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onegirl

  • one year ago

Find an equation of the tangent line at the given value of x.

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  1. onegirl
    • one year ago
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  2. tkhunny
    • one year ago
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    What's your plan? \(\int\limits_{0}^{x}f(x)\;dx = F(x) - F(0)\) where \(F(x)\) is an antiderivative of \(f(x)\). \(\dfrac{d}{dx}(F(x)=F(0)) = F'(x) = f(x)\) Isn't that interesting?

  3. tkhunny
    • one year ago
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    * "=" in the parentheses should be "-". F(x) - F(0)

  4. onegirl
    • one year ago
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    okay

  5. tkhunny
    • one year ago
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    * Wow! I must be sleeping. I used 'x' two different ways, too! Okay, now that I've confused you. let's just show you. \(\dfrac{d}{dx}\int\limits_{0}^{x}\sin(\sqrt{t^{2}-\pi^2})\;dt = \sin(\sqrt{x^{2}-\pi^2})\) Ponder on it.

  6. onegirl
    • one year ago
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    hmm ok

  7. tkhunny
    • one year ago
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    Well, what do you do with a derivative when you want the slope of a tangent line?

  8. agent0smith
    • one year ago
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    \[\Large \dfrac{d}{dx}\int\limits\limits_{0}^{x}\sin(\sqrt{t^{2}-\pi^2})\;dt = \sin(\sqrt{x^{2}-\pi^2})\] You remember how to do that part from earlier, right @onegirl ?

  9. agent0smith
    • one year ago
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    We need a tangent line y = mx+b So to find the tangent line at x=0, we need the slope of the line m... we've already differentiated it above, now we need to find m by plugging in x=0. Then we find b by using the original function in the question.

  10. onegirl
    • one year ago
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    Yes okay

  11. onegirl
    • one year ago
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    okay i think i got it

  12. agent0smith
    • one year ago
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    @tkhunny you put the equation in wrong (- where a + should be), and i copied it wrongly... should be \[\large \dfrac{d}{dx}\int\limits\limits_{0}^{x}\sin(\sqrt{t^{2}+\pi^2})\;dt = \sin(\sqrt{x^{2}+\pi^2})\]

  13. onegirl
    • one year ago
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    yea i saw that too

  14. tkhunny
    • one year ago
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    Ah, well that's silly. Why would I do that? Good call!

  15. agent0smith
    • one year ago
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    @onegirl did you get the slope m by putting x= 0 into \[\large m = \sin(\sqrt{x^{2}+\pi^2})\] Then find b by using \[\large \dfrac{d}{dx}\int\limits\limits\limits_{0}^{0}\sin(\sqrt{t^{2}+\pi^2})\;dt = 0\] (since the area from 0 to 0 is zero) So since b=0 it'll be y = mx

  16. onegirl
    • one year ago
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    yes

  17. onegirl
    • one year ago
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    thanks!

  18. agent0smith
    • one year ago
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    Looks like m=0 too, since \[\large m = \sin(\sqrt{0^{2}+\pi^2}) = \sin \sqrt {\pi^2} = \sin \pi = 0\] so y=0

  19. onegirl
    • one year ago
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    okay

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