### Video Transcript

Sketch the graph of the
vector-valued function π of π‘ equals five cos π‘ π plus five sin π‘ π.

Letβs begin by thinking about what
this vector-valued function actually tells us. It takes a real number π‘ and the
output is a position vector. The horizontal component is five
cos π‘ and its vertical component is five sin π‘. We could say that in the
π₯π¦-plane, the π₯-value of any coordinate on our graph would be given by five cos
π‘ and the π¦-value would be given by five sin π‘. And then, once weβre armed with
this information, we could do one of two things. We could construct a table and try
inputting various values of π‘ and plotting the π₯- and π¦-coordinates. Alternatively, we can try
manipulating our equations to eliminate π‘ and see if we get something we
recognise.

Letβs look at that latter
method. We begin by spotting that we have a
function in cos of π‘ and one in sin of π‘. Now, we know that cos squared π‘
plus sin squared π‘ equals one. So weβre going to square each
expression for π₯ and π¦ and find their sum. We obtain π₯ squared to be equal to
five squared cos squared π‘ or 25 cos squared π‘. And similarly, π¦ squared is 25 sin
squared π‘. Then, we see that π₯ squared plus
π¦ squared is 25 cos squared π‘ plus 25 sin squared π‘. We factor by 25. And on the right-hand side, our
expression becomes 25 times cos squared π‘ plus sin squared π‘. Of course, cos squared π‘ plus sin
squared π‘ is equal to one. So we find that π₯ squared plus π¦
squared equals 25.

And at this stage, you might
recognise this equation. Itβs the equation for a circle
whose centre is at the origin and whose radius is the square root of 25 or five
units. And now, we have enough information
to be able to sketch our graph. Itβs going to look a little
something like this. Now, we arenβt actually quite
finished. Notice how we created an equation
in π₯ and π¦. These are parametric equations. And when we plot a parametric
graph, we must consider the direction in which the curve is sketched. So weβre going to take a couple of
values of π‘. Letβs take π‘ equals zero and π‘
equals one.

When π‘ is equal to zero, we know
that π₯ is equal to five times cos of zero, which is just five. Similarly, π¦ is equal to five sin
of zero, which is zero. And so, we begin by plotting the
coordinate of five, zero. Similarly, when π‘ is equal to one,
π₯ is equal to five times cos of one, which is zero, and π¦ is equal to five sin of
one, which is five. So we move from five, zero to zero,
five. This tells us weβre moving along
this circle in a counterclockwise direction. And so, we add the arrows, as
shown.