The Second Derivative and
Introducing Motion Applications
Learning Goal: To calculate the second derivative and use derivatives to
determine the position, velocity and acceleration functions.
Success Criteria: Students can…
• determine the second derivative of a function.
• determine the velocity and acceleration functions given a position
function.
• understand how the concept of direction is conveyed in a motion
problem.
Given a function 𝑦 = 𝑓 𝑥 , we observe that the derivative 𝑦 = 𝑓′(𝑥) is also
a function. From now on, let us refer to 𝑦 = 𝑓′(𝑥) as the first derivative.
Since the first derivative is a function, one might wonder whether we can
take its derivative.
• We define the second derivative of a function 𝑦 = 𝑓 𝑥 to be the
derivative of the first derivative function 𝑦 = 𝑓′(𝑥).
• Prime Notation: 𝒚 = 𝒇′′(𝒙) “f double prime at x”
• Leibniz Notation:
𝒅𝟐 𝒚
𝒅𝒙𝟐
“d two y by d x squared”
• Since the second derivative is also a function, it is also possible to find the
third derivative, fourth derivative, etc. These will not be needed for our
course.
• What information does the second derivative function tell us? For now, we
can think of the output values of the second derivative corresponding to
the slopes of the tangents on the graph of the first derivative function.
Ex: If 𝑓 𝑥 =
𝑥
,
𝑥−1
evaluate 𝑓′′(2).
1 𝑥−1 − 𝑥 1
−1
𝑥 =
=
2
𝑥−1
𝑥−1
2
′′
−3
𝑓 𝑥 =2 𝑥−1
1 =
𝑥−1 3
𝑓′
𝑓 ′′
2
2 =
2−1
3
=2
2
=− 𝑥−1
−2
Motion
Consider a position function s(t) that describes the position of an object
along a straight line as a function of time.
• In physics, you will know that position is defined to be a vector quantity.
You would be taught that vectors are quantities that include a magnitude
and direction. Direction can be rather confusing to describe in general and
will depend on the context of the question. (We will see different
contexts in the vectors portion of the course). At our level of calculus,
since we are assuming the object must move along a straight line (i.e. in
one dimension), then there are only TWO possible directions to consider.
The sign (+ or −) will distinguish between the two directions.
Up (+)
Left (−)
Right (+)
Down (−)
• We define the (instantaneous) velocity to be the function v t = 𝑠′(𝑡).
• Recall: velocity is the rate of change of position
• We define the (instantaneous) acceleration to be the function
𝑎 𝑡 = 𝑠 ′′ 𝑡 = 𝑣′(𝑡)
• Acceleration is the rate of change of velocity
Ex: A common formula taught in physics is for the height of an object thrown
in the air (assume no air or wind resistance). An example of such a function
might be 𝑠 𝑡 = −4.9𝑡 2 + 8𝑡 + 15 where 𝑠 is the height in meters and 𝑡 is the
time in seconds. You might notice that the leading coefficient is almost always
− 4.9. Determine the acceleration function and explain why this is the case.
𝑣 𝑡 = −9.8𝑡 + 8
𝑎 𝑡 = −9.8
Note: −9.8 𝑚/𝑠 2 is approximately the known constant representing the
acceleration due to gravity on earth.