Question

Find the length of the curve. x=1+3t^2 y=4+2t^3 0 13 years ago

Answer 1

you should have a formula...it works out nicely

Answer 2

I got sqrt of 76t^2

Answer 3

Oh, I see what I did wrong

Answer 4

the length of the curve given parametric equations:\[L = \int\limits_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}}dt\]\[a \le t \le b\]

Answer 5

No, I don't get it

Answer 6

I got 7 as an answer, but that's not right.

Answer 7

what do you get for your derivatives ?

Answer 8

Not 7, 5

Answer 9

dx/dt=6t dy/dt=6t^2

Answer 10

good

Answer 11

then you squared them and added them

Answer 12

Yes

Answer 13

what did you get?

Answer 14

\[36t^2+36t^4\]

Answer 15

all under a square root correct

Answer 16

yes

Answer 17

simplify it

Answer 18

6t+6t^2

Answer 19

\[\sqrt{36t^2+36t^4}\]

\[\sqrt{36t^2(1+t^2)}=\cdots\]

Answer 20

oh, I see. \[6t \sqrt{1+t^2}\] u=1+t^2 du=2tdt

Answer 21

yes...you have it

Answer 22

I think I'm doing something wrong I get the answer to be 3 and the book says it's 4sqrt(2)-1

Answer 23

that is what your book gives you?

Answer 24

I get \[4\sqrt{2}-2\]

Answer 25

It's right I checked on a graphing calc.

Answer 26

Something is wrong.

Answer 27

You are right

Answer 28

I wrote it wrong, my bad.

Answer 29

\[\int\limits_{0}^{1}6t\sqrt{1+t^2}dt\]

\[u=1+t^2\]
\[du=2tdt\]

\[\frac{du}{2}=tdt\]

\[\int\limits_{1}^{2}3\sqrt{u}du\]

correct?

Answer 30

Yeah

Answer 31

\[=3\frac{2}{3}u^{3/2}\Big|_1^2\]

\[=2u^{3/2}\Big|_1^2=2(2^{3/2})-2(1^{3/2})\]

Answer 32

\[=4\sqrt{2}-2\]

Answer 33

I must have put something else in because I just got 4, thanks!