## onegirl Find an equation of the tangent line at the given value of x. 11 months ago 11 months ago

1. onegirl

2. 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?

3. tkhunny

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

4. onegirl

okay

5. 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.

6. onegirl

hmm ok

7. tkhunny

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

8. 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 ?

9. 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.

10. onegirl

Yes okay

11. onegirl

okay i think i got it

12. agent0smith

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

13. onegirl

yea i saw that too

14. tkhunny

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

15. 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

16. onegirl

yes

17. onegirl

thanks!

18. 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

19. onegirl

okay