 0.75 = 3.75 miles.
This is not only the change in position on [0.25, 1.5], but also precisely the distance
traveled on [0.25, 1.5], which can also be computed by finding the area under the velocity
curve over the same interval. There are profound ideas and connections present in this
example that we will spend much of the remainder of Chapter 4 studying and exploring.
For now, it is most important to observe that if we are given a formula for a velocity
function v, it can be very helpful to find a function s that satisfies s ′ = v. In this context, we
say that s is an antiderivative of v. More generally, just as we say that f ′ is the derivative of
f for a given function f , if we are given a function g and G is a function such that G ′ = g,
we say that G is an antiderivative of g. For example, if g(x) = 3x 2 + 2x, an antiderivative
of g is G(x) = x 3 + x 2 , since G ′ (x) = g(x). Note that we say “an” antiderivative of g
rather than “the” antiderivative of g because H(x) = x 3 + x 2 + 5 is also a function whose
derivative is g, and thus H is another antiderivative of g.
Activity 4.2.
A ball is tossed vertically in such a way that its velocity function is given by v(t) =
32 − 32t, where t is measured in seconds and v in feet per second. Assume that this
function is valid for 0 ≤ t ≤ 2.
(a) For what values of t is the velocity of the ball positive? What does this tell you
about the motion of the ball on this interval of time values?
(b) Find an antiderivative, s, of v that satisfies s(0) = 0.
(c) Compute the value of s(1) − s(
1
2 ). What is the meaning of the value you find?
(d) Using the graph of y = v(t) provided in Figure 4.6, find the exact area of the
region under the velocity curve between t =
1
2 and t = 1. What is the meaning
of the value you find?
(e) Answer the same questions as in (c) and (d) but instead using the interval [0, 1].
(f) What is the value of s(2) − s(0)? What does this result tell you about the flight
of the ball? How is this value connected to the provided graph of y = v(t)?
Explain.
⊳
3 Here we are making the implicit assumption that s(0) = 0; we will further discuss the different possibilities
for values of s(0) in subsequent study.