Question

Find the length of the curve x=1+9t^2, y=3+6t^3, (0 10 years ago

Answer 1

hey

Answer 2

can u help?

Answer 3

\[\text{ we have to use this } \int\limits_{a(t)}^{b(t)} \sqrt{(x')^2+(y')^2} dt \text{ or try to use } \\ \int\limits_{a(x)}^{b(x)} \sqrt{ 1+(y')^2} dx\]

Answer 4

oh okay great

Answer 5

do you want to try to write y in terms of x?

Answer 6

yes please

Answer 7

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

Answer 8

this question is a tad weird for me
it is 0<x<6 right?

Answer 9

yea it says Find the length of the curve X=1+9t^2, and y+3+6t^3 (0 <t<6)

Answer 10

that should clarify

Answer 11

oh that makes the problem tons easier
you said 0<x<6

Answer 12

yeah sorry

Answer 13

then just use that one parametric formula I gave you

Answer 14

\[\int\limits\limits_{0}^{6}\sqrt{((1+9t^2)')^2+((3+6t^3)')^2} dt\]

Answer 15

i did but my answer wasn't right can u help me check the final answer again?

Answer 16

ok we can check you work
what did you get for (1+9t^2)'?
and for (3+6t^3)'?

Answer 17

i got 461 for the final answer, but know thats wrong what did u get?

Answer 18

ok but what did you get for the derivatives of those two expressions I asked you above

Answer 19

i did but just wanted to check my final answer first

Answer 20

is there anyway you can tell me what you got for the derivative of (1+9t^2) and (3+6t^3)

Answer 21

did you want me to check your problem?

Answer 22

In order for me to do that I need to see where you went wrong if anywhere

Answer 23

yes i wanted to review the retriceer first then go over questions if possible

Answer 24

18t and 18t^2 is the detractive

Answer 25

i just like to see once i have the answer
if i can work back and see my error
it helps my learning

Answer 26

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?

Answer 27

yes

Answer 28

then do i solve from there?? because i got 29200?

Answer 29

notice both terms in the sqrt( ) have a t^2 in common
\[\text{ since } t>0 \text{ then we can write }\\ 18 \int\limits_0^6 \sqrt{t^2} \sqrt{1+t^2} dt \\ \\ 18 \int\limits_0^6 t \sqrt{1+t^2} dt \\ \text{ now use substituion }\]

Answer 30

let u=1+t^2

Answer 31

i have to go soon because my battery is about to die, could you share the answer?

Answer 32

is mine right?

Answer 33

please let me know mat 1%

Answer 34

I haven't done the problem. I was hoping we could do it together.
If u=1+t^2
then du/dt=?

Answer 35

do you know power rule?

Answer 36

constant rule says (1)'=0
power rule says (t^2)=?

Answer 37

use wolframalpha if you just want answers