Welcome to a video on the velocity, speed and acceleration using vector-valued functions.
If r of t is the position vector value function
and r of t is differentiable. Then the first derivative
of r of t will give us the velocity of vector valued function.
The magnitude of the velocity of the vector value function will give us the speed.
v of t will give us the acceleration vector-valued function.
And if we have a plane curve, it'll only involve the x and y components
and if we have a space curve it will involve the x,y and z. components.
And you should be familiar with this idea from Calc 1.
Let's go and take a look at some examples, and we'll also look at the examples geometrically.
So if we want to determine the velocity, vector, speed and acceleration vector
of an object when t is equal to 1.
We first need to determine the velocity of vector value function and the acceleration vector valued function.
So the velocity of vector valued function will be equal to the derivative
of the given position vector valued function.
So we'll take the derivative of each component.
So the derivative of t would be 1 and the derivative of 0.5 t squared plus 4. We would multiply by 2.
That would give us -1 t or -t for the y component.
To determine the speed we would need to find the magnitude,.
So we would have the square root of 1 squared plus
negative t squared. It'll give us a square-root of
1 plus t squared. Let's go ahead and find the acceleration
vector-valued function. Remember the acceleration
is going to be equal to either the second derivative
of the position function or
the derivative of the velocity function.
We'll here is the first derivative or the velocity
vector value function. So lets go and take a derivative
of each of these components. The derivative of 1 would be 0.
The derivative of -t would be -1. So this is what we needed to find
in order to determine the velocity, vector, speed and acceleration when t is
equal to 1. So to find the velocity vector.
We'll replace t with 1 in the velocity vectored valued function.
So we'll have 1 and then the y component would be -1.
The speed, would be equal to
the square root of 1 plus t squared or 1
plus 1 square that will be the square root of 2.
Which is approximately 1.4.
And then our acceleration vector. Where a
of 1 its gonna be equal to the component vector 0 -1. Notice the acceleration does not change
when the value of t changes. Let's go and take a look at this graphically.
So what we see here in blue is the curve traced by the position vector valued function.
And then in red we have the
velocity vector at the given value of t and then this purple vector here is the acceleration vector.
Notice that it does not change as the value of t changes.
In the green vector is the vector that traces out the curve. One thing we
should notice right away, is that the acceleration vector
and the velocity of the vector are normally not orthogonal or perpendicular.
So if we pause when t is equal to 1.
Roughly right here, the velocity vector that we found is this red vector here,
its magnitude or its length is approximately 1.4. Which represents the speed,
and the acceleration vector that we found is this purple vector here.
Let's go and take a look at
one more example. Here we want to do the same thing we want to find the velocity
vector, speed and acceleration vector of an object when t is equal to 2
given by this vector-valued function. Let's start by determining the
velocity vector value function. Which will be the derivative
of the given position function. Well the derivative of cosine pi t.
Here we will have to apply the chain rule.
The derivative of cosine u was going to be negative sine u times u prime.
That's gonna give us negative sine
pi t times pi or negative pie
sine pi t. For the y component the derivative of
sine pi t is going to be cosine pi t times pi or pi
cosine pi t. And then the derivative
t square divided by 2 is just gonna be t.
Now to find the speed, we have to find the magnitude of this vector
valued function. Which is going to give us the square root
of negative pi sine
pi t squared, plus
pi cosine pi t squared plus t squared.
Now notice that both of these are going to have pie squared, so you factor out pi squared.
From these first two terms, were going to end up having sine squared pi t, plus cosine squared pi t
and that's gonna give us 1. So this is gonna simplify nicely to
the square root of pi squared plus t squared.
Now to find the acceleration vector-valued function.
We need to find the derivative of the velocity vector valued function.
Well negative pi is a constant in the derivative sine
pie t is going to be ,cosine pi t times pi.
So we' re going to have negative pi squared times cosine pi t
Then the derivative of pi cosine pi t is gonna be pi
times negative sine pi t times pi. So we have another
negative pi squared sine pi t
and derivative of t would be 1.
Okay with this is all the information we need now to find
the velocity and acceleration vector as well as the speed
when t is equal to 2 .When t is equal to 2 we're going to have negative pi times sine
2 pi sine 2 pi 0. For the y component
cosine of 2 pi is gonna be 1. So we'll have 1 times pi or pi.
And then for t we'll have 2, and to determine the speed
when t is equal to 2 we already found the equation for the magnitude
of the velocity vector so when t is 2
the speed is going to be pi squared plus
2 squared which would be 4. This is approximately 3.7.
And then lastly to find the acceleration vector.
When t is 2 we'll evaluate this value function.
So when t is 2 we'll have cosine
2 pi here that's going to be 1. So the x component will be negative pi squared.
Here we're going to have assigned 2 pi that's 0
and then here we're going to have just 1. So here's our velocity vector
when t is 2. Here's our acceleration vector when t is 2
and here's the speed when t is equal to 2. Let's go and take a look at this graphically as well.
Okay, what we're seeing here in light blue is the space curve,
defined by the position vector valued function.
In gray we have the vector that traces out the curve.
In yellow we have the velocity vector
and in red we have the acceleration vector and right now it is set where t is equal to 2.
So this yellow vector is our specific velocity vector that we found
and this red vector here's is a specific acceleration vector that we found.
If we rotate this we get a nice idea of its orientation in space.
Let's go ahead and trace out a part of this curve so we can see the relationship between
the velocity and acceleration vectors.
Okay that's gonna do for this video. I hope you found these examples helpful!