Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a homework question that is as follows:

If $C$ is the curve given by $X ( t ) = ( 1 + 3 \sin(t)) I + ( 1 + 3 \sin^2(t) ) J + ( 1 + 5\sin^3(t) ) K$, $0 ≤ t ≤ π/2$, and $F$ is the radial vector field $F ( x, y, z ) = xI + y J + z K$, compute the work done by $F$ on a particle moving along $C$.

Can someone explain the recipe without actually doing it for me?

I'll respond ASAP.

Thank you!

EDIT:

I'm just kind of stuck because I don't know what $x$, $y$, and $z$ are in terms of $t$. Does $x = 1 + 3\sin(t)$?

EDIT 2:

If $x = 1 + 3\sin(t), y = 1 + 3\sin^2(t), z = 1+5\sin^3(t)$

Then $dx = 3\cos(t), dy = 3\sin(2t), dz = 15\sin^2(t)\cos(t)$

share|improve this question
    
Should $3\sin(2)$ be $3\sin(t)$ in the ${\bf j}$ component? –  Bill Cook Dec 14 '11 at 20:47
1  
Yes, to your question, since you are only interested in points on the curve. Write $F$ in terms of $t$. Then write $F\cdot X'$ in terms of $t$ and integrate over $[0,\pi/2] $. –  David Mitra Dec 14 '11 at 20:48
    
@Bill: Fixed that error Mr. Cook, thanks! David: Alright cool, I'll get to work on it –  user13327 Dec 14 '11 at 20:50

1 Answer 1

up vote 1 down vote accepted

To compute a line integral (using a parametrization and not calling on some "big" theorem) you simply compute the derivative of your parametrization and then plug everything in.

If your parametrization is $X(t) = x(t){\bf i}+y(t){\bf j}+z(t){\bf k}$ where $a \leq t \leq b$, then...

Every $x$ in your vector field's component formulas should be replaced by $x(t)$. Each $y$ with $y(t)$ and $z$ with $z(t)$. Next, replace $d{\bf X}$ with ${\bf X}'(t)\,dt$. Then compute the dot product $F({\bf X}(t)) {\bf \cdot} {\bf X}'(t)$ and integrate with respect to $t$ from $a$ to $b$.

$$ \int_C F {\bf \cdot} d{\bf X} = \int_a^b F({\bf X}(t)) {\bf \cdot} {\bf X}'(t)\;dt = \int_a^b F(x(t),y(t),z(t)) {\bf \cdot} {\bf X}'(t)\;dt $$

If your line integral involves things like "$P\;dx+Q\;dy+R\;dz$", you simply plug-in $x(t)$ for $x$, $y(t)$ for $y$, $z(t)$ for $z$ then $x'(t)dt$ for $dx$, $y'(t)dt$ for $dy$, and $z'(t)dt$ for $dz$.

$$ \int_C Pdx+Qdy+Rdz = \int_a^b \Big(P(x(t),y(t),z(t))x'(t) + \qquad \qquad \quad\qquad \qquad \qquad \quad\qquad \quad$$ $$ \qquad \qquad \qquad Q(x(t),y(t),z(t))y'(t) + R(x(t),y(t),z(t))z'(t) \Big)dt $$

share|improve this answer
    
You win at life sir –  user13327 Dec 14 '11 at 21:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.