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.

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.