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

A particle following the parameterization π‘₯ = cos (2πœ‹6), 𝑦 = sin (2πœ‹π‘‘) of the unit circle starts at (1, 0) and moves counterclockwise. At what values of 0 ≀ 𝑑 ≀ 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œ‹ multiplied by 𝑑 and 𝑦 is equal to the sin of two πœ‹ multiplied by 𝑑 of the unit circle starts at the point one zero and moves counterclockwise. At what values of 𝑑 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œ‹ multiplied by 𝑑 gives us the π‘₯-coordinate of the position of the particle at 𝑑. And our second equation of 𝑦 is equal to the sin of two πœ‹ multiplied by 𝑑 gives us the 𝑦-coordinate of the position of the particle at 𝑑. So finding the values of 𝑑 where a particle is at the point zero, one is equivalent to finding the values of 𝑑 at which the π‘₯-coordinate is equal to zero and the 𝑦-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œ‹ multiplied by 𝑑. 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œ‹ multiplied by 𝑑. Where we must be careful because the question restricts the values of 𝑑 to be between zero and four inclusive. So we have to solve these two equations for values of 𝑑 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 π‘₯. We want to know where this will be equal to zero. Since this will help us determine where the cosine of two πœ‹ multiplied by 𝑑 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œ‹ over two, three πœ‹ over two, five πœ‹ over two, seven πœ‹ over two, and so on. We can see that these are just multiples of πœ‹ added to or subtracted from πœ‹ over two. Or, more formally, πœ‹ over two plus 𝑛 multiplied by πœ‹, where 𝑛 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œ‹ plus πœ‹ over two. We can then replace the π‘₯ in this statement with two πœ‹ multiplied by 𝑑. To give us that if the cos of two πœ‹ multiplied by 𝑑 is equal to zero, then two πœ‹ multiplied by 𝑑 is equal to πœ‹ over two plus an integer multiple of πœ‹. We can then solve this by dividing both sides of the equation by two πœ‹, giving us that 𝑑 is equal to a quarter plus 𝑛 divided by two, where 𝑛 is any integer. Since the question tells us that 𝑑 is to be restricted between zero and four inclusive, let’s try substituting in 𝑛 as the positive integers.

So we get that 𝑑 is equal to a quarter, a quarter plus a half, a quarter plus two over two, etcetera. If we then calculate these values for 𝑑, we get that 𝑑 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 𝑑 must be less than or equal to four. So what we have just shown is that these are the values of 𝑑 between zero and four inclusive where the cos of two πœ‹ multiplied by 𝑑 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 𝑑 where the 𝑦-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œ‹ over two or five πœ‹ 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œ‹. So in particular, the sin of π‘₯ is equal to one when π‘₯ is equal to πœ‹ over two plus an even integer multiple of πœ‹. Therefore, what we have shown is that sin of two πœ‹ multiplied by 𝑑 is equal to one whenever two πœ‹ multiplied by 𝑑 is equal to πœ‹ over two plus an even integer multiple of πœ‹.

Next, we can divide both sides of this equation by two πœ‹ to give us an expression for 𝑑. Which we can calculate to give us that 𝑑 is equal to a quarter plus any integer 𝑛. Just as we did before, we want the values of 𝑛 which give us a 𝑑 between zero and four inclusive. So we pick 𝑛 is equal to zero, one, two, and three to get that 𝑑 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 𝑦-coordinate is equal to one.

What we have done is we have found all the values of 𝑑 between zero and four inclusive which give an π‘₯-coordinate of zero and all the values of 𝑑 between zero and four inclusive which gives a 𝑦-coordinate of one. So any values of 𝑑 which are in both of these lists will give us an π‘₯-coordinate of zero and a 𝑦-coordinate of one. So we just check which values of 𝑑 are in both of these lists. And we get that if 𝑑 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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