Match the vector equations in Exercises 1-8 with the graphs (a)-(h) given here. $$\mathbf{r}(t)=(2 \cos t) \mathbf{i}+(2 \sin t) \mathbf{k}, \quad 0 \leq t \leq \pi$$

you were asked to match a vector equation with one of eight drafts. Let's notice a few things. First, Our victory question is given the in terra metric form. The Variable T plays an indirect role and establishing a relationship between the vector variables I and J. Also, when we look at the possible graphs, you see that they're in three dimensions. We've got an X axis, the Y axis and Z axis. So even though we don't see a variable K in the vector equation, we need to imagine it's It's there. So I will add a turn for the Vector SE zero times K to remind us that were and three dimensions. Now let's write the vector equation as scale or equations. We'll have three of them, one for its, which is the component of the eye vector. Why, which is the component of the J vector and see we'll always be zero. So let's take, um, it's plus points and put them on a graph and see what we come up with. So we have our tea parameter and our X y and Z variables. Notice that are variable t parameter T is to find between 0 to 2 pi so we could take some sample points where t is equal to zero I over to hi. Three piles or two and then two pi where t equals zero. We've got co signing t equals one Well played by two gives us X equals two, um, for why we have scientific, which is equal to zero two time 00 She is always equal to zero. So if a Spence for that if I over to co sign of tea people zero so x will be zero Sign of high over two is 12 times one is two at T equals pi We have co sign equal to minus one giving this minus two Sinus zero again that three pi over to co sign a secret zero again and sign is equal to minus one so that it was a minus two and a two pi Everything is like it was zero. So let's spot those points They're all either on the x axis of the y axis. So this was our T equals zero on the X axis to you will supply over to equals pi t equals reply over to and T was to buy. It's the same point is t equals zero. So what do you suppose is in between those points? Well, maybe we're going to suspect that it's circle after all these equations. X equals to co sign t y equals two signed Teoh If you take the twos away, those are exactly the equations they're used to define. Triggered a metric functions sine and co sign on the unit circle. Let's see if we could verify that using a little algebra so X squared plus y squared can be written as to co sign T. I swear it was two scientists where and that could be simplified to for co sign swear Teoh us science Where t and using the I treat a metric identity that's four times wondered or four. So now we have are general form for the equation of a circle at sport plus y squared equals R squared where R is the radius. So the radius is too and this center will be the origin because there are no extra wide terms here and we're actually in three dimensions. Although this is all on the ex wife plane so well, But in the Z component there So all we have to do now is matchup of our graph here with the choice that is given and that will be choice G.

## Discussion

## Video Transcript

you were asked to match a vector equation with one of eight drafts. Let's notice a few things. First, Our victory question is given the in terra metric form. The Variable T plays an indirect role and establishing a relationship between the vector variables I and J. Also, when we look at the possible graphs, you see that they're in three dimensions. We've got an X axis, the Y axis and Z axis. So even though we don't see a variable K in the vector equation, we need to imagine it's It's there. So I will add a turn for the Vector SE zero times K to remind us that were and three dimensions. Now let's write the vector equation as scale or equations. We'll have three of them, one for its, which is the component of the eye vector. Why, which is the component of the J vector and see we'll always be zero. So let's take, um, it's plus points and put them on a graph and see what we come up with. So we have our tea parameter and our X y and Z variables. Notice that are variable t parameter T is to find between 0 to 2 pi so we could take some sample points where t is equal to zero I over to hi. Three piles or two and then two pi where t equals zero. We've got co signing t equals one Well played by two gives us X equals two, um, for why we have scientific, which is equal to zero two time 00 She is always equal to zero. So if a Spence for that if I over to co sign of tea people zero so x will be zero Sign of high over two is 12 times one is two at T equals pi We have co sign equal to minus one giving this minus two Sinus zero again that three pi over to co sign a secret zero again and sign is equal to minus one so that it was a minus two and a two pi Everything is like it was zero. So let's spot those points They're all either on the x axis of the y axis. So this was our T equals zero on the X axis to you will supply over to equals pi t equals reply over to and T was to buy. It's the same point is t equals zero. So what do you suppose is in between those points? Well, maybe we're going to suspect that it's circle after all these equations. X equals to co sign t y equals two signed Teoh If you take the twos away, those are exactly the equations they're used to define. Triggered a metric functions sine and co sign on the unit circle. Let's see if we could verify that using a little algebra so X squared plus y squared can be written as to co sign T. I swear it was two scientists where and that could be simplified to for co sign swear Teoh us science Where t and using the I treat a metric identity that's four times wondered or four. So now we have are general form for the equation of a circle at sport plus y squared equals R squared where R is the radius. So the radius is too and this center will be the origin because there are no extra wide terms here and we're actually in three dimensions. Although this is all on the ex wife plane so well, But in the Z component there So all we have to do now is matchup of our graph here with the choice that is given and that will be choice G.

## Recommended Questions

Match the vector equations in Exercises $1-8$ with the graphs (a)-(h)

given here.

$$

\mathbf{r}(t)=(2 \cos t) \mathbf{i}+(2 \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi

$$

Match the vector equations in Exercises $1 - 8$ with the graphs $( a ) - ( h )$ given here.

$$

\mathbf { r } ( t ) = ( 2 \cos t ) \mathbf { i } + ( 2 \sin t ) \mathbf { j } , \quad 0 \leq t \leq 2 \pi

$$

Match the vector equations in Exercises $1 - 8$ with the graphs $( a ) - ( h )$ given here.

$$

\mathbf { r } ( t ) = t \mathbf { i } + t \mathbf { j } + t \mathbf { k } , \quad 0 \leq t \leq 2

$$

Match the vector equations in Exercises $1 - 8$ with the graphs $( a ) - ( h )$ given here.

$$

\mathbf { r } ( t ) = t \mathbf { j } + ( 2 - 2 t ) \mathbf { k } , \quad 0 \leq t \leq 1

$$

Match the vector equations in Exercises $1 - 8$ with the graphs $( a ) - ( h )$ given here.

$$

\mathbf { r } ( t ) = \left( t ^ { 2 } - 1 \right) \mathbf { j } + 2 t \mathbf { k } , \quad - 1 \leq t \leq 1

$$

Match the vector equations in Exercises $1 - 8$ with the graphs $( a ) - ( h )$ given here.

$$

\mathbf { r } ( t ) = t \mathbf { i } + ( 1 - t ) \mathbf { j } , \quad 0 \leq t \leq 1

$$

Match the vector equations in Exercises $1 - 8$ with the graphs $( a ) - ( h )$ given here.

$$

\mathbf { r } ( t ) = t \mathbf { i } , \quad - 1 \leq t \leq 1

$$

Match the vector equations in Exercises $1 - 8$ with the graphs $( a ) - ( h )$ given here.

$$

\mathbf { r } ( t ) = \mathbf { i } + \mathbf { j } + t \mathbf { k } , \quad - 1 \leq t \leq 1

$$

Match the vector equations with the graphs (a)-(h) given here. (GRAPH CANT COPY)

$$\mathbf{r}(t)=(2 \cos t) \mathbf{i}+(2 \sin t) \mathbf{k}, \quad 0 \leq t \leq \pi$$

Match the vector equations with the graphs (a)-(h) given here. (GRAPH CANT COPY)

$$\mathbf{r}(t)=(2 \cos t) \mathbf{1}+(2 \sin t) \mathbf{J}, \quad 0 \leq t \leq 2 \pi$$