### Video Transcript

Consider the curve π« of π is the vector-valued function sin of two π , two sin
squared of π , two cos of π . Determine π« prime of π and find the tangent line πΏ to the curve when π is equal
to zero.

Weβre given the vector-valued function π« of π . And weβre asked to find two things. First, weβre asked to find π« prime of π . Thatβs the derivative of π« with respect to π . Next, we need to find the tangent line πΏ to our curve when π is equal to zero.

First, letβs start by finding π« prime of π . To do this, we need to differentiate our vector-valued function π« of π
component-wise. So, we need to differentiate each component separately. Letβs start with our first component. So, we need to differentiate the first component of π« of π . Thatβs the derivative of the sin of two π with respect to π . To do this, we can recall the following standard trigonometric derivative result. For any real constant π, the derivative of the sin of ππ with respect to π is
equal to π times the cos of ππ . In our case, the value of π is two, so we set π equal to two. And this gives us the derivative of the first component of π« of π with respect to
π is equal to two cos of two π .

But we need to differentiate all three of our component functions separately. Letβs now differentiate our second component function. So, we need to evaluate the derivative of two sin squared of π with respect to
π . Thereβs a few different ways we could do this. For example, we could use the chain rule or the general power rule. And both of these options will work; however, weβll do this by using the double angle
formula for cosine. First, recall the cos of two π is equivalent to one minus two times the sin squared
of π .

And we can rearrange this to find an equation for two sin squared of π . We can subtract one from both sides of this equivalence and then multiply through by
negative one. This gives us two sin squared of π is equivalent to negative the cos of two π plus
one. We can substitute this into our derivative. This means instead we need to differentiate negative the cos of two π plus one. And we can evaluate this derivative term by term by recalling the following standard
trigonometric derivative result. For any real constant π, the derivative of negative the cos of ππ with respect to
π is equal to π times the sin of ππ .

And in this case, our value of π is two, so the derivative of our second component
function with respect to π is two times the sin of two π . And we could leave our answer like this. However, weβll rewrite this by using our double angle formula for sine. We recall the double angle formula for sine tells us the sin of two π is equivalent
to two times the sin of π multiplied by the cos of π . So, we can replace the sin of two π in our derivative with two times the sin of π
multiplied by the cos of π . And if we do this and then simplify, we get four times the sin of π multiplied by
the cos of π .

Now, thereβs only one more component function we need to differentiate. We need to differentiate two cos of π with respect to π . And to do this, weβll recall our final trigonometric derivative result. For any real constant π, the derivative of the cos of ππ with respect to π is
equal to negative π times the sin of ππ . Of course, this time our value of π is one, so we just get negative two sin of
π .

And now that weβve differentiated all of the component functions of π« of π , weβre
ready to find an expression for π« prime of π . π« prime of π will be a vector-valued function. And each of our components will just be the derivative of the corresponding component
function of π« of π . So, weβve shown π« prime of π is two cos of two π , four sin π times the cos of π ,
negative two sin π . And we donβt need to simplify this expression anymore, so weβll leave it as it
is.

Letβs now move on to the second part of this question. We need to find the tangent line πΏ to our curve when π is equal to zero. And since our curve is defined as a vector-valued function, weβll give our tangent
line as a vector-valued function. To do this, letβs start by recalling the general vector equation for a line. To define a line by using a vector equation, we need two vectors. First, we need the vector πΏ zero. πΏ zero can be any point which lies on our line. It doesnβt matter which one we choose. We also need a second vector π― to determine the direction our line travels. We call π― the direction vector.

And to find suitable vectors in this case, we need to recall what we mean by the
tangent line to a curve when π is equal to zero. First, remember, a tangent line to a curve must touch the curve at that point. In other words, π« of zero must be a point on our line. So, our point πΏ zero can be the vector π« evaluated at zero. Next, remember, our tangent line must have the same slope as our curve at this
point. In other words, the slope of our line should be π« prime of zero. And since we have an expression for π« of π and π« prime of π , we can find both of
these vectors.

Letβs start with π« evaluated at zero. We need to substitute π is equal to zero into our expression for π« of π . This gives us the vector sin of two times zero, two sin squared of zero, two cos of
zero. And if we evaluate each of these components separately, we get the vector zero, zero,
two. We now want to do exactly the same to find π« prime of zero. We need to substitute π is equal to zero into the expression we found for π« prime
of π . This gives us the vector two cos of two times zero, four times the sin of zero
multiplied by the cos of zero, negative two sin of zero.

And if we evaluate each of these components separately, we get the vector two, zero,
zero. Now, all we need to do is substitute our expressions for π« of zero and π« prime of
zero into our equation for a line. Doing this gives us the equation of the tangent line πΏ is the vector zero, zero, two
plus π‘ times the vector two, zero, zero. And this is our final answer.

Therefore, we were able to show for the vector-valued curve π« of π is sin of two
π , two sin squared of π , two cos of π , then we could find π« prime of π by
differentiating π« of π component-wise. This gave us that π« prime of π is equal to two cos of two π , four sin of π times
cos of π , negative two sin of π . And we were also able to find the tangent line πΏ to our curve π« of π when π was
equal to zero. We got the vector equation for this line is zero, zero, two plus π‘ times two, zero,
zero.