A dead body was found within a closed room of a house where the temperature was a constant 65 f. At the time of discovery the core temperature of the body was determined to be 85 f. One hour later a second measurement showed that the core temperature of the body was 80 f. Assume that the time of death corresponds to t = 0 and that the core temperature at that time was 98.6 f. Determine how many hours elapsed before the body was found. [hint: let t1 > 0 denote the time that the body was discovered.] (round your answer to one decimal place.)
Answers
Answer:
Math 16B: Homework 5 Solutions
Due: July 30
1. Find the nth Taylor polynomial for each of the following functions at the given point:
(a) n = 2: f(x) = x
1/3 at x = 27:
We have f
′
(x) = 1
3x2/3 and f
′′(x) = −2
9x5/3
. Since f(27) = 3, f
′
(27) = 1
27 and
f
′′(27) = −2
2187 , the 2nd Taylor polynomial of f centered at x = 27 is
P(x) = 3 + (
1
27 )
1
(x − 27) + (
−2
2187 )
2
(x − 27)2
= 3 +
x − 27
27
−
(x − 27)2
2187
.
(b) n = 2: f(x) = tan(x) at x =
π
4
:
We have f
′
(x) = sec2
(x) and f
′′(x) = 2 sec(x)(sec(x) tan(x)) = 2 sec2
(x) tan(x).
Since f(π/4) = 1, f
′
(π/4) = (√
2)2 = 2 and f
′′(π/4) = 2(2)(1) = 4, the 2nd
Taylor polynomial of f centered at x =
π
4
is
P(x) = 1 + (2)
1
(
x −
π
4
)
+
(4)
2
(
x −
π
4
)2
= 1 + 2 (
x −
π
4
)
+ 2 (
x −
π
4
)2
.
(c) n = 3: f(x) = 1
7−x
at x = 5:
We have f
′
(x) = 1
(7−x)
2
, f
′′(x) = 2
(7−x)
3 and f
′′′(x) = 6
(7−x)
4
. Since f(5) = 1
2
,
f
′
(5) = 1
4
, f
′′(5) = 2
8 =
1
4
and f
′′′(5) = 6
16 =
3
8
, the 3rd Taylor polynomial of f
centered at x = 5 is
P(x) = 1
2
+
(
1
4
)
1
(x − 5) + (
1
4
)
2
(x − 5)2 +
(
3
8
)
3! (x − 5)3
=
1
2
+
(x − 5)
4
+
(x − 5)2
8
+
(x − 5)3
16
.
(d) n = 4: f(x) = ln(x) at x = 1:
We have f
′
(x) = 1
x
, f
′′(x) = −1
x2
, f
′′′(x) = 2
x3 and f
(4)(x) = −6
x4
. Since f(1) = 0,
f
′
(1) = 1, f
′′(1) = −1, f
′′′(1) = 2 and f
(4)(1) = −6 the 4th Taylor polynomial of
f centered at x = 1 is
P(x) = 0 + (1)
1
(x − 1) + (−1)
2
(x − 1)2 +
(2)
3! (x − 1)3 +
(−6)
4! (x − 1)4
= (x − 1) −
1
2
(x − 1)2 +
1
3
(x − 1)3 −
1
4
(x − 1)4
.