Question

(Line Integral) integrate F over the given curve:
f(x,y)=x^3/y C: y=x^2/2 0<=x<=2

Answer 1

my problem is coming up with r(t)

Answer 2

i dont really understand what you want. the equation of the curve is y=0.5*x^2.

the intergration needs to take place between x = 0 and 2. what does the function have to do with anything?

Answer 3

it is a line integral over a given curve related to vector fields

Answer 4

Calculus 3

Answer 5

its 5 years since ive done calculus, and i dont integrate on a daily basis anymore, so i cant remember everything anymore..hope someone else can help

Answer 6

r(t)=t^2/2 with 0<=t<=2

Answer 7

x=t, y=t^2/2

Answer 8

If you're desperate about using t anyway.

Answer 9

THomas I thought so too, but I can't get the right answer. I think it has to do with Cos t and Sin t

Answer 10

I don't think it's sine or cosine, because the curve isn't a circle or an ellipse of some kind.

Answer 11

can you work out the problem with x=t y=.5t^2 and see if we get smae answer. The anser is 10(5^1/2)-2/3

Answer 12

i get something totally different

Answer 13

You get:
\[\int\limits_{0}^{2}2tdt\]
right?

Answer 14

yep

Answer 15

2root2

Answer 16

you have to take vector and multiply |v|

Answer 17

f(x,y)=int a to b (g(t,h(t))|v(t)|dt

Answer 18

I don't know how this works anymore I'm afraid. I think you're right that the integrand isn't just 2t, but I'm not sure what to add. What do you mean by |v(t)|?

Answer 19

|r '(t)|?

Answer 20

if v=ti+tj+0k v=sqrt 1^2+1^2= sqrt 2=|v|

Answer 21

\[|v|=\sqrt{2t^{2}}=\sqrt{2}t\]
right? but what's v in this context?

Answer 22

v is the length of the curve from my understanding, in the formula |v(t)| is a constant that we take out of the integral and multiply at the end. It is derived from the vector. I got the same as you, but I cannot figure out how they arrived at 10root 5 something!

Answer 23

http://www.wolframalpha.com/input/?i=line+integral&lk=1&a=ClashPrefs_*MathWorld.LineIntegral-

This made things clearer for me.

Answer 24

Wait, we don't have a vector field, we have a scalar field.

Answer 25

Yes scalar, vector field is the next topic, but I have to get a handle on this first :)

Answer 26

http://www.wolframalpha.com/input/?i=path+integral

This applies.

Answer 27

yes that is the formula

Answer 28

So we need to compute |r '(t)|.

Answer 29

yep

Answer 30

r(t)=(t,t^2/2), so what's r '(t)?

Answer 31

1^2 1/2^2?sqroot?

Answer 32

or is it 1^2 1^2=sqrt2?

Answer 33

wait is that r prime t?

Answer 34

yes, the derivative.

Answer 35

my book gives me V(t)

Answer 36

not prime

Answer 37

what's prime?

Answer 38

derivative no?

Answer 39

Right, we need to compute the derivative of r(t), which is a vector. Just derive both components to t.

Answer 40

oh crap Velocity is the derivative of R(t) isn't it.

Answer 41

Oh, probably.

Answer 42

and acceleration is double prime of r(t) or prime of V

Answer 43

Even if we accept V(t) is the derivative, I still do not get the right answer though.

Answer 44

derivative would be 1^2 +1^2 all under sqroot= sqrt 2 yes?

Answer 45

v(t)=(1,t)

Answer 46

r(t)=(t,t^2/2), so for v(t) you derive both components to t.

Answer 47

er, right sqrt 1+t how does that help?

Answer 48

\[\sqrt{1+t^{2}}\]

Answer 49

yep again i erred, so we tack it onto the end of t^3/t^2?

Answer 50

yes, so we get \[\int\limits_{0}^{2}2t\sqrt{1+t^{2}}\]

Answer 51

im with you so far.

Answer 52

integration by parts?

Answer 53

no u substitution

Answer 54

that's right.

Answer 55

\[\int\limits_{0}^{\sqrt{2}}\sqrt{1+u}du\]

Answer 56

right?

Answer 57

no 1/2sqrtu

Answer 58

from 1 to 5

Answer 59

let me work this out i think your on right track

Answer 60

\[2\int\limits_{0}^{2}t ^{3}/t ^{2}\sqrt{1+t ^{2}}\] yes?

Answer 61

right.

Answer 62

My substitution is wrong, I'm not sure why.

Answer 63

ll =1 ul=5 u sub =1+t^2 dr us ub =2t

Answer 64

ll=0+1=1 ul=2^2+1=5

Answer 65

I just worked out the whole problem Its correct!!!! THANK YOU!!!!

Answer 66

You're welcome.

Answer 67

I've been working on this crap for 2 days, taking a break good job mate!

Answer 68

http://tutorial.math.lamar.edu/Classes/CalcIII/CalcIII.aspx

Here is a good Calc 3 supplement.