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anonymous

  • 5 years ago

HELP:find the first partial derivative of f(x,y) integral from x to y cos(t^9) dt

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  1. anonymous
    • 5 years ago
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    Can you restate the question? You want to find the first order partial derivative of a function f(x,y) with respects to both x and y correct? What exactly is your function?

  2. anonymous
    • 5 years ago
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    \[f(x,y)= \int\limits_{y}^{x} \cos (t^9) dt\]

  3. anonymous
    • 5 years ago
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    this is what i got

  4. anonymous
    • 5 years ago
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    \[1/x^9 \sin(x^9)+1/y^9 \sin(y^9) +C\]

  5. anonymous
    • 5 years ago
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    not sure if its right though

  6. anonymous
    • 5 years ago
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    ohhh..

  7. anonymous
    • 5 years ago
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    i am suppose to integrate by parts right?

  8. anonymous
    • 5 years ago
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    1/9sinx^9-1/9sinx^9+c

  9. anonymous
    • 5 years ago
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    In said situation, given theorized, ideal Pythagorean standards, one could, when, concluded extremities parametize area, fortitude of x with respect to inifinite limit gives, under mathematical conditions, that, stated integral has an unreal answer if eliminated factors cancel in a forthright fashion. All is explained by science

  10. anonymous
    • 5 years ago
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    i totally spaced out on that

  11. anonymous
    • 5 years ago
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    1/9sinx^9-1/9siny^9+c

  12. anonymous
    • 5 years ago
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    Ok lets see, I believe your integral is right, although there's no need for the +c because you have definite bounds...

  13. anonymous
    • 5 years ago
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    yes you are right no need +c

  14. anonymous
    • 5 years ago
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    so we don't integrate this by u substitution?

  15. anonymous
    • 5 years ago
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    or some other integration method?

  16. anonymous
    • 5 years ago
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    no need substitution

  17. anonymous
    • 5 years ago
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    hmm.......... actually

  18. anonymous
    • 5 years ago
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    Definite C, when referring to tangency, is, without question, derivized by cascadation of three bounds in the three dimensional coordinate plane, as defined by Cartesian math as well as Mendel, as, with respect to z, a common factor is pulled, given that a trifactor of polynomic trigonometry is frozen in intervals of, cosine, when not shown to be aligned, yields a C of zero which can, ultimately, not need be shown unless graphing is necessary.

  19. anonymous
    • 5 years ago
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    So we got 1/x^9 sin(x9)+1/y^9 sin(y9) when evaluated between x and y.... This means our integral without evaluating it, and in terms of t (aka indefinite integral) was.... 1/t^9 cos(t^9)... if you derive that in terms of t you do not get the original equation...

  20. anonymous
    • 5 years ago
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    so what does that mean?

  21. anonymous
    • 5 years ago
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    You are either missing part of the problem or your teacher is an retriceole. You can't solve this easily, if at all, because you are integrating for cosine with a function inside and nothing outside to return upon integrating. Try using a trig identity if you can think of one, or www.wolframalpha.com/. Otherwise you can just say unable to solve because it seems unreasonable for you to be asked to solve that if your teacher hasn't shown you how.

  22. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=integral%28cos%28t^9%29dt%29

  23. anonymous
    • 5 years ago
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    Hmmm i actually had a problem just like this on my vector calc test a few weeks ago... I didn't get the problem but i think it has to do with cos(x^2) being an odd function... I'm looking for some sort of trig identity or trick now because neither u sub or by parts will work for something like this, and wolfram gives me a rediculious answer

  24. anonymous
    • 5 years ago
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    yeah we haven't learned this and there isn't an example in a book

  25. anonymous
    • 5 years ago
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    http://how-to-spell-ridiculous.com/

  26. anonymous
    • 5 years ago
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    You are going to need an incomplete gamma or knowledge of the Heaviside to solve this.

  27. anonymous
    • 5 years ago
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    hahahaha thanks ScienceWiz

  28. anonymous
    • 5 years ago
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    <3

  29. anonymous
    • 5 years ago
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    So I think I know the answer... If we use the Fundamental Theorem of Calculus we can state that the Integral of your original function then derivative in respects to t is just the original function right?

  30. anonymous
    • 5 years ago
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    So if you take the Integral of your original function f(x) then evaluate with respects to x and y. You'll get some F(x)-F(y) right.

  31. anonymous
    • 5 years ago
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    yes

  32. anonymous
    • 5 years ago
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    where F(x)= integral of x

  33. anonymous
    • 5 years ago
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    then we take the partial with respects to x of F(x)-F(y). F(y) goes to 0 because there are no x terms, while F(x) goes to the original function just in terms of x

  34. anonymous
    • 5 years ago
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    the same applies for your partial derivative of y... So you should just get for your\[f _{x}=\cos(x^9)\] and\[f _{y}=\cos(y^9)\]

  35. anonymous
    • 5 years ago
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    Thoughts?

  36. anonymous
    • 5 years ago
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    yeah I am kind of confused because I am a bit rusty on the FTC

  37. anonymous
    • 5 years ago
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    I'm having a hard time following exactly what you are saying. The main issue that I see is there is no way to get around integrating for t at somepoint. And you can't do that at the level of knowledge you are at right now. Also this link says no. And if you delete "from x to y" you can look at the indefinite integral. http://www.wolframalpha.com/input/?i=integral%28cos%28t^9%29dt%29+from+x+to+y

  38. anonymous
    • 5 years ago
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    FTC just states The derivative of the Integral is your original equation..

  39. anonymous
    • 5 years ago
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    Ya but you can't apply that theorem to this situation unless you were to integrate again, which would cause the same problem.

  40. anonymous
    • 5 years ago
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    hmmm? the initial problem states "find the first partial derivative of f(x,y) integral from x to y cos(t^9) dt" doesn't that mean you want to find the partial derivatives? Or do you just want the integral?

  41. anonymous
    • 5 years ago
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    I need to find the first partial derivatives fx and fy

  42. anonymous
    • 5 years ago
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    hmmm.. ok let me try explaining my logic better... So FTC states...\[d/dx(\int\limits_{a}^{x}f(t)dt) :=F'(t)= f(x)\] this means after you take the integral if you take the derivative you get your original function... So our function is f(x,y)=∫yxcos(t9)dt\[f(x,y)=\int\limits_{y}^{x}\cos(t^9) dt= \int\limits_{y}^{x}f (t)dt\] We can just rewrite our\[ \int\limits_{y}^{x}f (t)dt= F(x)-F(y) \] where F(x) is equal to the integral of our original equation evaluated at x... So now we take out function F(x)-F(y) and take the partials in respects to x and y...\[f _{x}(F(x)-F(y))= f(x)\] and \[f _{y}(F(x)-F(y))= -f(y)\] This is because when we take the partial of F(y) with respects to x it goes to 0...

  43. anonymous
    • 5 years ago
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    so f(x) = cos (x ^9) and -f(y)= -cos(y^9)? because of FTC

  44. anonymous
    • 5 years ago
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    Pretty much.

  45. anonymous
    • 5 years ago
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    that seems much easier :) I appreciate your help!! THANKS

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