5 Steps to a 5 AP Calculus BC 2019

(Marvins-Underground-K-12) #1

36 STEP 2. Determine Your Test Readiness



  1. 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


.



  1. 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.


  1. 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.
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