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stoydgen

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Hi guys whats happening? Umm ok here it is...

The curve "y=tan x", Between x=0 and x="pi/4", is rotated about the x axis. Find the volume of the resulting solid of revolution.

I know i gotta use the y^2dx thingy, but how do i differentiate tan^2 x?
 

CM_Tutor

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Originally posted by CrashOveride
That's one step closer to suicide.
:confused:

a = 3, b = -4 leads to |15 + c| / 5 = 2, and hence c = -5 or -25, and hence 3x - 4y - 5 = 0 and 3x - 4y - 25 = 0 are the lines.

a = -3, b = 4 leads to |c - 15| / 5 = 2, and hence c = 25 or 5, and hence -3x + 4y + 25 = 0 and -3x + 4y + 5 = 0 are the lines.

These are the same answers.

Note, you really should use a = 3, b = -4, as the definition of general form requires a > 0.
 

CM_Tutor

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Stoydgen, you don't want to differentiate tan<sup>2</sup>x, you want to integrate it. This is done by rewriting it using the Pythagorean identities - in this case, using tan<sup>2</sup>x + 1 = sec<sup>2</sup>x.
 

CrashOveride

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Yeah i know man, i didnt even write out the equations or notice that values of c were similar heh :)

Ah and thanks for the note.
 

stoydgen

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Umm my bad, typo hehe :p
Anyways i got it sorted now, comes out to be something like .67 or so. Umm thanks anyways :D
 

CM_Tutor

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Originally posted by ToO LaZy ^*
umm..how do you simplify these again?..

In(2/3) - In(1/3)
Do you mean ln(2 / 3) - ln(1 / 3)?

If so, either:

ln(2 / 3) - ln(1 / 3) = (ln 2 - ln 3) - (ln 1 - ln 3) = ln 2, noting that ln 1 = 0

or

ln(2 / 3) - ln(1 / 3) = ln[(2 / 3) / (1 / 3)] = ln[(2 / 3) * (3 / 1)] = ln 2

If not, what do you mean?
 

CM_Tutor

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So, looking at the attachment, you mean ln (as in little L) not In (as in big I), in which case the answer is ln 2 by either of the methods I posted above. :)
 
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i mean Ln as in log

isn't your method derived fron
Ln2/Ln3

EDIT: actually, don't worry, i found out :D
 
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how do you solve these problems plz?

1. If m= square root of 2 - 1 (stupid computer), find the value of m + 1/m

and

2. If f(x) = x^2 + 3x + 2 then
(a) find f(x+h)

(b) find f(x+h)-f(x)/h

(c) solve f(x) = 12
 

CM_Tutor

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Originally posted by **freakstar87**
1. If m= square root of 2 - 1 (stupid computer), find the value of m + 1/m
m = sqrt(2) - 1

1 / m = 1 / [sqrt(2) - 1] = {1 / [sqrt(2) - 1]} * {[sqrt(2) + 1] / [sqrt(2) + 1]}
= [sqrt(2) + 1] / {[sqrt(2) - 1][sqrt(2) + 1]}
= [sqrt(2) + 1] / {[sqrt(2)]<sup>2</sup> - [1]<sup>2</sup>}
= [sqrt(2) + 1] / (2 - 1)
= sqrt(2) + 1

So, m + (1 / m) = [sqrt(2) - 1] + [sqrt(2) + 1] = 2 * sqrt(2)
2. If f(x) = x^2 + 3x + 2 then
(a) find f(x+h)

(b) find f(x+h)-f(x)/h

(c) solve f(x) = 12
(a) f(x + h) = (x + h)<sup>2</sup> + 3(x + h) + 2

(b) [f(x + h) - f(x)] / h = [(x + h)<sup>2</sup> + 3(x + h) + 2 - (x<sup>2</sup> + 3x + 2)] / h
= (x<sup>2</sup> + 2xh + h<sup>2</sup> + 3x + 3h + 2 - x<sup>2</sup> - 3x - 2) / h
= (2xh + h<sup>2</sup> + 3h) / h
= h(2x + h + 3) / h
= 2x + h + 3

(c) f(x) = 12
x<sup>2</sup> + 3x + 2 = 12
x<sup>2</sup> + 3x - 10 = 0
(x + 5)(x - 2) = 0

So, x = -5 or x = 2
 

Slidey

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You do realise we can't answer them until you post them? :)
 

flower_farie

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CM_Tutor said:
Stoydgen, you don't want to differentiate tan<sup>2</sup>x, you want to integrate it. This is done by rewriting it using the Pythagorean identities - in this case, using tan<sup>2</sup>x + 1 = sec<sup>2</sup>x.
Isn't it just the identity of sinx/cosx since its not squared...then u integrate the sinx/cosx using logs? or am i confused?
 

withoutaface

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flower_farie said:
Isn't it just the identity of sinx/cosx since its not squared...then u integrate the sinx/cosx using logs? or am i confused?
Yes, the integral of tanx is -ln(cosx)+c
 

bono2cool

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I NEED HELP QUICKLY!!!!!!!!!!!!PLZZZZZZ

john borrowed 2000 in 1st jan 1980. he agreed to pay back 300 on ist jan in each succeeding year and to add 6% percent per annum interest on the amount owing during the year completed
find amount still owing immediately afta 1st jan 1985?
FIND THE NUMBER OF YEARS NECESSARY TO FREE HIIMSLEF OFF DEBT?
 

bono2cool

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^^ anotha thin if deres any1 kind enough to explain to meh how to find an oblique aymptope dat will be really great
e.g. y=x^2 / 2x+3
y=2x+3 / x^2+4
 

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