3.4, #10.
Change in income =∫120 r(t)dt = ∫120 40(1.002)t dt = $485.80
3.4, #14.
Notice that each square on the graph represents 10/6, or 5/3
miles. At t=1/3 hours, v=0. The area between the v graph
and the t-axis over the interval 0 < t < 1/3 is
−∫01/3v dt which is about one square or 5/3 miles.
At t=1/3 she is about 5−5/3 or about 10/3 miles from the lake.
v is positive on the interval 1/3 ≤ t ≤ 1, so she
is moving away from the lake on that interval. About 8 squares
are between the v graph and the t−axis. At t=1,
⌠ ⌡
1
0
v dt =
⌠ ⌡
1/3
0
v dt+
⌠ ⌡
1
1/3
v dt ≈ −
5
3
+8·
10
6
=
35
3
,
and the cyclist is about 5+[35/3]=[50/3] = 162/3 miles from the lake. Since, starting from the
moment t=1/3, she moves away from the lake, the
cyclist will be farthest away from the lake at t=1.
The maximal distance is 162/3 miles.
3.5, #2.
lim
t→∞
t3e−t=0
A viewing window with 0 ≤ x ≤ 20
(or larger x) and 0 ≤ y ≤ 2 suggests these limits.
(In fact, tne−1→ 0 as t→∞ for any
n > 0. This expresses in another way the fact
that exponential functions dominate power functions.)
3.5, #3.
lim
x→ ∞
sinx
x
=0
A viewing window with 0 ≤ x ≤ 20
(or larger x) and −2 ≤ y ≤ 2 suggests these limits.
It is easy to give a direct argument why the limit is 0.
The numerator sinx is always between −1 and 1 while the
denominator x gets arbitrarily large as x→∞.
3.5, #5.
If the exponent were a fixed number, like 3, we could bring
the limit inside the expression that is raised to the exponent. However,
the exponent n is not fixed, but rather is going to infinity
at the same time that the 1+1/n is going to one.
So we cannot bring the limit inside the exponent.
Now, if we were to ask the calculator to evaluate
(1+1/n)n for n= say 1020, it will try to evaluate
1+[1/(1020)] and it will have no choice but to round off the
answer to exactly 1. Then it will take 11020, which is 1.
File translated from
TEX
by
TTH,
version 4.05. On 15 Sep 2014, 12:58.