I'm slightly confused on this one...?
$$\large If~~f(x) = \int\limits_{9}^{3x}\sqrt{2t^3+1}dt$$determine the equation of the line tangent to f at x = 3

Question

I'm slightly confused on this one...?
$$\large If~~f(x) = \int\limits_{9}^{3x}\sqrt{2t^3+1}dt$$determine the equation of the line tangent to f at x = 3

jigglypuff314 · Answer

If you're not too busy
would this be "no answer" or something like that? :/

@zepdrix @mathslover @surjithayer help? :3

anonymous · Answer

That's an interesting problem, and it has a pretty neat answer. What have you tried?

jigglypuff314 · Answer

f(3) = intergral from 9 to 9...    which already suggests something's weird >,<

jigglypuff314 · Answer

truthfully, my first attempt was using second fundamental theorem to get f'(x)
then got  f'(3) = 114.5906...
but still hit the same problem with the   f(3) = ?

anonymous · Answer

I really want to answer but I feel like @zzr0ck3r will ninja me.

zzr0ck3r · Answer

well now im not sure...

jigglypuff314 · Answer

@mathmale could you help me?

anonymous · Answer

Does that mean I can answer without being ninja'd?

zzr0ck3r · Answer

$f'(x)=\sqrt{2(3x)^3+1}\f'(3)=\sqrt{1459}\f(3)=0$
so
$0=\sqrt{1459}*3+b \b=-3\sqrt{1459}$
so
$y=\sqrt{1459}x-3\sqrt{1459}$ is what I think

anonymous · Answer

^yep.

anonymous · Answer

's what I got.

jigglypuff314 · Answer

>.> so that's why I only got one point off XD
ok, so that's whut I answered on my practice test
but had forgot f(3) = ?  :P
mk thanks!

jigglypuff314 · Answer

side note @zzr0ck3r forgot chain rule :P
3 sqrt(1459)  not just  sqrt(1459)

mathmale · Answer

As you probably realize, $$\large If~~f(x) = \int\limits\limits_{9}^{3x}\sqrt{2t^3+1}dt$$ then f(x) is defined as an unfinished definite integral.

Anyone know how to find the derivative of the following?
$$f(x) = \int\limits\limits_{9}^{x}\sqrt{2t^3+1}dt?$$

Look carefully!  I made I small change to the definite integral.

zzr0ck3r · Answer

$\sqrt{2x^3+1}$ by 
FTC

zzr0ck3r · Answer

thats how I got $f'(x)=\sqrt{2(3x)^3+1}$

jigglypuff314 · Answer

chain rule     derivative of  3x  =  3
so   f'(x) = 3 sqrt (2(3x)^3 + 1)

zzr0ck3r · Answer

ahh well thats going to change some stuff

zepdrix · Answer

So what's the part you're stuck on?
Evaluating f(x) from 9 to 9?

jigglypuff314 · Answer

yeah, I had a brain fart :P  thought it was und  rather than 0

zepdrix · Answer

0 seems right :)
You can think of some area under a curve which has no "width" dimension.

mathmale · Answer

Remember, we're finding the derivative of a function defined as a definite integral.  In this particular case, the lower limit of integration, 9, simply disappeasrs.  And yes, we must use the chain rule (the derivative of 3x is, indeed, 3).

What is the significance/purpose of the result we now have?

mathmale · Answer

In other words, why do we want the derivative of f(x) in the first place?

jigglypuff314 · Answer

equation of line tangent requires slope
f'(x) gives slopes of f(x)

mathmale · Answer

Cool.   And, at x=3, that slope is ... ?

jigglypuff314 · Answer

3 sqrt(1459)   or   114.5906...

mathmale · Answer

Evaluate $$f'(x)=\sqrt{2(3x)^3+1}$$ at x=3.  Why?

How would we find f(3)?

jigglypuff314 · Answer

so equation I got was   y = 3 sqrt(1459) (x-3)

mathmale · Answer

I'm not checking that, but it looks reasonable.  Why not leave this as $$3\sqrt{1459}$$without bothering to evaluate it?  More accurate to take that approach.

mathmale · Answer

Good for @zzr0ck3r !  I see that he/she has taken exactly this approach.

jigglypuff314 · Answer

I was crunched for time in class while doing it so I left it as what my calculator gave as 114.5906  which was marked correct
I understand what I did wrong now,
Thank you for all your help! :)

mathmale · Answer

I thought this was a very good discussion and appreciate all the contributions each of you has made.  :)