5 Steps to a 5 AP Calculus BC 2019

(Marvins-Underground-K-12) #1

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


x

y
10

6

x^2 – 36

Figure 11.9-4

Step 2: Find the minimum value ofP.
Enter
yt=x+ 10 ∗x/

(√
(x∧ 2 − 36 )

)
.
Use the [Minimum] function of
the calculator and obtain the
minimum point (9.306, 22.388).
Step 3: Verify with the First Derivative
Test.
Entery 2 =(y1(x),x) and
observe. (See Figure 11.9-5.)

9.306

rel. min.

decr. incr.









y 1 = f

y 2 = f′

Figure 11.9-5

Step 4: Check endpoints.
The domain ofxis (6,∞).

Sincex= 9 .306 is the only
relative extremum, it is the
absolute minimum.
Thus, the maximum length of the
pipe is 22.388 feet.



  1. limx→− 1
    1 +cosπx
    x^2 − 1
    =limx→− 1
    −πsinπx
    2 x




0


− 2


= 0


32.


dx
dt
=− 4 t
dy
dt
= 2 t. The speed of the object

is


(− 4 t)^2 +(2t)^2 =

16 t^2 + 4 t^2 =


20 t^2 = 2 t



  1. Evaluated


att=

1


2


, the speed is 2

(
1
2

)√
5 =


5.


  1. Integrate



2 x


x+ 3 dxby parts with
u= 2 x,du= 2 dx,dv=


x+ 4 dx, and

v=

2


3


(x+4)^3 /^2. Then

2 x


x+ 3 dx=

4


3


x(x+3)^3 /^2 −

4
3


(x+3)^3 /^2 dx=

4


3


x(x+3)^3 /^2 −

4
3

(
2
5
(x+3)^5 /^2

)

. Simplifying this


expression, we get


2 x


x+ 3 dx

=


4


5


(x+3)^3 /^2 (x−2).
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