anonymous
  • anonymous
Find the length of the curve x=1+9t^2, y=3+6t^3, (0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
hey
anonymous
  • anonymous
can u help?
freckles
  • freckles
\[\text{ we have to use this } \int\limits_{a(t)}^{b(t)} \sqrt{(x')^2+(y')^2} dt \text{ or try to use } \\ \int\limits_{a(x)}^{b(x)} \sqrt{ 1+(y')^2} dx\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
oh okay great
freckles
  • freckles
do you want to try to write y in terms of x?
anonymous
  • anonymous
yes please
freckles
  • freckles
i'm asking that because if we try to write the x limits in terms of t limits we will end up with 4 numbers 2 of them being imaginary
freckles
  • freckles
this question is a tad weird for me it is 0
anonymous
  • anonymous
yea it says Find the length of the curve X=1+9t^2, and y+3+6t^3 (0
anonymous
  • anonymous
that should clarify
freckles
  • freckles
oh that makes the problem tons easier you said 0
anonymous
  • anonymous
yeah sorry
freckles
  • freckles
then just use that one parametric formula I gave you
freckles
  • freckles
\[\int\limits\limits_{0}^{6}\sqrt{((1+9t^2)')^2+((3+6t^3)')^2} dt\]
anonymous
  • anonymous
i did but my answer wasn't right can u help me check the final answer again?
freckles
  • freckles
ok we can check you work what did you get for (1+9t^2)'? and for (3+6t^3)'?
anonymous
  • anonymous
i got 461 for the final answer, but know thats wrong what did u get?
freckles
  • freckles
ok but what did you get for the derivatives of those two expressions I asked you above
anonymous
  • anonymous
i did but just wanted to check my final answer first
freckles
  • freckles
is there anyway you can tell me what you got for the derivative of (1+9t^2) and (3+6t^3)
freckles
  • freckles
did you want me to check your problem?
freckles
  • freckles
In order for me to do that I need to see where you went wrong if anywhere
anonymous
  • anonymous
yes i wanted to review the retriceer first then go over questions if possible
anonymous
  • anonymous
18t and 18t^2 is the detractive
anonymous
  • anonymous
i just like to see once i have the answer if i can work back and see my error it helps my learning
freckles
  • freckles
great! now inputting those in we get: \[\int\limits_{0}^{6}\sqrt{(18t)^2+(18t^2)^2} dt \\ \sqrt{18^2} \int\limits_0^6\sqrt{t^2+t^4} dt \\ 18 \int\limits_0^6 \sqrt{t^2+t^4} dt\] did you get to this part?
anonymous
  • anonymous
yes
anonymous
  • anonymous
then do i solve from there?? because i got 29200?
freckles
  • freckles
notice both terms in the sqrt( ) have a t^2 in common \[\text{ since } t>0 \text{ then we can write }\\ 18 \int\limits_0^6 \sqrt{t^2} \sqrt{1+t^2} dt \\ \\ 18 \int\limits_0^6 t \sqrt{1+t^2} dt \\ \text{ now use substituion }\]
freckles
  • freckles
let u=1+t^2
anonymous
  • anonymous
i have to go soon because my battery is about to die, could you share the answer?
anonymous
  • anonymous
is mine right?
anonymous
  • anonymous
please let me know mat 1%
freckles
  • freckles
I haven't done the problem. I was hoping we could do it together. If u=1+t^2 then du/dt=?
freckles
  • freckles
do you know power rule?
freckles
  • freckles
constant rule says (1)'=0 power rule says (t^2)=?
freckles
  • freckles
use wolframalpha if you just want answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.