Welcome to a video on the derivative of a vector-valued function.
We can see here that to determine the derivative of a vector-valued function, we need to determine the derivative of each component of the vector-valued function.
Let's go ahead and take a look at an example and then we'll look at this geometrically as well.
So if we want to find the derivative of this vector-valued function, the first thing we need to do is rewrite the functions in terms of "t" in a form that we can easily apply the power rule.
So we're gonna go ahead and rewrite this as five t to the negative two times the unit vector i
minus four t to the power of one-half times the unit vector j.
And now we can apply the power rule to find the derivative of this vector-valued function.
So we're gonna have negative ten t to the negative three times i minus
here we'll have one-half times the four that'll give us two
and then one-half minus one, that'll be t to the negative one-half.
Let's go ahead and simplify this.
We're gonna have negative ten over t to the third.
And here we're gonna have negative two. And we'll move this down to the denominator, it's gonna be square root of t, times the unit vector j.
Now remember, when we found derivatives before, the value of the derivative function gave us a slope at a given value of X.
So if we evaluate the derivative of this vector-valued function at let's say t equals one,
this will give us a tangent vector to the curve at t equals one.
So the tangent vector would be negative ten i minus two j.
Before we take a look at this graphically, remember that this means the initial point of this vector would be at the origin.
So this vector is in component form
But what we're gonna do is actually move this vector to the point on the curve when t is equal to one to get a better feel for this tangent vector.
So what I mean by that is this. In pink we have the curve generated by the vector-valued function r of t.
This red vector is the vector generated by the vector-valued function when t is equal to one,
and this blue vector is the tangent vector when t equals one.
Again remember, in our example we had it in component form meaning it would be down here at the origin.
We just moved it up to the point on the curve when t is equal to one and you can see now that it is a tangent vector.
Let's go and take a look at another example.
Notice this vector has an X, Y, and Z component so we'll have to find three derivatives.
Let's go ahead and do that.
The derivative of this vector-valued function would be:
The derivative of cosine t times the unit vector i
well the derivative of cosine t would be negative sine t so we'll have minus three sine t
times the unit vector i plus
The derivative of two sine t is going to be two cosine t
times the unit vector j
minus two t times the unit vector k.
Now, for an illustration, let's go ahead and evaluate this derivative at t equals pi over two.
Well the sine of pi over two would be equal to one, so we'd have negative three i
plus
The cosine of pi over two would be zero so we'd have zero j.
minus
two times pi over two would be pi so we'd have pi times the unit vector k.
Or if we wanted to, in component form it would be negative three, zero, negative pi.
Notice this vector is in component form, but what we'll do graphically is take a look at this vector when the initial point is at the terminal point of the vector
generated by the original vector-valued function at pi over two.
So again in purple, we see the curve generated by the original vector-valued function.
The red vector is the vector generated by the original vector-valued function when t is equal to pi over two.
And then in blue, we see the tangent vector at the terminal point of that red vector.
Again we found it in component form, we just moved the initial point to the terminal point of the red vector
to give a better illustration of actually what's happening when we evaluate the derivative of a vector-valued function at a given value of t.
Let's see if we can find the derivative of this vector-valued function.
So we first need to find the derivative of three tangent seven t.
This is gonna require the chain rule, where u is going to be equal to seven t which means u prime is going to be equal to seven.
So the derivative of three tangent u is going to be three secant squared u, which is actually seven t, times u prime so we have another factor of seven.
Plus, now here it might be helpful to think of this as sine t squared because this will require the chain rule where sine t is actually u.
So the derivative of this is going to be two
times sine t to the first power times u prime. Well the derivative of sine t is going to be cosine t.
And then we're going to have minus four times the derivative of natural log t. Well the derivative of natural log t is going to be one over t.
Let's go ahead and clean this up.
I'm going to go ahead and write this in component form to save some space.
So we're gonna have three times seven, that's twenty-one secant squared seven t for the X component.
The Y component is going to be two sin t cosine t.
And the Z component is going to be negative four over t.
I think we'll go ahead and stop here. I hope you found these examples helpful.