### Written Homework 10 Solutions

```Written Homework 10 Solutions
MATH 1310 - CSM
Assignment: pp 217-210; problems: 1(abcd), 3, 4(ab), 5, 7. pp 238-240; problems: 3, 4, 5,
10. pp 48; problem 27.
1. Solutions:
1
(a) y = 1
=
( 5 − t)2
0
1
1
( 5 − t)

2
= y2.
3
1
1
 = y3.
= q
3/2
1
− 2t)
25 − 2t

4
1
1
 = y4.
(c) y 0 = 1
= q
4/3
1
3
( 125 − 3t)
− 3t
(b) y 0 =
1
( 25
125
(d) So the general case should look like: y(t) =
1
r
n−1
1
5n−1
3. Solutions:
(a) x0 = − sin(t) = −(sin t) = −y.
y 0 = cos t = x.
(b) x0 = − sin(t + π/2) = −(sin(t + π/2)) = −y.
y 0 = cos(t + π/2) = x.
4. Solutions:
(a) y 0 = ln(10) · 5 · 10t ≈ 2.3 · 5 · 10t = 2.3y.
(b) y 0 = ln(10) · C · 10t ≈ 2.3 · C · 10t = 2.3y.
5. Solutions:
(a) C = 286.
(b) C = 1/5.
(c) C = −7/3.
1
− (n − 1)t
.
7. Solutions:
1
ln 2 t/0.69
(a) dy/dt = ln 2 · 2
·
=
·2
≈ 1 · 2t/0.69 = y.
0.69
0.69
k ln 2 t/0.69
k
kt/0.69
=
(b) dy/dt = ln 2 · 2
·
·2
≈ k · 2t/0.69 = ky.
0.69
0.69
k ln 2
k
=A
2t/0.69 ≈ A·k·2t/0.69 = k·A·2t/0.69 = kP .
(c) dP/dt = A·ln 2·2kt/0.69
0.69
0.69
t/0.69
PP 238-240:
3. Solutions:
(a)
(b)
(c)
(d)
(e)
(f)
d 3x
7e = 7e3x · 3 = 21e3x .
dx
d
Cekx = Cekx · k = kCekx .
dx
d
1.5et = 1.5et .
dt
d
1.5e2t = 3e2t .
dt
d 3x
2e − 3e2x = 6e3x − 6e2x = 6e2x (ex − 1).
dx
d cos t
e
= − sin t · ecos t .
dt
4. Solutions:
(a) fx = y · exy .
fy = x · exy .
(b) fx = 6x · e2y .
fy = 3x2 · e2y · 2.
(c) fu = eu sin v.
fv = eu cos v.
(d) fu = eu sin v · sin v.
fv = eu sin v · 3 sin v · 3 cos v.
5. Solutions:
(a) exp(2x + 3) = e2x+3 = e2x · e3 .
2
(b) (exp(x))2 = ex · ex = e2x 6= ex .
(c) exp(17x)/exp(5x) =
e17x
= e17x−5x = e12x .
e5x
2
radium is twice the size of a second lump, then the larger sample will produce twice as
much lead as the second in any given time. In other words, the rate of decay is proportional
to the amount of radium present. Measurements show that 1 gram of radium decays into
lead at the rate of 1/2337 grams per year. Write an equation that links the decay rate to
the size of the radium sample. How does your equation indicate that the process involves
decay rather than growth?
Solution: Since the rate of decay is proportional to the among of radium present and
1 gram of radium decays into lead at the rate of 1/2337 grams per year, we can write
the following equation where R represents the size of the radium sample and R0 gives the
decay rate:
R0 =
−R
,
2337
R(t) = R0 · e−t/2337
(1)