Determine the first and second
derivative of the given vector valued
function. Then determine r prime of PI
over 2 and r double prime of PI over 2.
Before we do this let's look at this
graphically. The plane curve given by r
of t is graphed here in black. The point
on the curve when t equals negative 1 is
this point here. Notice from here we have
a red vector a blue vector and a purple
vector. The red vector is given by R of
negative 1. The blue vector is the vector
given by r prime of negative 1 and the
purple vector is the vector given by r
double prime of negative 1. If we think
of r of t as the position function,
then r prime of negative 1 is the
velocity vector, which notice how is
tangent to the curve and r double
prime of negative 1 is the acceleration
vector. Now let's animate these three
vectors as t increases from negative 1
to positive 4. We will notice the red
vector traces out the curve. The blue
vector remains tangent to the curve
which is the velocity vector and the
purple vector is the acceleration vector.
Let's do that again.
In this question we're asked to find r
prime of PI over 2 and r double prime
of PI over 2. When T is equal to PI over
2, this is the point on the curve. PI over
2 is approximately one point five seven,
which is right about here. So r prime of
PI over two is going to be this tangent
vector, which we can think of as the
velocity vector. And r double prime of PI
over two, we can think of as the
acceleration vector which is this purple
vector. Going back to our work, let's
begin by determining r prime of t. To do
this we differentiate the x and y
components of r of t with respect to t.
The derivative of 6t plus two sine t is
equal to six plus two cosine t. Now we
have times the unit vector i. And we have
plus the derivative of 5t plus seven
cosine t is equal to five minus seven
sine t, and then we have times the unit
vector j. Now to find r double prime of
t, we find the second derivative of the
x and y-components of r of t or the
derivative of the x and y-components of
r prime of t. The derivative of six
plus two cosine t is negative 2 sine t,
giving us negative 2 sine t times the
unit vector i. And then we have plus the
derivative of five minus seven sine t
with respect to t, which gives us
negative seven cosine t or minus seven
cosine t times the unit vector j. And now
we need to find r prime of PI over 2,
which gives us six plus two cosine PI
over two times i plus five minus seven
sine PI over 2 times j. Cosine PI over
two is equal to zero which gives us six
times the unit vector i.
Sine PI over 2 is equal to 1 giving us 5
minus 7 which is negative 2, giving us
minus 2 times the unit vector j. So this
is r prime of PI over 2 and now let's
find our double prime of PI over 2.
Performing substitution, we have negative
2 times sine PI over 2 times i, then
minus 7 times cosine PI over 2 times j.
Again sine PI over 2 is equal to 1
giving us negative 2 times i,  cosine PI
over 2 is 0 giving us plus 0 times the unit
vector j. Therefore r double prime of
PI over 2 is just negative 2i. Let's go
back to the graph one last time to check
our work. This is the point on the curve
when t equals PI over 2. Notice how the
tangent vector also the velocity vector,
does look like it has an X component of
6 and a Y component of negative 2. And
r double prime of PI over 2,
this purple vector which can also be
viewed as the acceleration vector, does
look like it has an X component of
negative 2 and a Y component of 0,
verifying our work is correct. I hope you
found this helpful.
