AP Calc Notes: ID – 5 Derivative Interpretations
Ex: Suppose f(t) represents the height of a person in centimeters as a function of his/her age in years.
a. What does f(5) = 110 mean?
When the kid is (exactly) five years old, he is 110 cm tall.
b. What is the sign of f '(t)?
Positive, until approx age = 18, then zero until old age, then possibly negative
c. What are the units of f '(t)?
df
cm
has units of
dt
yr
d. What does f '(5) = 7 mean?
When the kid is (exactly) five years old, his height is increasing at a rate of 7 cm/yr.
e. What does f '(85) = -1 mean?
When the person is (exactly) eighty-five years old, her height is decreasing at a rate of 1
cm/yr.
f. Using f(5) = 110 and f '(5) = 7, estimate f(5.5).
We will assume the kid grows linearly during that time interval. In other words, we’ll use the
tangent line to f at t = 5 to approximate f.
f = 7(t – 5) + 110 → f(5.5) ≈ 7(5.5 – 5) + 110 = 113.5 cm
g. What are the units of f "(t)?
d2f
cm
has units of
per year = cm/yr2
2
yr
dt
h. What does f "(5) = -1.5 mean?
When the kid is (exactly) five years old, his rate of growth is decreasing by 1.5 cm/yr/yr.
i. Using the value f "(5) = -1.5, is the answer to part f above most likely an under- or over-estimate?
f
If f" stays negative over the interval [5, 5.5], then
f’ will be decreasing, f will be concave down, and
tangent line approximations will be overestimates.
5
5.5
Definition of the Derivative (Review)
f ' ( a ) = lim
x →a
or
f(x) - f(a)
f(x) - f(a)
≈
if x “close to” a
x-a
x-a
f(a + h) - f(a) f(a + h) - f(a)
if h “close to” 0
=
h→0
h
h
= lim
To get an estimate of f ',
f '( a) ≈ =
f(a + h) - f(a)
h
(h > 0)
“Right hand difference quotient.”
a
or
f '( a) ≈ =
f(a) - f(a - h)
h
a+h
(h > 0)
“Left hand difference quotient.”
a-h
or
f '( a) ≈ =
a
f(a + h) - f(a - h)
(h > 0)
2h
“Symmetric difference quotient.”
a-h
a
a+h
Ex: The table gives the temperatures in Slopetown, Canada over a 24 hour period on a recent day.
t (hours since midnight )
T (°C)
0
6.5
3
4.9
6
4.0
9
8.3
12
14.3
15
18.2
18
16.0
21
10.2
24
7.0
a. What is the average rate of change of temperature from 6 AM to 3 PM? What does this mean?
ΔT
18.2 - 4
=
= 1.578/hr
Δt
15 - 6
Between 6 AM and 3 PM, the temp increased 1.578 degrees C per hour on average.
b. Estimate the rate of change of temperature at
1) Midnight (0 hours) Must use RH difference quotient; take h as small as possible.
T'(0) ≈
4.9 - 6.5
= -0.533/hr
3-0
2) Midnight (24 hours) Must use LH difference quotient; take h as small as possible.
T'(24) ≈
7.0 - 10.2
= -1.067 /hr
24 - 21
3) Noon Can use any of the three difference quotients; take h as small as possible.
Using the symmetric difference quotient:
T'(12) ≈
18.2 - 8.3
= 1.65/hr
15 - 9
What does each of these values mean?
*Use the words increasing and decreasing with the absolute value of the rate of change.
At 0 hours, the temp was decreasing approx. 0.533 degrees C per hour.
At 24 hours, the temp was decreasing approx. 1.067 degrees C per hour.
At noon, the temp was increasing approx. 1.65 degrees C per hour.