BAM! Mr. Tarrou. In this video we're going to do an example of a particle in motion
and describing its position with a position vector which makes this
technically a calc 2 or calc about the bc concept and if you're studying for
the AP test here in 2020 you know that it is not going to be on this year's
test but it will be on future years and maybe we're just looking at this for
learning something in our calculus classes and it sounds complicated in the
fact that I'm talking about position vectors here but really all this
question is involving is our understanding of that change theorem
that when you integrate velocity you get a displacement or net change from your
initial position so let's go ahead and get going
we have a particle moves in the XY plane so that it's velocity vector at time T
is V of T is equal to and depend on your textbook you'll have this notation
differently like in IB our vectors are going to be stacked probably in a
vertical orientation where it'd be T squared and this is the sine of PI T but
we're using this this sort of arrow notation here that vector V of T is
equal to or described by T squared for the movements in the horizontal
direction along the x axis and sine of PI T for our vertical movement and the
particles position at T equals 1 is described by the position vector of 2 0
what's a position vector of the particle when T is equal to 5 ok so basically
this really is a review of some concepts you learn in calc a B or calc 1 which is
we're just going to be analyzing the position of a particle both horizontally
and vertically so it's really kind of just two separate problems of net change
theorem basically so at time is equal to 1 the horizontal position is 2 so as far
as horizontally where's the position where's the particle at T is equal to 5
that is going to be equal to 2 Plus now we're going to integrate the velocity
function describing that horizontal movement so because our initial position
of this particle is described when T is equal to 1 we're going to then use that
independent value of 1 as our lower limit because that's going to be its
position its position horizontally was 2 when T is equal to 1 and now we're going
to then move or look at that velocity from time is equal to 1 to time is equal
to 5 so we're looking at 2 plus the definite integral from 1 to 5 of T
squared DT so again this is the horizontal location and we have 2 plus
1/3 T cubed with an upper limit of 5 and 1 so that's going to be 2 plus 1/3 5
cubed minus 1/3 of 1 cubed so that's going to be 2 plus 125 over 3 minus now
that's going to just be one third so two plus 124 over 3 which is going to be 2
plus what is that going to be 41 and 1/3 or 43 and a third
okay now what is that vertical location going to be so because right now we have
the first part of our vector which is forty three and one-third what where is
the vertical locate location well our initial position is at zero I'm going to
do this in a different color off of that we're going to move again from that time
period of one to five and this is our velocity function for the vertical
velocity and that means if we integrate a rate of change if we integrate that
velocity we get a displacement or an amount of change from position off of
zero so as we integrate this we have a function of PI T inside the sine
function we can do a u-substitution which of course is totally appropriate
or we can just go well with U is equal to PI T then u prime is going to be
equal to PI this plus zero is not really going to do anything here and if we
introduce a multiplication of Pi and our integrand to finish the chain rule then
we need to balance that with the division of Pi so we have 1 over pi and
now we're just going to integrate we have 1 over pi is going to get
multiplied to our result of integrating here and the integral of the sine of U u
prime is equal to negative cosine of U or PI T an upper limit of 5 and 1 so we
just have to be careful with the negative or we can just bring it out
front and have negative 1 over pi times
cosine of PI T with our 5 and 1 moving that negative up front is going to make
our work a little bit easier and minimize our chance of a sign error
okay so the cosine of 5pi is equal to negative one and the cosine of pi is
once again negative one no so we just have negative one plus one which is zero
or vertical location is zero so our position function for our particle is
going to be forty three and one third and zero okay so we know the horizontal
position of the particle at T is equal to five we know the vertical position of
the particle at T is equal to five and answer is always zero seems so
unrewarding but make sure that you if this is a free response question that
you state your answer and context the problem with that specific time period
so the position vector of the particle when T is equal to five is forty three
and one-third and zero I miss true BAM go to your homework
you