Welcome to another lesson
on integrating vector valued functions.
In this lesson we'll determine the specific antiderivative
of a vector valued function given an initial condition.
Let's take a look at our first example.
We want to find the antiderivative of r prime of t
that satisfies the initial condition r of zero.
So to find the antiderivative,
we want to integrate r prime of t.
So we'll have the integral
of three t squared times i
minus six square root t j.
And we learn from our first lesson,
we now integrate each component individually.
So now we'll have the integral of three t squared dt
times i and then minus
the integral of six, for square root t we'll write
t to the 1/2 dt times j.
And now we'll integrate.
So we'll have three times
the integral of t squared would be t to the third
divided by three plus a constant of integration
which we'll call C sub one times i.
And then we'll have minus
six times, if we add one to 1/2 that's 3/2,
so we'll have t to the 3/2 divided by 3/2
which is the same as multiplying by the reciprocal of 2/3
plus C sub two times j.
And now we'll simplify.
Three over three simplifies to one,
so we have t to the third plus C sub one i
minus six and three simplify one and two,
so we have four t to the 3/2
plus C sub two times j.
Notice how here we have a family of vector valued functions
because of the C sub one and C sub two.
But because we're given the initial condition r of zero,
we can find the value of C sub one and C sub two
for the specific antiderivative vector valued function.
So if r of zero
equals one i
plus two j,
notice how this tells us when t is zero,
the x component must be one and the y component must be two.
So if t equals zero, looking at the x component,
we would have zero cubed plus C sub one must equal one,
so we know C sub one must equal one.
But we have to be careful about the y component.
Again, when t is zero, this tells us that
minus four times zero to the 3/2
plus C sub two must equal positive two.
So notice how this is zero, but the result here is
negative C sub two equals two
and therefore C sub two is equal to negative two.
Which means the vector valued function that satisfies
this initial condition is r of t equals
t to the third
plus one times i
and then minus
four t to the 3/2 minus two times j.
Let's take a look at a second example.
We want to find the antiderivative of r prime of t
given the initial condition r of zero.
So again, the first step is to integrate
the given vector valued function.
So we'll have the integral of sine two t i
minus cosine t j
plus one divided by the quantity one plus t squared k.
Now we'll integrate each component individually.
Notice to integrate sine two t we have to perform
u substitution where u is equal to two t
and differential u is equal to two dt.
So if we divide both sides by two,
notice how we'll have an extra factor of 1/2
when writing in terms of u.
And the integral of sine u is negative cosine u.
So we'll have negative 1/2 cosine two t
plus C sub one times I
minus the integral of two cosine t
which will be minus two sine t
plus C sub two times j.
Then the integral of one divided by
the quantity one plus t squared
will be equal to arctangent t.
So we'll have plus arctangent t
plus C sub three times k.
Notice here we have a family of vector valued functions,
so now we'll use the initial condition here
to find the values of C sub one, C sub two, and C sub three
to find our particular antiderivative.
Let's go ahead and do this on the next slide.
So now looking at the x component when t is zero,
negative 1/2 cosine zero
plus C sub one must equal three.
Looking at the y component,
notice that minus two sine zero
plus C sub two must equal negative two.
Looking at the z component, notice that arctangent zero
plus C sub three must equal positive one.
Well, cosine zero is equal to one,
so here we have negative 1/2 plus C sub one equals three.
Adding 1/2 to both sides that would have C sub one
equals 3 1/2 or 7/2.
Need to be a little careful here.
Sine zero is zero, because of this negative sign here,
it's going to be negative C sub two equals negative two.
So C sub two equals positive two.
And then finally over here on the right,
arctangent zero is zero.
So we have C sub three equals one.
Which means our specific antiderivative
vector valued function is r of t equals
negative 1/2 cosine two t
plus 7/2
times i
minus two sine t
plus two times j plus
arctangent t
plus one times k.
Okay that's going to do it for these two examples.
Next we'll take a look at definite integration
involving vector valued functions.
I hope you found this helpful.