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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.

05:01

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.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy