Question

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Answer 1

this one looks hard :/

Answer 2

YEa this one I am not sure on, didn't deal with much in terms of differentials in these kinds of story problems :/ sorry hopefully jziggy can help!

Answer 3

ok ill give u another one

Answer 4

Well first thing you can do is fill our the top row by placing your x value of your point into the f(x) so find f(2) for 1.9 1.99 1.999 2 2.1 ,etc.

Answer 5

ok :)

Answer 6

wait dont we just put the top row of numbers into the csc(x) for example csc(1.9)

Answer 7

oh yes sorry xD the point is given to help when we get a tagent line equation

Answer 8

so i got that done now we need the tang line

Answer 9

yea we do so before we go on, do you know perhaps how we can get one?

Answer 10

think of what a derivative is

Answer 11

take the first derivative of the f(x)=csc x?

Answer 12

yes! which ,if figure correctly, gives us which part of a linear equation?

Answer 13

hmm the slope?

Answer 14

yep! so take the 1st derv of csc x (that I do not know off top my head so :P)

Answer 15

−cot(x)⋅csc(x)

Answer 16

of course it needs to be a complicated one like cscx but wht ever haha we can still do it!

Now what do you suppose we do next with f'(x) = -cot(x)csc(x)

Answer 17

Since we need a actual eveluated value for a slope.

Answer 18

umm put it in the y=mx+b format?

Answer 19

nope not quite yet, we need to find a m value first from this derivative. We would do this f'(2) since they gave us 2 as the x value in our given point

Answer 20

and then evauluate for the values given
@jziggy we are working on this :)

Answer 21

oh i see

Answer 22

so calculate what f'(2) equals (-cot(2)csc(2))

Answer 23

working on it :P

Answer 24

Looks like you've got this one so i'll go take a look at the first problem

Answer 25

sounds good!

Answer 26

here i got one for ya
i solved that one already

Answer 27

@jziggy

Answer 28

(−cot(2)csc(2)) « »
= 0.50330897

Answer 29

ok so that is the slope at that particular point so now we use ur y = mx +b equation :P and we gotta find b

Answer 30

by plugging in (2, csc2)

Answer 31

ok so y=0.5033x+b

Answer 32

yeap y would be csc 2 and the x is 2 again

Answer 33

solve for b

Answer 34

so csc(2)=0.5033(2)

Answer 35

yeap and do csc(2) - .5033(2) = b

Answer 36

0.09313227

Answer 37

so y=0.5033x+0.0931

Answer 38

so now ur tangent line equation (as ugly as it is) is ur T(x) so now do same thing again

T(1.9) T(1.99) etc.

Answer 39

grrr lol

Answer 40

and hopefully that is all good, havent done a tangent line problem in while especially one to that degree :P

Answer 41

ill check it by graphing when im done :)

Answer 42

I would do that for you but i seem to have left my calc at school so

Answer 43

Anyways I must go anything else you have jziggy will have to help!

Answer 44

dam i think we got off by 0.0001 but whatever :P

Answer 45

did you graph it?

Answer 46

is that what u mean?

Answer 47

well the values at 2 for T(x) is different from csc x by 0.0001

Answer 48

lol that isnt so much only 1/10000 difference

Answer 49

yeah it will be fine

Answer 50

They will be different, the tangent line would be just an estimation of it

Answer 51

i probably rounded wrong

Answer 52

yea if you used more didgits for the slope value liek we found or y-intercept or even the csc(2) for the y in y=mx + b might change it

Answer 53

ooh wait at 2 nvm ignore that :D

Answer 54

but yeah, with error that small it's most likely rounding error

Answer 55

yeah i checked it its right!

Answer 56

rounding can change answer by that much, it isnt significant tho as we said! :P

Answer 57

ok next problem im putting up

Answer 58

Alright, i think i've got your other problem solved as well you ready?

Answer 59

ok good luck jziggy im out! :P

Answer 60

Yay for trig equations.... lol
f(x) = 2sec(x) + tan(x)
f'(x) = 2sec(x)tan(x) + sec^2(x)

Set this equal to zero to get the critical numbers
2sec(x)tan(x) + sec^2(x) = 0

You can easily factor out sec(x) to get
sec(x)(2tan(x) + sec(x)) = 0 which gives us 2 equations

sec(x) = 0 which isn't possible since sec(x) is never equal to zero

2tan(x) + sec(x) = 0
by converting to sines and cosines we get

(2sin(x))/cos(x) + 1/cos(x) = 0
(2sin(x) + 1)/cos(x) since only the numerator matters for finding zeroes we now have

2sin(x) + 1 = 0
2sin(x) = -1
sin(x) = -1/2

Solve for x to get that our critical numbers are 7pi/6 and 11pi/6.

Answer 61

@Valkarie70