Question

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

Answer 1

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?

Answer 2

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

Answer 3

okay

Answer 4

* 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.

Answer 5

hmm ok

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

Yes okay

Answer 10

okay i think i got it

Answer 11

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

Answer 12

yea i saw that too

Answer 13

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

Answer 14

@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

Answer 15

yes

Answer 16

thanks!

Answer 17

Looks like m=0 too, since \[\large m = \sin(\sqrt{0^{2}+\pi^2}) = \sin \sqrt {\pi^2} = \sin \pi = 0\]

so y=0

Answer 18

okay