Welcome to part two of
the properties of the derivative of
a vector valued function.
The goal of this video is to verify
the property involving the derivative of
the cross product of two vector valued functions.
So in the previous video we briefly discussed
these six properties and then we actually
showed the proof for property four.
For this video we're not going to prove
property five, but we are going to verify it
using two different vectors.
This tells us that the derivative of
the cross product of vector valued function r and u
is equal to r crossed with the derivative of u
plus the derivative of r crossed with u.
So again, we're not going to show the proof
but we will show verification with
an example in this video.
What we'll do on this screen is
find the derivative of the cross product
using these two vectors and then on the next screen
we'll determine the right side of this equation
and make sure that they're both equal to each other.
Let's start by finding r crossed with u.
We'll do this by evaluating a three by three determinant
where the first row will be the ijk vectors.
The second row will be the components from
the vector valued function r, so we'll have
t squared, two t cubed, and negative t.
The third row will come from the vector valued function u
so we'll have t, t to the fourth, and four.
We'll evaluate this by using the cofactor expansion method.
Remember this first two by two determinant
will be formed by eliminating row one and column one
so we'll have two t cubed, negative t,
t to the fourth and four.
This determinant will be formed by eliminating
row one and column two, so we'll have
t squared, negative t, t, and four.
The last two by two determinant will be formed by
eliminating row one and column three,
so t squared, two t to the third,
t and t to the fourth.
I'm going to write the result in component form,
so for the x component we'll have eight t to the third
minus negative t to the fifth.
That will become eight t to the third
plus t to the fifth.
Then we'll have four t squared minus negative t squared.
That will be four t squared plus t squared,
that will be five t squared but we are subtracting it
so it will be a negative five t squared.
Here we'll have t to the sixth minus two t to the fourth.
Now we need to determine the derivative of
this cross product.
So we have 24 t squared plus five t to the fourth.
We'll have negative 10 t, then we'll have six t to the fifth
minus eight t to the third.
The derivative of the cross product
or the left side of this equation.
Now what we'll do is determine
the right side of the equation and make sure
that it's equal to what we have here in blue.
Let's go ahead and take the necessary
information over to the next slide and continue.
Notice to evaluate the right side
we are going to have to find the derivative
of the vector valued function r and u.
Let's go ahead and do that now.
We have two t, six t squared, and negative one.
For u prime we have one, four t to the third, and zero.
Let's go ahead and find this cross product first
so again the first row will be ijk.
Second row will be t squared, two t cubed,
and negative t.
The third row will be from u prime
so we'll have one, four t to the third and zero.
Let's go ahead and evaluate this.
The first two by two determinate
will be from eliminating row one and column one
so we'll have two t cubed, negative t,
four t cubed, and zero.
The second two by two determinant will be
formed by eliminating row one and column two
so we'll have t squared, negative t, one, zero.
Then eliminate row one and column three.
We'll have t squared, two t cubed,
one, and four t to the third.
Let's go ahead and write this in component form.
We'll have zero minus negative four t to the fourth.
That will be positive four t to the fourth.
Here we'll have zero minus negative t.
It will become positive t but we are subtracting it
so it will become negative t
and here we'll have four t to the fifth
minus two t to the third.
Now let's go ahead and find r prime crossed with u.
First row will be from r prime so we'll have
two t, six t squared, negative one
and then u will give us t, t to the fourth, and four.
So eliminate row one column one, we'll have
six t squared, negative one, t to the fourth, and then four.
Eliminating row one column two,
we'll have two t, negative one, t and four.
Row one column three, we'll have
two t, six t squared, t, and t to the fourth.
So in component form, we'll have
24 t squared minus negative t to the fourth.
That'll be 24 t squared plus t to the fourth.
Here we'll have eight t minus negative t
or eight t plus t.
That will be nine t but we are subtracting it
so it will be negative nine t
and then here we have two t to the fifth
minus six t to the third.
Okay so what we're trying to find now
is the sum of these two cross products.
Let's go ahead and figure out what that would be.
We're going to go ahead
and add the corresponding components.
Let's go ahead and color code them.
We're going to add this with this,
this with this, and this with this.
So we're going to have 24 t squared
plus five t to the fourth.
We're going to have negative t
plus negative nine t, that's negative 10 t,
and then combining the z components
we're going to have six t to the fifth
minus eight t to the third.
Now we have found the right side of this equation
to be this vector valued function.
This should be the same as what we found
on the previous screen.
Let's go ahead and check it.
Here's the derivative of the cross of r and u.
Here's the sum of r crossed with u prime
and r prime crossed with u and it's a perfect match.
It's not a proof, but we have verified
that property is valid.
That will do it for this video.