Question

Help with this disgusting integral?

Answer 1

\[\int\limits_{0}^{2} \sqrt{(2t^2)+(3t^2)^2}\]

Answer 2

@Astrophysics

Answer 3

factor out a \(t^2\) from under the square root then use u-substitution

Answer 4

Is the \(2t^2\) term squared as well?

Answer 5

My apologies the 2t term is being squared. Must have accidentally mistyped. So it's actually (2t)^2 or 4t^2

Answer 6

\[\int\limits_{0}^{2} \sqrt{(2t)^2 + (3t^2)^2}\]

Answer 7

Use the method suggested by Zarkon.

Answer 8

Ah. So it'd be integral from 0 to 2 of 4t^2 + 9t^4 we'd factor t^2 out then split the radicals into multiplication then u-sub the 4 + 9t^2 to get du = 18t dt then divide both sides by 18 change the boundaries and then integrate sqrt u?

Answer 9

sounds about right.

inside you have sqr(u)
outside you have t dt
which becomes, using du = 18 t dt
t dt = 1/18 du
so
\[ \frac{1}{18} \int \sqrt{u} \ du\]

for the limits, u = 9t^2+4. at t=0 that corresponds to u = 4
similar for the upper limit

Answer 10

right. thanks

Answer 11

would it be possible to integrate this if we were to add +t's to the x and y instead? so x = t^2 + t and y = t^3+t?

Answer 12

that integration seems like theres no real way to solve for it unless you use approximation. :o

Answer 13

But why?

Answer 14

Because the question itsself may have been presented incorrectly by our instructor and we were advised to add the extra t. I was just speculating on whether or not it was solvable with the original problem we were given (x = t^2, y = t^3). However, when I did solve it that way the feedback that was given included that i should've added an extra t to both even though it explicitly says on the paper x = t^2 and y = t^3. I really dont know why considering this way is solvable.

Answer 15

Apparently it was his mistake of using the original x and y but apparently adding in those extra t's makes the problem unsolvable with calc 2 methods lol... i dont know

Answer 16

I assume there is more to the original problem?
I can't respond without a lot more detail on what you are solving

Answer 17

The total problem states: Find the exact length of the curve where x = t^2 and y = t^3 from the bounds 0 (less than equal to) t (less than equal to) 2

Answer 18

I apologize if i'm not making any sense, just been trying to solve these for awhile now. :P

Answer 19

ok, and that is how you get your integral

Answer 20

I think you need to take the derivative first but I am not sure.

Answer 21

I use
http://tutorial.math.lamar.edu/Classes/CalcII/ArcLength.aspx
to remind me.

if we use his formula
\[ \int \sqrt{1 - \left(\frac{dy}{dx}\right)^2} \ dx\]
with x= t^2 dx = 2 t dt
y= t^3 dy = 3 t^2 dt
dy/dx = 3/2 t

replace dy/dx with 3/2 t and replace dx with 2 t dt
and we get
\[ \int \sqrt{1 - \frac{9t^2}{4}} \ 2 t \ dt \]

Answer 22

* that should be 1 + 9t^2/4 ***

if we factor ¼ out of the root we get
\[ \int \sqrt{4 + 9t^2} \ t \ dt \]
which is easily integrated

Answer 23

adding "t" e.g. x= t^2 + t and y= t^3+t
seems like that would make it difficult

Answer 24

http://tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx
\[\int\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\]

Answer 25

yes, that looks more direct

Answer 26

right, that's what i used thomas. both of those are correct though right?

Answer 27

the formulas i mean

Answer 28

yes, they are variations. the parametric version is better suited here

Answer 29

Did you take the derivative though?

Answer 30

thomas' link shows how to go from the first version to the parametric version

Answer 31

Paul's Math Notes is love, Paul's Math Notes is life.

Answer 32

yeah i took the derivaitve. dx/dt = 2t and dy/dt = 3t^2

Answer 33

from x = t^2 and y = t^3

Answer 34

anyway thanks you guys. <3

Answer 35

I don't know why they suggest adding the "t"
wolfram seems to use numerical integration to get an answer.

Answer 36

neither do i. i'm just sitting here looking @ the feedback and it says "add extra t's" lol......i dont even.... oh well. i'll be sure to ask him during next lecture to see whats up.

Answer 37

ty for the help though, appreciate it.

Answer 38

have you done a substitution yet?

Answer 39

you only need an algebraic substitution

Answer 40

\[\int\limits \sqrt{4+9t^2} t dt \\ \text{ Let } u=4+9t^2\]

Answer 41

but scrolling up @phi has already told you to use this sub
he even showed you how to change the limits

Answer 42

yeah thanks already figured it out mostly. :D

Answer 43

Refer to the attachment from Mathematica v9.