Transcript Slide 1

Part (a)
The integral
is equal to the area
the(in
graph. In
This represents
total under
distance
this case,meters)
we’re adding
twobytriangles
a rectangle.
traveled
the car and
during
the first 24 seconds.
24
v(t) dt =
½(4)(20)
+
(12)(20) +
½(8)(20)
0
24
0
v(t) dt = 40 + 240 + 80
=
360 meters
Part (b)
m=5
m=0
m=-5/2
v’(4) does not exist. The reason why is that the
limit
t approaches
4 from
left is We
5, while
the
v’(20)asequals
-5/2 meters
perthe
second.
get this
limit
as tfrom
approaches
4 of
from
is as
zero.
Since
answer
the slope
thethe
v(t)right
graph
it passes
the left- and right-hand
through limits
t=20. are different, no
overall limit exists at t=4. Therefore, there is no
derivative there either.
Part (c)
Acceleration is the derivative (slope)
of the velocity graph.
a(t) =
5
0
-5/2
DNE
when
when
when
when
0 < t < 4
4 < t < 16
16 < t < 24
t=4 or t=16
Part (d)
(20,10)
(8,20)
Since this function is piecewise, the only way to find
the average rate of change is to use the slope formula
from Algebra 1.
Avg. rate of change =
10-20
20-8
=
-5
6
m/sec
Normally, the Mean Value Theorem would tell us that
the slope of the curve is -5/6 somewhere between t=8
and t=20. From Part (c), however, we know that the
slope is never actually equal to -5/6 anywhere.
Part (d)
(8,20)
(20,10)
The reason that the Mean Value Theorem can’t
guarantee us a slope of -5/6 is because the MVT can
only be used on those portions of functions that are
fully differentiable. Remember, our function has no
derivative at t=16. Therefore, the MVT can’t be used.