36 STEP 2. Determine Your Test Readiness
- See Figure DS-14.
10–1 2 x
y y=x+ 2
y=x^2
Figure DS-14
To find points of intersection, setx^2 =x+ 2
⇒x^2 −x− 2 = 0 ⇒x=2orx=−1.
Area of cross section=((x+2)−x^2 )^2.
Volume of solid,V=
∫ 2
− 1
(
x+ 2 −x^2
) 2
dx.
Using your calculator, you obtain:V=
81
10
.
- Separate and simplify
dP
dt
=. 35 P
(
1 −
P
4000
)
.
1
P
(
1 −
P
4000
)dP=. 35 dt
4000
P(4000−P)
dP=. 35 dt
Integrate with a partial fraction
decomposition.
∫
4000
P(4000−P)
dP=
∫
. 35 dt
∫
dP
P
+
∫
dP
4000 −P
=
∫
. 35 dt
ln|P|−ln
∣∣
4000 −P
∣∣
=. 35 t+C 1
ln
∣∣
∣
∣
P
4000 −P
∣∣
∣
∣=.^35 t+C^1
P
4000 −P
=C 2 e.^35 t
Population att=0 is 100, so
100
4000 − 100
=
100
3900
=
1
39
=C 2.
The population model is
P
4000 −P
=
e.^35 t
39
⇒ 39 P=e.^35 t(4000−P)
⇒ 39 P+e.^35 tP= 4000 e.^35 t
⇒P
(
39 +e.^35 t
)
= 4000 e.^35 t
⇒P=
4000 e.^35 t
39 +e.^35 t
⇒P=
4000
39 e−.^35 t+ 1
.
Whent=5,P=
4000
39 e−.35(5)+ 1
⇒P=
4000
39 e−^1.^75 + 1
≈ 514 .325.
- If
dy
dx
=
−y
x^2
andy=3 andx=2, approximate
ywhenx=3. Use Euler’s
Method with an increment of 0.5.
y(2)=3 and
(
dy
dx
)
x=2,y= 3
=
− 3
4
so
y(2.5)=y(2)+ 0. 5
(
dy
dx
)
x=2,y= 3
= 3 + 0 .5(− 0 .75)= 2. 625
(
dy
dx
)
x= 2 .5,y= 2. 625
=
− 2. 625
(2.5)^2
=
− 21
8(6.25)
=
− 21
50
=− 0. 42
y(3)=y(2.5)+ 0. 5
(
dy
dx
)
x= 2 .5,y= 2. 625
= 2. 625 + 0 .5(− 0 .42)
= 2. 625 − 0. 21 = 2. 415.