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

1. onegirl Group Title

2. tkhunny Group Title

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 Group Title

* "=" in the parentheses should be "-". F(x) - F(0)

4. onegirl Group Title

okay

5. tkhunny Group Title

* 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 Group Title

hmm ok

7. tkhunny Group Title

Well, what do you do with a derivative when you want the slope of a tangent line?

8. agent0smith Group Title

$\Large \dfrac{d}{dx}\int\limits\limits_{0}^{x}\sin(\sqrt{t^{2}-\pi^2})\;dt = \sin(\sqrt{x^{2}-\pi^2})$ You remember how to do that part from earlier, right @onegirl ?

9. agent0smith Group Title

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 Group Title

Yes okay

11. onegirl Group Title

okay i think i got it

12. agent0smith Group Title

@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 Group Title

yea i saw that too

14. tkhunny Group Title

Ah, well that's silly. Why would I do that? Good call!

15. agent0smith Group Title

@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 Group Title

yes

17. onegirl Group Title

thanks!

18. agent0smith Group Title

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 Group Title

okay