Question Video: Unit Circle Parameterization to Find When a Paritcle Is at a Point | Nagwa Question Video: Unit Circle Parameterization to Find When a Paritcle Is at a Point | Nagwa

# Question Video: Unit Circle Parameterization to Find When a Paritcle Is at a Point Mathematics • Higher Education

A particle following the parameterization π₯ = cos (2π6), π¦ = sin (2ππ‘) of the unit circle starts at (1, 0) and moves counterclockwise. At what values of 0 β€ π‘ β€ 4 is the particle at (0, 1)? Give exact values.

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### Video Transcript

A particle following the parameterization π₯ is equal to the cos of two π multiplied by π‘ and π¦ is equal to the sin of two π multiplied by π‘ of the unit circle starts at the point one zero and moves counterclockwise. At what values of π‘ 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 π₯ is equal to the cosine of two π multiplied by π‘ gives us the π₯-coordinate of the position of the particle at π‘. And our second equation of π¦ is equal to the sin of two π multiplied by π‘ gives us the π¦-coordinate of the position of the particle at π‘. So finding the values of π‘ where a particle is at the point zero, one is equivalent to finding the values of π‘ at which the π₯-coordinate is equal to zero and the π¦-coordinate is equal to one.

So if the π₯-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 π‘. And if the π¦-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 π‘. Where we must be careful because the question restricts the values of π‘ to be between zero and four inclusive. So we have to solve these two equations for values of π‘ between zero and four inclusive.

There are a multiple different ways of solving equations like this. Weβre going to use a sketch. So we start by sketching a graph of the cos of π₯. 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 π‘ will be equal to zero. We see that our graph of the cos of π₯ is equal to zero when π₯ 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 π multiplied by π, where π is an integer.

So what we have shown is that if the cos of π₯ is equal to zero, then π₯ is an integer multiple of π plus π over two. We can then replace the π₯ in this statement with two π multiplied by π‘. To give us that if the cos of two π multiplied by π‘ is equal to zero, then two π multiplied by π‘ 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 π‘ is equal to a quarter plus π divided by two, where π is any integer. Since the question tells us that π‘ is to be restricted between zero and four inclusive, letβs try substituting in π as the positive integers.

So we get that π‘ is equal to a quarter, a quarter plus a half, a quarter plus two over two, etcetera. If we then calculate these values for π‘, we get that π‘ 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 π‘ must be less than or equal to four. So what we have just shown is that these are the values of π‘ between zero and four inclusive where the cos of two π multiplied by π‘ is equal to zero. And so, the π₯-coordinate of the position of our particle is equal to zero.

We can now do the same to find the values of π‘ where the π¦-coordinate of the position of our particle is equal to one. So we sketch the graph of the sign of π₯. We see that the graph of the sign of π₯ is equal to one when π₯ 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 π₯ is equal to the sin of π₯ plus or minus two π. So in particular, the sin of π₯ is equal to one when π₯ 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 π‘ is equal to one whenever two π multiplied by π‘ 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 π‘. Which we can calculate to give us that π‘ is equal to a quarter plus any integer π. Just as we did before, we want the values of π which give us a π‘ between zero and four inclusive. So we pick π is equal to zero, one, two, and three to get that π‘ is equal to a quarter, five-quarters, nine-quarters, or thirteen-quarters. Our particle will be at the point zero, one when both the π₯-coordinate is equal to zero and the π¦-coordinate is equal to one.

What we have done is we have found all the values of π‘ between zero and four inclusive which give an π₯-coordinate of zero and all the values of π‘ between zero and four inclusive which gives a π¦-coordinate of one. So any values of π‘ which are in both of these lists will give us an π₯-coordinate of zero and a π¦-coordinate of one. So we just check which values of π‘ are in both of these lists. And we get that if π‘ is equal to a quarter, five-quarters, nine-quarters, or thirteen-quarters, then the position of the particle will be zero, one.

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