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Geek Culture / Calculus From Hell!!! I need help

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jblack
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Posted: 22nd Dec 2009 00:11
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Alright, I'm having hell with my calculus homework and would like some assistance please. My teacher is gone this week, so I have to come to the internet!

Ok so I have 3 questions.

My first question is this:
Find the tangent line of the curve at (0,-2) for the equation

-x^4 + xy + y^3 = -8

Please walk me through that one.

The second and third questions involve the same process.

I have to find the derivative of

e^x^3

and the next question I have to find the derivative of

(1+cos^2(7x))^3

I'm supposed to use the chain rule to figure this stuff out. My teacher has not taught us these things, however, and I have a test over problems similar to these on Wednesday.

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Zotoaster
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Posted: 22nd Dec 2009 00:33
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The first one is tricky because you have two variables in it, x and y, which means you are describing a surface, rather than a single curve. So the tangent just depends on which direction you are moving. Are you moving along the X or the Y axis, or along some arbitrary vector? Or is 'y' supposed to be a constant?

Second one is easy. You just use the chain rule.
Multiply by the exponent, so you have (x^3)*(e^(x^3)) then multiply by the derivative of the exponent, so you have 3(x^2)(x^3)*(e^(x^3)).

Third one, again, chain rule.

3(1+cos^2(7x))^2
Then
-3(7sin^2(7x))^2
=-21(sin^2(7x))^2

"everyone forgets a semi-colon sometimes." - Phaelax
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NeX the Fairly Fast Ferret
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Posted: 22nd Dec 2009 00:34
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Uh... yeah... simple... right.

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Outscape
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Posted: 22nd Dec 2009 00:37
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Zotoaster
wait..
what?


Huge List of Ways to Increase FPS
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jblack
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Posted: 22nd Dec 2009 00:39 Edited at: 22nd Dec 2009 02:01
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*nix

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AndrewT
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Posted: 22nd Dec 2009 00:41
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Zotoaster:

I might be missing something but I'm fairly certain that -x^4 + xy + y^3 = -8 is a curve, not a surface, and it looks like this:



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demons breath
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Posted: 22nd Dec 2009 00:46 Edited at: 22nd Dec 2009 00:46
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It's been a while since I've really done Calculus (had a module at uni this term but I failed pretty bad and did no work and ended up finishing an exam with "0=1. Oh surely that's not right. Oh poo.")

However if I recall arights, you would just differentiate the equation. Differentiate using the product rule, and terms of y are treated like you would treat terms of x, except when you differentiate a y you times that by dy/dx. Therefore

d/dx (-x^4 + xy + y^3 = -8)
-4x^3 + x*(dy/dx) + y + (3y^2)*(dy/dx) = 0

Now collect terms of (dy/dx) so

(dy/dx)*(x+3y^2) = 4x^3 + y

divide this across so you end up with just dy/dx on one side

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

This gives the gradient of the curve

Therefore the gradient of the tangent line at (0,-2) is

dy/dx = (0^3 - 2)/(0 + 3((-2)^2))
dy/dx = -2/12
dy/dx = -1/6


I think Has been a while though.

EDIT: And yeah it's just a curve given in terms of x and y, it's not a plane...

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jblack
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Posted: 22nd Dec 2009 00:54 Edited at: 22nd Dec 2009 01:51
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Thank you all very much, even the person that was hella sarcastic. Zotoaster you are so kind and demon's breath i am pretty sure you're right with the first problem.

EDIT: I think you may have messed up on one part demon's breath.

I think it should be 4x^3 - y, rather than + y

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Xarshi
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Posted: 22nd Dec 2009 03:31
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Use implicit differentiation for the first one. It's a curve, not a surface. For a surface it'd be like:
f(x, y) = 6xy^2. Then you just differentiate with regards to one of the variables and treat the other ones as constants. Which is called partial differentiation, but afaik you won't have to do that until calculus 3 in college? Maybe? I'm not sure...

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