anonymous
  • anonymous
Find an equation of the tangent line at the given value of x.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
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tkhunny
  • tkhunny
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?
tkhunny
  • tkhunny
* "=" in the parentheses should be "-". F(x) - F(0)

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anonymous
  • anonymous
okay
tkhunny
  • tkhunny
* 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.
anonymous
  • anonymous
hmm ok
tkhunny
  • tkhunny
Well, what do you do with a derivative when you want the slope of a tangent line?
agent0smith
  • agent0smith
\[\Large \dfrac{d}{dx}\int\limits\limits_{0}^{x}\sin(\sqrt{t^{2}-\pi^2})\;dt = \sin(\sqrt{x^{2}-\pi^2})\] You remember how to do that part from earlier, right @onegirl ?
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
Yes okay
anonymous
  • anonymous
okay i think i got it
agent0smith
  • agent0smith
@tkhunny you put the equation in wrong (- where a + should be), and i copied it wrongly... should be \[\large \dfrac{d}{dx}\int\limits\limits_{0}^{x}\sin(\sqrt{t^{2}+\pi^2})\;dt = \sin(\sqrt{x^{2}+\pi^2})\]
anonymous
  • anonymous
yea i saw that too
tkhunny
  • tkhunny
Ah, well that's silly. Why would I do that? Good call!
agent0smith
  • agent0smith
@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
anonymous
  • anonymous
yes
anonymous
  • anonymous
thanks!
agent0smith
  • agent0smith
Looks like m=0 too, since \[\large m = \sin(\sqrt{0^{2}+\pi^2}) = \sin \sqrt {\pi^2} = \sin \pi = 0\] so y=0
anonymous
  • anonymous
okay

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