### Video Transcript

For the curve defined by the
vector-valued equation π« of π‘ is equal to three π‘ squared minus five π‘ and
square root of five π‘, find the value of π« prime of two.

Weβre given a curve which is
defined by a given vector-valued function. And we need to determine the value
of π« prime evaluated at two. First, π« is a function of π‘. So this will be when π‘ is equal to
two. And we know that π« prime will be
the derivative of our vector-valued function with respect to π‘. So we first need to recall how we
differentiate vector-valued functions. In particular, this is given as a
column vector. In fact, we differentiate these in
all the same way. We just differentiate it
component-wise.

So letβs start by differentiating
our first component function with respect to π‘. Thatβs three π‘ squared minus five
π‘. And this is just a polynomial, so
we can differentiate this term by term by using the power rule for
differentiation. We get six π‘ minus five.

We want to do the same for our
second component function. We need to differentiate the square
root of five π‘ with respect to π‘. And we can do this by using our
laws of exponents to rewrite root five π‘ as root five times π‘ to the power of
one-half. We can then differentiate this by
using the power rule for differentiation. And we do this once again by
multiplying by our exponent of π‘ and then reducing this exponent by one. This gives us root five over two
times π‘ to the power of negative one-half. And we can then simplify this by
using our laws of exponents. We get root five divided by two
root π‘.

Now that weβve differentiated both
of our component functions, we can find an expression for π« prime of π‘. Remember, to find π« prime of π‘,
we just need to differentiate π« of π‘ component-wise with respect to π‘. This gives us that π« prime of π‘
is equal to six π‘ minus five and then root five divided by two root π‘. But remember, the question is
asking us to evaluate π« prime of π‘ at two. So we need to substitute in π‘ is
equal to two. Substituting in π‘ is equal to two,
we get π« prime of two is equal to six times two minus five and then root five
divided by two root two.

We just need to evaluate and
simplify this expression. In our first component, six times
two minus five is equal to seven. And in our second component, we can
simplify by rationalizing our denominator. Weβll multiply both the numerator
and the denominator through by root two. Doing this and then simplifying
gives us root 10 divided by four.

Therefore, we were able to show if
π« of π‘ is the vector-valued function three π‘ squared minus five π‘ and then root
five π‘, we were able to show that π« prime of two is equal to seven root 10 over
four.