Velocity & Acceleration Vectors Example 3


Uploaded by TheIntegralCALC on 25.01.2011

Transcript:
Hi everyone. Welcome back to integralcalc.com. Today we’re going to be doing another velocity
and acceleration problem. This one is the function r of t equals 3 cos t times i plus
3 sin of t times j minus 4tk. Remember that when we are looking at velocity
and acceleration vectors the velocity vector is going to be equal to the coefficients associated
with i, j and k of the first derivative of this function and the acceleration vector
is going to be made up of the coefficients of i, j and k of the 2nd derivative of this
function. And really all it boils down to is taking
the first derivative and the second derivative, pulling the coefficients out of it, and then
we're going to calculate speed as well. So you could see what that looks like. The first
derivative is going to be r prime of t and that's going to be equal to velocity. so we're
just taking the derivative. Remember that the derivative of cos is negative
sin. So that’s why we have this negative sin. Coming out in front here, we end up with
negative 2 sin of t times i and remember that when you're doing these problems, you can
basically ignore completely i, j and k. They're still there but you could almost like cover
them up and pretend that they're not and just take the derivative with respect to t, leaving
i totally alone. So that’s what we did. We took the derivative
of 3cos of t, ignored i and then plugged it right where it was. For the term that includes
j here, the derivative of sin is cos so we end up with 3 cos t and we just leave that
j alone. And then the derivative of 4t k, again ignore
the k completely, just look at the 4t. And of course the derivative of 4t is simply 4
so we end up with simply minus 4k. Then to find the velocity vector, we take
the coefficients in front of i, j and k, which are really easy to see in this problem. You
can tell right away the coefficient on i is negative 3 sin of t, the coefficient on j
is 3 cos of t and the coefficient on k is negative 4. So all we did was grab those coefficients,
and obviously that if you have a negative in front of one of these terms, you're going
to include that sign. It’s not just a 4, it's negative 4. Anything outside i, j and
k, we grab as the coefficient. These are the three direction numbers that are going to
make up our velocity vector. Now to find acceleration, we're going to take
the second derivative which is simply the derivative of this function that we had here.
Again, same thing, forget about i, j, and k. Just take the derivative with respect to
t. So the derivative of sin here is cos, so we're going to end up with negative 3 cos
t. The derivative of cos is negative sin so we
bring this derivative in front of the three and end up with negative 3 sin of t times
j. And the derivative of 4 is 0.
It's a constant so the derivative is 0. That means we're going to have 0 times k and that
whole term is going to go away. And so then we'll just grab the coefficients again. Negative
cos t negative 3 sin t and because we have no k term, that means that the coefficient
must be 0 as we knew from our previous step. So those are the three vectors that we've
grabbed from our acceleration vector and then to find speed, forget about acceleration for
a second, go back to velocity and we're going to look at the velocity vector here.
We're going to take these three coefficients and we’re going to square each of them,
add them all together and then put them inside the square root sign.
You can see that we've grabbed each one of these, we squared them so negative 3 sin of
t squared 3 cos of t squared and negative 4 squared. We add them all together and we
just throw a square root symbol over to the top.
That is the equation that gives us our speed.  So once we’ve done that, all we have to
do is simplify as far as we can and then we're done. So this simplification is going to look
like this. We'll get negative 3 times negative 3 is a
positive 9 and we'll end up with sin squared of t, 9 cos squared t and a negative 4 squared
is obviously going to give you a plus 16. And then that's as far as we can really simplify
this one. So we take our velocity vector, our acceleration
vector and our function here for speed and we put them all together and that is our final
answer. And one quick thing you need to know, if they
ask you find all of these three things and they say find velocity, acceleration and speed
at time 5 or 5 seconds after the person started running or whatever the word problem is, you
would just plug in 5 for t to each of these and that will give you velocity, acceleration,
and speed. So I hope that helped you guys and I’ll
see you in the next problem. Bye.