Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Line integrals

:frown:

I must integrate

\int_C \vec{F} \cdot d\vec{r}

where C is given by parametric equations

x = t^4

,

y = 0

,

z = t^2

and

\vec{F} = xz\hat{i} +z^2 \hat{j} + x^2 \hat{k}

Haven't got a clue :s-smilie:

:s-smilie:

What's your definition of a line integral?
Also, where does t vary between?

OP

Original post by Glutamic Acid

What's your definition of a line integral?
Also, where does t vary between?

Well since the vector field isn't conservative, I'm not sure.

What t varies between can only be seen from the fact the line integral is between (0, 0, 0) and (1, 0, 1) which I forgot to mention for some reason

OP

This is some of my working:

\frac{dr}{dt} = <3t^2, 0, 1>

Thus

\int _C \vec{F} . d \vec{r} = \int _C <xy, z^2, x^2> . <3t^2, 0, 1> dt

Is this near right?

OP

Anyone understand where I'm going wrong?

$Avatar for Get me off the £\?%!^@ computer$

Get me off the £\?%!^@ computer

Original post by Dagnabbit

This is some of my working:

\frac{dr}{dt} = <3t^2, 0, 1>

Thus

\int _C \vec{F} . d \vec{r} = \int _C <xy, z^2, x^2> . <3t^2, 0, 1> dt

Is this near right?

You said "C is given by parametric equations

x = t^4

,

y = 0

,

z = t^2

" so I'd check this bit <3t^2, 0, 1> if I were you.

Once you've done that, express x, y and z in terms of t, do the dot product and integrate between your limits.

(edited 13 years ago)

Original post by Dagnabbit

:frown:

I must integrate

\int_C \vec{F} \cdot d\vec{r}

where C is given by parametric equations

x = t^4

,

y = 0

,

z = t^2

and

\vec{F} = xz\hat{i} +z^2 \hat{j} + x^2 \hat{k}

Haven't got a clue :s-smilie:

:s-smilie:

You were wrong with the

d\vec{r}

\int_C \vec{F} \cdot d\vec{r}

As

\vec{F} = xz\vec{i} +z^2 \vec{j} + x^2 \vec{k}

\vec{r}= t^4\vec{i} +0 \vec{j} + t^2\vec{k}

So your integral

\int_C (t^6\vec{i}+t^4\vec{j}+t^8\vec{k})\cdot d(t^4\vec{i}+0\vec{j}+t^2\vec{k})=

\int_C (t^6\vec{i}+t^4\vec{j}+t^8\vec{k})\cdot (4t^3\vec{i}+0\vec{j}+2t\vec{k})\ dt=

\int^{t2}_{t1} (4t^9+2t^9\ dt=\int^{t2}_{t1} 6t^9\ dt

$Avatar for Get me off the £\?%!^@ computer$

Get me off the £\?%!^@ computer

Original post by ztibor

You were wrong with the

d\vec{r}

\int_C \vec{F} \cdot d\vec{r}

As

\vec{F} = xz\vec{i} +z^2 \vec{j} + x^2 \vec{k}

\vec{r}= t^4\vec{i} +0 \vec{j} + t^2\vec{k}

So your integral

\int_C (t^6\vec{i}+t^4\vec{j}+t^8\vec{k})\cdot d(t^4\vec{i}+0\vec{j}+t^2\vec{k})=

\int_C (t^6\vec{i}+t^4\vec{j}+t^8\vec{k})\cdot (4t^3\vec{i}+0\vec{j}+2t\vec{k})\ dt=

\int^{t2}_{t1} (4t^9+2t^9\ dt=\int^{t2}_{t1} 6t^9\ dt

I also worked out the answer but I knew better than to post it.

(edited 13 years ago)

Original post by Get me off the £\?%!^@ computer

I also worked out the answer but I knew better than to post it.

Surely

$Avatar for Get me off the £\?%!^@ computer$

Get me off the £\?%!^@ computer

Original post by ztibor

Surely

Indubitably.

Quick Reply

Latest

Trending

Trending

Articles for you

LaTex

What is an LLM and how could it benefit your career?

What is an LLM and how could it benefit your career?

Powerful questions business students forget to ask their career advisers

Powerful questions business students forget to ask their career advisers

Why I chose a law degree and what studying law was like

Why I chose a law degree and what studying law was like