The force exerted by an electric charge
at the origin on a charged particle at the point x,y,z
with position vector r is F of r equals K times r
divided by the magnitude of r cubed, where K is constant.
Assume K equals 10, find the work done as the particle
moves along a straight line from the point 2,0,0
to the point 2,2,3.
The work done as the particle moves along the straight line
is equal to line integral along the curve C of F
dot differential r.
To evaluate the line integral and the find the work done,
we'll write the integral in this form here
where we write F as a function of t
using the parametrization of the curve C,
and differential r is equal to r prime of t dt.
So let's first find r of t that would trace out the line
from this point to this point.
We'll notice how the x coordinate stays at two
and so the F component of r would just be two.
The y coordinate goes from zero to two,
so let's let the y component be equal to 2t.
The z coordinate goes from zero to three,
let's let the z component of r be equal to 3t.
So this will trace out the straight line
from this point to this point
when t is on the close interval from zero to one.
Looking at our formula for F, we'll also have to find
the magnitude of r, let's go ahead and do that now.
The magnitude of r of t is going to be equal to
the square root of two squared plus 2t squared
plus 3t squared.
So simplifying, here we're going to have the square root
of four plus 4t squared plus 9t squared,
that's plus 13t squared.
Let's go ahead and write this as the quantity four plus 13
t squared to the 1/2.
We're also going to need our prime of t,
let's find that now as well.
Our prime of t is going to have an x component,
this derivative of two which is zero.
The y component which is a derivative of 2t which is two.
And the z component which is a derivative of 3t
which should be three.
Now let's find F of r of t.
We're going to have K which is 10
times r of t which is here,
is we have an x component of two,
a y component of 2t, a z component of 3t
divided by the magnitude of r of t cube.
Well, if we cube this we're going to have the quantity
four plus 13t squared to 3/2 power.
Let's write this as an x component of 20
divided by the quantity four plus 13t squared to the 3/2,
a y component of 20t divided by
the quantity four plus 13t squared to the 3/2,
and the z component would be 30t divided by
the quantity four plus 13t squared raised to 3/2.
Which means the work done as the particle moves
from this point to this point along a straight line
is equal to the integral, and can we have F of r of t
or F of x of t, y of t, z of t which we have here.
Dotted with our prime of t which we have here,
we have x component of zero, a y component of two,
a z component of three dt.
[mumbles] integration for t as the find here
are going to be from zero to one.
Our next step will be to find the dot product
and then we'll evaluate the integral.
Let's do this on the next slide.
So we'll have the integral from zero to one
of this fraction times zero, of course is zero.
And this fraction times two is going to be 40t
divided by the quantity four plus 13t squared to the 3/2
and then plus this fraction times three is going to be 90t
divided by the quantity four plus 13t squared to the 3/2.
Notice how we can add this two fractions
because we have a common denominator.
So our new numerator is going to be 130t,
let's write this as 130 times integral from zero to one of t
divided by the quantity four plus 13t squared
raised to 3/2 dt.
And I'll perform mu substitution over that u
be equal to four plus 13t squared
and therefore differential u is equal to 26t dt,
let's divide both sides by 26.
We have 1/26 du equals t dt.
So in terms of u, we can think of all of this,
as again, t dt is equal to 1/26 du
and this should be one over u to the 3/2
or u to the negative 3/2.
So the integral to this vector u would be equal to 130
divided by 26 times u to the negative 1/2
divided by negative 1/2 or 13/26 times negative two,
times u to the negative 1/2.
Which means the respect to t would have 130 divided by 26
times negative two and u to the negative 1/2
is really four plus 13 t squared to the negative 1/2.
Let's simplify here 130 divided by 26 is five,
five times negative two is negative 10,
so we have negative 10.
Let's write this as one over the square root of four
plus 13t squared.
So when t is one, we're going to have one over,
when t is one this is going to be the square root of 17,
minus when t is zero, I'm going to have one
over the square root of four.
Of course the square root of four simplifies to two
so let's write the exact value as negative 10
divided by the square root of 17
and then we have plus 10 divided by two or plus five.
So this will be the exact amount of work
done as the particle moves along the straight line
from this point to this point.
Where the decimal approximation,
this should be approximately 2.5746.
I hope you found this helpful.