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onegirl
Group Title
Find an equation of the tangent line at the given value of x.
 one year ago
 one year ago
onegirl Group Title
Find an equation of the tangent line at the given value of x.
 one year ago
 one year ago

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tkhunny Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.2
* "=" in the parentheses should be "". F(x)  F(0)
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.2
* 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.
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.2
Well, what do you do with a derivative when you want the slope of a tangent line?
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.1
\[\Large \dfrac{d}{dx}\int\limits\limits_{0}^{x}\sin(\sqrt{t^{2}\pi^2})\;dt = \sin(\sqrt{x^{2}\pi^2})\] You remember how to do that part from earlier, right @onegirl ?
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
okay i think i got it
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.1
@tkhunny you put the equation in wrong ( where a + should be), and i copied it wrongly... should be \[\large \dfrac{d}{dx}\int\limits\limits_{0}^{x}\sin(\sqrt{t^{2}+\pi^2})\;dt = \sin(\sqrt{x^{2}+\pi^2})\]
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
yea i saw that too
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.2
Ah, well that's silly. Why would I do that? Good call!
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.1
@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
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.1
Looks like m=0 too, since \[\large m = \sin(\sqrt{0^{2}+\pi^2}) = \sin \sqrt {\pi^2} = \sin \pi = 0\] so y=0
 one year ago
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