### Video Transcript

In this video, weโre going to look
at how we can use our understanding of calculus to find the derivative of a vector
function in both two or three dimensions. Weโll also look at how to find the
derivative of dot and cross products of two vector functions, find unit tangent
vectors, and even consider higher-order derivatives of these functions.

The derivative ๐ซ prime of a vector
function, ๐ซ, is to find much in the same way as for real-valued functions. Itโs the limit as โ approaches zero
of ๐ซ of ๐ก plus โ minus ๐ซ of ๐ก over โ, provided that limit exists. But what does this actually
mean? We know that we can find the
difference between two vectors by simply finding the difference of their component
parts. So, if the vector ๐ซ is given by ๐
of ๐ก, ๐ of ๐ก, โ of ๐ก, then ๐ซ of ๐ก plus โ minus ๐ซ of ๐ก is ๐ of ๐ก plus โ
minus ๐ of ๐ก, ๐ of ๐ก plus โ minus ๐ of ๐ก, โ of ๐ก plus โ minus โ of ๐ก.

We also know that we can divide a
vector by a constant by dividing each of the component parts by that constant. And of course, โ itself is a
constant. And this is great news as it means
we can find the derivative of a vector function by taking the derivative of each of
its component parts.

In other words, if ๐ซ of ๐ก is some
vector function ๐ of ๐ก, ๐ of ๐ก, โ of ๐ก, where ๐, ๐, and โ are differentiable
function, then ๐ซ prime of ๐ก is ๐ prime of ๐ก, ๐ prime of ๐ก, โ prime of ๐ก. We can also see that if โ of ๐ก is
equal to zero, โ prime of ๐ก is equal to zero. So, this method holds perfectly for
vector functions in two dimensions, too. Letโs have a look at an
example.

Find the derivative of a
vector-valued function ๐ซ of ๐ก equals one plus ๐ก cubed ๐ข plus five ๐ก squared
plus one ๐ฃ plus ๐ก cubed plus two ๐ค.

Remember, we can find the
derivative of a vector-valued function by taking the derivative of each
component. That means weโre individually going
to differentiate, with respect to ๐ก, one plus ๐ก cubed, five ๐ก squared plus one,
and ๐ก cubed plus two. We then recall that to
differentiate polynomial terms, we multiply the entire term by the exponent and then
reduce that exponent by one. The derivative of one is zero, and
the derivative of ๐ก cubed is three ๐ก squared. So, differentiating our component
for ๐ข and we get three ๐ก squared.

Next, weโre going to differentiate
the component for ๐ฃ. Thatโs 10๐ก plus zero, which is
simply 10๐ก. Finally, weโre going to
differentiate ๐ก cubed plus two with respect to ๐ก. Of course, the derivative of that
constant is zero, so we obtain the derivative of ๐ก cubed plus two to be three ๐ก
squared plus zero or just three ๐ก squared. So, the derivative ๐ prime of ๐ก
is three ๐ก squared ๐ข plus 10๐ก ๐ฃ plus three ๐ก squared ๐ค.

So, really quite a straightforward
process. And of course, the standard rules
for differentiation apply. We can still use the chain rule,
product rule, and quotient rule alongside the standard results for differentiating
special functions such as trigonometric or logarithmic ones. Armed with this information, it
follows that we should be able to find the equation of the tangent lines to these
vector functions. Letโs see what that might look like.

Calculate ๐ prime of ๐ and find
the vector form of the equation of the tangent line at ๐ of zero for ๐ of ๐
equals ๐ plus one, ๐ squared plus one, ๐ cubed plus one.

In this question, weโve been given
a rule for ๐ in terms of its ๐ฅ-, ๐ฆ-, and ๐ง-coordinates. We can alternatively think of this
as a vector-valued function such that the vector ๐ of ๐ is equal to ๐ plus one,
๐ squared plus one, ๐ cubed plus one, where ๐ plus one is the ๐ข-component, ๐
squared plus one is the ๐ฃ-component, and ๐ cubed plus one is the ๐ค-component.

Remember then, to find the
derivative of a vector-valued function, we simply take the derivative of each
component. This means weโre going to
individually differentiate with respect to ๐ , ๐ plus one, ๐ squared plus one, and
๐ cubed plus one. That gives us one, two ๐ , and
three ๐ squared, respectively. Taking this back into the rule for
the ๐ฅ-, ๐ฆ-, and ๐ง-coordinates, and we find ๐ prime of ๐ to be equal to one, two
s, three ๐ squared.

We do still need to find the vector
form of the equation of the tangent line at ๐ of zero. So, we recall that a line that
passes through a point ๐ฅ nought, ๐ฆ nought, ๐ง nought with the direction vector ๐ฏ
is given by the equation ๐ซ equals ๐ซ nought plus ๐ก times ๐ฏ. So, weโll need two things, a point
that our line passes through and its direction of travel. We can find the latter by
evaluating the derivative when the parameter value, ๐ , is equal to zero. So, letโs substitute ๐ equals zero
into our derivative. Thatโs one, zero, zero or one ๐ข
plus zero ๐ฃ plus zero ๐ค.

We can also find that the point our
tangent line passes through by substituting ๐ equals zero into the original
function. Thatโs ๐ of zero. When we do, we obtain ๐ of zero to
be zero plus one, zero plus one, zero plus one. Thatโs one, one, one. And so, the vector form of the
equation of the tangent line at ๐ zero, letโs call that ๐ฟ, is one, one, one plus
๐ก times one, zero, zero.

There might even be occasions where
weโre required to find a unit tangent vector to a line. Letโs think back to our previous
example. We will begin in much the same
way.

We begin by calculating the first
derivative. Thatโs ๐ prime of ๐ . Itโs one, two ๐ , three ๐
squared. We then work out the magnitude of
this derivative. Itโs the square root of the sum of
the squares of the component parts. So, itโs the square root of one
squared plus two ๐ squared plus three ๐ squared squared. Thatโs the square root of one plus
four ๐ squared plus nine ๐ to the fourth power. The unit tangent vector is the
first derivative divided by the magnitude of that derivative. Here, thatโs one over the square
root of one plus four ๐ squared plus nine ๐ to the fourth power, two ๐ over that
root, and three ๐ squared also over that root.

Okay, so far, so good. We know how to find the derivative
of a vector-valued function, how to find the equation of the tangent line, and the
unit tangent vector. But what about if weโre dealing
with a dot or a cross product of two vectors-valued functions? Well, the differentiation formula
for real-valued functions have counterparts for vector-valued functions.

Suppose ๐ฎ and ๐ฏ are
differentiable vector functions. We can say that the derivative of
the dot product of ๐ฎ and ๐ฏ is equal to the dot product of ๐ฎ prime and ๐ฏ plus the
dot product of ๐ฎ and ๐ฏ prime. Similarly, the derivative of the
cross product of ๐ฎ and ๐ฏ is equal to the cross product of ๐ฎ prime and ๐ฏ plus the
cross product of ๐ฎ and ๐ฏ prime. Similarly, we can apply the chain
rule to find the derivative of ๐ฎ of ๐ of ๐ก. Itโs ๐ prime of ๐ก times ๐ฎ prime
of ๐ of ๐ก. Letโs have a look at an example of
one of these in action.

Calculate the derivative of the dot
product of ๐ซ ๐ก and ๐ฌ ๐ก for the vector-valued functions ๐ซ of ๐ก equals sin ๐ก ๐ข
plus cos ๐ก ๐ฃ and ๐ฌ of ๐ก equals cos ๐ก ๐ข plus sin ๐ก ๐ฃ. We can extend the product rule we
know so well here and say that for differentiable vector-valued functions, ๐ฎ and ๐ฏ,
the derivative of the dot product of ๐ฎ and ๐ฏ is equal to the dot product of ๐ฎ
prime and ๐ฏ plus the dot product of ๐ฎ and ๐ฏ prime. In our case, we can say that the
derivative of the dot product of ๐ซ and ๐ฌ is equal to the dot product of ๐ซ prime
and ๐ฌ plus the dot product of ๐ซ and ๐ฌ prime.

So, weโll begin by calculating ๐
prime of ๐ก and ๐ prime of ๐ก, remembering that to differentiate a vector-valued
function, we differentiate each of its component parts. So, for ๐ prime of ๐ก, weโre going
to differentiate sin of ๐ก and cos of ๐ก. The derivative of sin ๐ก is cos ๐ก,
and the derivative of cos ๐ก is negative sin ๐ก. So, we find ๐ prime of ๐ก to be
equal to cos ๐ก ๐ข minus sin ๐ก ๐ฃ. To differentiate ๐ of ๐ก, weโre
going to differentiate cos of ๐ก and sin of ๐ก. That gives us a negative sin ๐ก ๐ข
plus cos ๐ก ๐ฃ.

Next, weโre going to find the dot
product of ๐ซ prime of ๐ก and ๐ฌ of ๐ก. Itโs cos ๐ก times cos ๐ก, which is
cos squared ๐ก plus negative sin ๐ก multiplied by sin ๐ก. So, thatโs negative sin squared
๐ก. We can rewrite this as simply cos
two ๐ก. Letโs find the dot product of ๐ซ of
๐ก and ๐ฌ prime of ๐ก. Itโs sin ๐ก times negative sin ๐ก,
which is negative sin squared ๐ก plus cos ๐ก times cos ๐ก. Thatโs cos squared ๐ก again. We see that this is equal to
negative sin squared ๐ก plus cos squared ๐ก, which, once again, is equal to cos two
๐ก.

The derivative of the dot product
of ๐ซ and ๐ฌ is the sum of these. Itโs cos two ๐ก plus cos two ๐ก,
which is equal to two cos two ๐ก. Now, itโs useful to know that weโd
actually get the same answer by computing the dot product and then taking the
derivative. Letโs check that. The dot product of ๐ซ and ๐ฌ is sin
๐ก times cos ๐ก plus cos ๐ก times sin ๐ก. Weโll find the derivative of sin ๐ก
cos ๐ก plus cos ๐ก sin ๐ก.

Applying the product rule to sin ๐ก
cos ๐ก and then cos ๐ก sin ๐ก, we get negative sin squared ๐ก plus cos squared ๐ก
plus negative sin squared ๐ก plus cos squared ๐ก. And each of these is equal to cos
of two ๐ก. So once again, we obtain the
derivative of the dot product of ๐ซ and ๐ฌ to be two cos two ๐ก.

In our final example, weโll see how
we can find the second derivative of a vector-valued function.

Find the second derivative of a
vector-valued function, ๐ซ of ๐ก equals sin two ๐ก ๐ข minus cos ๐ก ๐ฃ plus ๐ to the
power of ๐ก ๐ค.

Remember, we find the derivative of
a vector-valued function by taking the derivative of each component. This means we can find the first
derivative, ๐ซ prime of ๐ก, by individually differentiating sin of two ๐ก, negative
cos of ๐ก, and ๐ of ๐ก with respect to ๐ก. The derivative of sin of two ๐ก is
two cos two ๐ก. The derivative of negative cos ๐ก
is sin ๐ก. And the derivative of ๐ to the
power of ๐ก is really nice. Itโs just ๐ to the power of
๐ก. So, weโve worked out ๐ prime of
๐ก, the first derivative.

We want to find the second
derivative. Thatโs the derivative of the
derivative. And it follows that we can use the
same process to achieve this, by differentiating ๐ซ prime of ๐ก. And this means we can individually,
once again, differentiate each component. Weโll differentiate two cos two ๐ก,
sin ๐ก, and ๐ to the power of ๐ก with respect to ๐ก. The derivative of two cos two ๐ก is
negative four cos two ๐ก. The derivative of sin ๐ก is cos
๐ก. And the derivative of ๐ to the
power of ๐ก is ๐ to the power of ๐ก. So in vector form, the second
derivative of our vector-valued function is negative four cos two ๐ก ๐ข plus cos ๐ก
๐ฃ plus ๐ to the power of ๐ก ๐ค.

In this video, we saw that we can
find the derivative of a vector-valued function by simply taking the derivative of
each component. We saw that we could use the
equation of a line given in vector form to find the vector equation of the tangent
line to the function at a point, and that we can find the unit tangent vector by
dividing the derivative function by the magnitude of the derivative. Finally, we saw that we can extend
our usual rules for differentiating real-valued functions to vector-valued
functions, as shown.