# Question Video: Unit Circle Parameterization to Find When a Paritcle Is at a Point Mathematics • 12th Grade

A particle following the parameterization 𝑥 = cos (2𝜋6), 𝑦 = sin (2𝜋𝑡) of the unit circle starts at (1, 0) and moves counterclockwise. At what values of 0 ≤ 𝑡 ≤ 4 is the particle at (0, 1)? Give exact values.

05:01

### 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.