Video Transcript
A particle following the
parameterization 𝑥 is equal to the cos of two 𝜋 multiplied by 𝑡 and 𝑦 is equal
to the sin of two 𝜋 multiplied by 𝑡 of the unit circle starts at the point one
zero and moves counterclockwise. At what values of 𝑡 is between
zero and four inclusive is the particle at the point zero, one? Give exact values.
The question gives us a pair of
parametric equations for the position of the particle. This means that our first equation
𝑥 is equal to the cosine of two 𝜋 multiplied by 𝑡 gives us the 𝑥-coordinate of
the position of the particle at 𝑡. And our second equation of 𝑦 is
equal to the sin of two 𝜋 multiplied by 𝑡 gives us the 𝑦-coordinate of the
position of the particle at 𝑡. So finding the values of 𝑡 where a
particle is at the point zero, one is equivalent to finding the values of 𝑡 at
which the 𝑥-coordinate is equal to zero and the 𝑦-coordinate is equal to one.
So if the 𝑥-coordinate of the
position of the particle is zero, this is the same as saying that zero is equal to
the cos of two 𝜋 multiplied by 𝑡. And if the 𝑦-coordinate of the
position of the particle is equal to one, this is the same as saying that one is
equal to the sin of two 𝜋 multiplied by 𝑡. Where we must be careful because
the question restricts the values of 𝑡 to be between zero and four inclusive. So we have to solve these two
equations for values of 𝑡 between zero and four inclusive.
There are a multiple different ways
of solving equations like this. We’re going to use a sketch. So we start by sketching a graph of
the cos of 𝑥. We want to know where this will be
equal to zero. Since this will help us determine
where the cosine of two 𝜋 multiplied by 𝑡 will be equal to zero. We see that our graph of the cos of
𝑥 is equal to zero when 𝑥 is equal to 𝜋 over two, three 𝜋 over two, five 𝜋 over
two, seven 𝜋 over two, and so on. We can see that these are just
multiples of 𝜋 added to or subtracted from 𝜋 over two. Or, more formally, 𝜋 over two plus
𝑛 multiplied by 𝜋, where 𝑛 is an integer.
So what we have shown is that if
the cos of 𝑥 is equal to zero, then 𝑥 is an integer multiple of 𝜋 plus 𝜋 over
two. We can then replace the 𝑥 in this
statement with two 𝜋 multiplied by 𝑡. To give us that if the cos of two
𝜋 multiplied by 𝑡 is equal to zero, then two 𝜋 multiplied by 𝑡 is equal to 𝜋
over two plus an integer multiple of 𝜋. We can then solve this by dividing
both sides of the equation by two 𝜋, giving us that 𝑡 is equal to a quarter plus
𝑛 divided by two, where 𝑛 is any integer. Since the question tells us that 𝑡
is to be restricted between zero and four inclusive, let’s try substituting in 𝑛 as
the positive integers.
So we get that 𝑡 is equal to a
quarter, a quarter plus a half, a quarter plus two over two, etcetera. If we then calculate these values
for 𝑡, we get that 𝑡 is equal to a quarter, three-quarters, five-quarters. And we carry this on all the way up
to 15 divided by four where we stop because we know that 𝑡 must be less than or
equal to four. So what we have just shown is that
these are the values of 𝑡 between zero and four inclusive where the cos of two 𝜋
multiplied by 𝑡 is equal to zero. And so, the 𝑥-coordinate of the
position of our particle is equal to zero.
We can now do the same to find the
values of 𝑡 where the 𝑦-coordinate of the position of our particle is equal to
one. So we sketch the graph of the sign
of 𝑥. We see that the graph of the sign
of 𝑥 is equal to one when 𝑥 is equal to 𝜋 over two or five 𝜋 over two, and so
on. A fact we know about the sine
function is that the sin of 𝑥 is equal to the sin of 𝑥 plus or minus two 𝜋. So in particular, the sin of 𝑥 is
equal to one when 𝑥 is equal to 𝜋 over two plus an even integer multiple of
𝜋. Therefore, what we have shown is
that sin of two 𝜋 multiplied by 𝑡 is equal to one whenever two 𝜋 multiplied by 𝑡
is equal to 𝜋 over two plus an even integer multiple of 𝜋.
Next, we can divide both sides of
this equation by two 𝜋 to give us an expression for 𝑡. Which we can calculate to give us
that 𝑡 is equal to a quarter plus any integer 𝑛. Just as we did before, we want the
values of 𝑛 which give us a 𝑡 between zero and four inclusive. So we pick 𝑛 is equal to zero,
one, two, and three to get that 𝑡 is equal to a quarter, five-quarters,
nine-quarters, or thirteen-quarters. Our particle will be at the point
zero, one when both the 𝑥-coordinate is equal to zero and the 𝑦-coordinate is
equal to one.
What we have done is we have found
all the values of 𝑡 between zero and four inclusive which give an 𝑥-coordinate of
zero and all the values of 𝑡 between zero and four inclusive which gives a
𝑦-coordinate of one. So any values of 𝑡 which are in
both of these lists will give us an 𝑥-coordinate of zero and a 𝑦-coordinate of
one. So we just check which values of 𝑡
are in both of these lists. And we get that if 𝑡 is equal to a
quarter, five-quarters, nine-quarters, or thirteen-quarters, then the position of
the particle will be zero, one.
