We want to find the position vector function,
r of t, for a particle with acceleration, initial velocity,
and initial position, given below.
So, we're given the acceleration vectored valued function,
the velocity vector at t equals zero,
and the position vector at t equals zero.
So, if we take a look at this graphically,
our ultimate goal is to find r of t
that would trace out this purple space curve.
And we're given the position vector at t equals zero,
this light green vector here.
And we're also given the velocity vector at t equals zero,
which would be this red vector.
This dark blue vector here,
is acceleration vector at t equals zero,
which we could find since we're given a of t.
And we can find the position vector function, r of t,
by using integration.
Since we're given the acceleration vectored valued function,
we'll first find the velocity vectored valued function
by taking the integral of a of t.
And then we can find the position vector function, r of t,
by finding the integral of v of t.
So, we'll first find v of t,
which is equal to the integral of a of t.
And we're given a of t, so we have the integral of the
vectored valued function with an x component of two t,
a y component of two sin t,
and a z component of cosine four t.
And, now, integrate each component with respect to t.
So, the s component would be the antiderivative of two t,
which would two times t squared, divided by two,
or just t squared, plus a constant,
which we'll call c sub one.
The y component would be the antiderivative of two sin t,
which would be negative two, cosine t,
plus, let's call it c sub two.
And the z component would be the antiderivative of
cosine four t, which does require u substitution,
u equals four t, differential u equals four dt.
So, if we divide both sides by four,
notice how 1/4 du equals dt,
which means you'll have an extra factor
of 1/4 when integrating.
So, the antiderivative of cosine four t is going to be
1/4 sin, four t,
plus, let's call it c sub three.
And, now, because we're given the velocity vector,
v of zero, we can determine the constants of integration.
So, if v of zero
equals
the vector
one comma negative three comma two,
if we substitute t equals zero into the x component,
notice how we would have zero squared plus c b one,
must equal one.
I get zero squared,
plus c sub one must equal one,
therefore, c sub one equals one.
Looking at the y component, if we substitute zero for t,
cosine zero is equal to one.
So, we'd have negative two times one, or negative two,
plus c sub two equals negative three.
Adding two to both sides.
We know that c sub two equals negative one.
And, then, finally, for the z component.
When t equals zero, notice how we have sin zero,
which is zero, so, we'd have c sub three equals two.
Which means now we know that v of t
has an x component of t squared, plus one,
a y component of negative two, cosine t, minus one,
and a z component of 1/4 sin four t,
plus two.
And, now, that we have v of t, we can integrate to find
our position vector function, r of t.
Let's do this on the next slide.
So, now r of t
is equal to the integral of v of t,
which we just found.
So, we'll have the integral of, again, the x component is
t squared, plus one, the y component is negative two,
cosine t, minus one,
and the z component is 1/4 sin, four t,
plus two.
And, once again, we'll find the antiderivative
of the xy and z components.
So, for the x component we'd have t to the third,
divided by three, plus t, plus a constant,
which we'll call c sub four.
For the y component, the antiderivative of negative two,
cosine t, is going to be negative two, sin t.
The antiderivative of one is t, so minus t, plus c sub five.
And, now, for the z component, notice how, again,
we have to perform u substitution, where u equals four t.
So, we'll have an extra factor of 1/4 again,
and since 1/4 times 1/4 is 1/16,
the antiderivative of 1/4 sin four t would be negative 1/16,
cosine four t, and then I have plus two t,
and then plus c sub six.
So, if you need to you can go back and take a look at the
u substitution when we found v of t.
And, now, because we're given r of zero,
we can determine the constants,
c sub four, c sub five, and c sub six,
which will give us our position vector function, r of t.
Again, we know that r of zero has an x component of two,
a y component of four, and a z component of negative one.
So, let's substitute t equals zero into each component
to determine our constants.
So, if we substitute t equals zero into the x component,
we'd have zero, plus zero, plus c sub four,
which must equal two.
So, we know c sub four equals two.
Looking at the y component.
T is zero.
We have sin zero here, which is zero.
So, we have zero, minus zero, plus c sub five,
which must equal four, so c sub five equals four.
And, finally, looking at the z component.
This one is a little bit more involved.
When t is zero we have cosine zero, which is one.
So, it'd have the equation negative 1/16,
plus two times zero, which is zero, plus c sub six,
which must equal negative one.
So, if we add 1/16 to both sides we'd have c sub six equals
negative one, or negative one, over one, plus 1/16.
Of course negative one is equivalent to
negative 16/16, and therefore,
c sub six equals negative 15/16.
So, finally, now we know our position vector function,
r of t,
has an x component of, let's say 1/3,
t to the third, plus t, plus c sub four,
which we know is two, so plus two.
The y component is negative two, sin t,
minus t, plus c sub five, which is four, so plus four.
And a z component of negative 1/16, cosine four t,
plus two t, plus c sub six,
which would be negative 15/16, or minus 15/16.
Quite a long process, but from the given information
we were able to determine the position vector function,
r of t,
that, once again, would trace out
the purple space curve that we see here.
I hope you found this helpful.