5 Steps to a 5 AP Calculus BC 2019

340 STEP 4. Review the Knowledge You Need to Score High

eln

∣∣y 1 / 2

x+ 1

∣∣

=eln 2

y^1 /^2

x+ 1

= 2

y^1 /^2 = 2 (x+ 1 )

y=( 2 )^2 (x+ 1 )^2

y= 4 (x+ 1 )^2.

Step 4. Verify result by differentiating:

dy

dx

= 4 ( 2 )(x+ 1 )=8(x+1).

Compare with

dy

dx

=

2 y

x+ 1

=

2

(

4 (x+ 1 )^2

)

(x+ 1 )

= 8 (x+ 1 ).

17. y(t)=y 0 ekt

y 0 =750, 000

y( 22 )=(750, 000)e(^0.^03 )(^22 )

≈

⎧

⎪⎨

⎪⎩

1. 45109 E 6 ≈1, 451, 090 using

a TI-89,

1, 451, 094 using a TI-85.

18. Step 1. Separate variables:

4 ey=

dy

dx

− 3 xey

4 ey+ 3 xey=

dy

dx

ey( 4 + 3 x)=

dy

dx

( 4 + 3 x)dx=

dy

ey

=e−ydy.

Step 2. Integrate both sides:

∫

( 4 + 3 x)dx=

∫

e−ydy

4 x+

3 x^2

2

=−e−y+C

Switch sides:e−y=−

3 x^2

2

− 4 x+C.

Step 3. Substitute given value:y( 0 )= 0

⇒e^0 = 0 − 0 +c⇒c=1.

Step 4. Take ln of both sides:

e−y=−

3 x^2

2

− 4 x+ 1

ln(e−y)=ln

(

−

3 x^2

2

− 4 x+ 1

)

y=−ln

(

1 − 4 x−

3 x^2

2

)

.

Step 5. Verify result by differentiating:

Enterd(−ln(1− 4 x− 3

(x−∧^2 )/2),x) and obtain

− 2 ( 3 x+ 4 )

3 x^2 + 8 x− 2

, which is equivalent

toey(4+ 3 x).

19. y(t)=y 0 ekt

k= 0 .0625,y( 10 )=50, 000

50, 000=y 0 e^10 (^0.^0625 )

y 0 =

50, 000

e^0.^625

⎧

⎪⎪⎨

⎪⎪⎩

$26763.1 using a TI-89,

$26763.071426≈$26763. 07

using a TI-85.

20. Setv(t)= 2 − 6 e−t=0. Using the [Zero]
    function on your calculator, compute
    t= 1 .09861.