Question

@jim_thompson5910

http://cyh.leeschools.net/UserContent/Documents/AP%20CalcBC%20SumAssign%2014-15.pdf

So if I was paying you Florida minimum wage for you helping me with AP Calc BC so far, I'd owe you probably around $50 :P

Add it to your invoice. Ha.

Answer 1

lol it's fine, I think we left off on 86?

Answer 2

Sisisi

Answer 3

able to get started on it? or no?

Answer 4

No way haha. I'm pretty much lost on the rest of all these questions hah

Answer 5

are you able to use a calculator on this part?

Answer 6

because it will help to get a visual of the region that forms the base

Answer 7

"Do the attached Multiple Choice Questions. Section I is with NO Calculator and you may use a
calculator for Section II."

But he told us in class you can make calculations but he doesn't want us graphing anything

Answer 8

ok, so you'll have to visually keep in mind what arctan looks like

Answer 9

So yes I can use one this is the 2nd section of part one (after 76 is section 2)

Answer 10

We just could not use one on section 1 part 1 (only for calculations)

But I can graph things in section 2 of part 1

Answer 11

can you help me with a question

Answer 12

oh so you're able to graph on 86?

Answer 13

Would you like me to graph the tan^-1 y=3 and x=1?

Answer 14

Yes I am

Answer 15

yes please

Answer 16

Okay I watched it

Answer 17

graphed it* oops haha

I was talking to my cousin about watching tv so I put watched instead of graphed... Shoulda had a V8.

Answer 18

well graphs are visual like tv, so you can watch a graph too

Answer 19

so do you see the region that forms the base?

Answer 20

This is true :P

Okay what to do next? Guide me O'wise 1

Answer 21

No I'm the worst visualizer haha

Answer 22

Is it where x and y meet?

Answer 23

At (1,3)?

Answer 24

you should have this

Answer 25

the region in green is the region that forms the base

Answer 26

Yes I see that

Answer 27

unfortunately I'm getting a function that can't be integrated by hand...hmm

Answer 28

Well you can use a calculator

Answer 29

What do you mean by "getting a function" ?

Answer 30

Pick any x value between 0 and 1. The vertical distance from the point (x,y) on y = arctan(x) to y = 3 is 3 - arctan(x)

So this is the side length of the square

The area of the cross section square is (3 - arctan(x))^2

Answer 31

So we will integrate (3 - arctan(x))^2 from 0 to 1

Answer 32

In other words, compute

\[\Large \int_{0}^{1} (3 - \arctan(x))^2 dx\]

Answer 33

Yeah but unfortunately my calculator is dead lol

Answer 34

I keep forgetting to buy triple A batteries

Plus I'm getting a new calculator soon, one I can actually charge and it up to date with the technology so yiy

Answer 35

& is up to date*

Answer 36

That's good. In the meantime, you can use a calculator like wolfram alpha since it's very good at calculus.

Answer 37

http://www.wolframalpha.com/input/?i=integral%28+%283-arctan%28x%29%29%5E2%2Cx%3D0..1+%29


I typed in "integral( (3-arctan(x))^2,x=0..1 )" without quotes

Answer 38

Am I supposed to get a number or something?

Answer 39

yeah it's at the very end/right side

Answer 40

Oh alright I see it. 6.612

Answer 41

yep

Answer 42

For 87 I have to take the 2nd derivative right?

Answer 43

In which case use the quotient rule

Answer 44

yes, so derive f ' to get f ''

Answer 45

This will be tedious haha one second

Answer 46

agreed, but you can do it

Answer 47

Just checking but I add exponents when I multiply right?

Answer 48

if you mean like x^2*x^3 = x^(2+3) = x^5, then yeah

Answer 49

Okay I'm going to upload a file, I'm stuck :P

Answer 50

go for it

Answer 51

Not sure how to combine the rest of the numerator lol

Answer 52

let me check your work

Answer 53

Okie dokie

Answer 54

it is incorrect

Answer 55

Ughhhhh what did I do wrong

Answer 56

still looking through and trying to spot it

Answer 57

You can ask me questions too

Answer 58

You should have this

\[\Large \frac{x^{1/2}}{1+x+x^3} = x^{1/2}(1+x+x^3)^{-1}\]

but I don't see that step

Answer 59

What why? I used the quotient rule?

Answer 60

oh gotcha

Answer 61

I was going to use the product rule

Answer 62

Oh I can do that

Answer 63

Probably would be easier lol

Answer 64

the good news is that your first line is correct

i think you just made a silly error in simplifying

Answer 65

Oh okay good because idk how to do this one with the product rule, only quotient

Answer 66

Okay so where is the mistake/up to where is right?

Answer 67

well any time you have a fraction like f(x)/g(x)

you can rewrite that as a product

f(x)/g(x) = f(x) * [g(x)]^(-1)

and then use the product rule

Answer 68

anyways, I'm checking your other lines

Answer 69

On the 2nd line I meant for it to be x^5/2/2

Answer 70

The numerator I should have: \[f''(x)=\frac{ 1 }{ 2 }x ^{-1/2}+\frac{ 1 }{ 2 }x ^{1/2}+\frac{ 1 }{ 2 }x ^{5/2}-x ^{1/2}-3x ^{3/2}\]

Answer 71

and that 3/2 should be 5/2

since 2 + 1/2 = 5/2

Answer 72

also, you should change that 5/2 to 3/2

somehow you got the two switched

Answer 73

Oh right I gotcha so I'm right and the last one should be -3x^5/2

Answer 74

here's all I could spot so far

Answer 75

OKAY SO\[\frac{ \frac{ 1 }{ 2\sqrt{x}}+\frac{\sqrt{x} }{ 2}+\frac{ x ^{5/2} }{ 2}-\sqrt{x}-3x ^{5/2}}{ (1+x+x^3)^2 }\]

Answer 76

No the 1st red one should be x^5/2/2 because -1/2+3 AKA -1/2+6/2=5/2

Answer 77

oh right, I guess it looked very close on my graph that I didn't think twice about it

Answer 78

Okay so now how would I simplify that?

Answer 79

I would multiply top and bottom by \(\Large \sqrt{x}\)

Answer 80

Can I ask why? :P

Answer 81

well you can multiply by 1 and it won't change the expression

Answer 82

and you have sqrt(x) in the denominator (of one fraction in the numerator) and you have other expressions that also have sqrt(x) in them

so the natural thing to do is to multiply everything by sqrt(x) to see how many radical terms go away

I'm taking advantage of the fact that

sqrt(x)*sqrt(x) = ( sqrt(x) )^2 = x

when x >= 0

Answer 83

and whenever you have a fraction within a fraction, one trick is to multiply every term by the inner LCD to clear out the inner fractions

Answer 84

Yeah I'll be honest, I don't know how to do that. Lol.

Answer 85

Example