Video Transcript
Find the domain of the vector-valued function 𝐫 of 𝑡 is equal to the sec of 𝑡 𝐢 plus the square root of three 𝑡 plus two 𝐣 plus one over five minus 𝑡 𝐤.
The question gives us a vector-valued function 𝐫 of 𝑡. And it wants us to find the domain of this function. Before we begin, let’s talk about notation for vectors. There’s a lot of different notations for vectors. For example, in this video, we’ve used half-arrow notation to represent vectors and the hat notation to represent unit vectors. However, you may also see bold notation, underlines, or full-arrow notation. It doesn’t matter which of these notations we use for vectors. It’s all personal preference.
The question wants us to find the domain of our vector-valued function. We can see that our function is a function in 𝑡. Recall, the domain of a function will be all of the possible inputs of our function. Remember, we evaluate a vector-valued function component-wise. In other words, for 𝑡 to be in the domain of our vector-valued function, all three of our component functions must be defined. In other words, 𝑡 must be a member of the domain of all three of these functions.
To see this, think what would happen if 𝑡 was not in the domain of one of these functions? Then we wouldn’t be able to define 𝐫 at 𝑡. So to find the domain of our function 𝐫 of 𝑡, we need to find the domain of each of these functions. Let’s start with the domain of the sec of 𝑡. To do this, recall that the sec of 𝑡 is defined to be one divided by the cos of 𝑡. And we know the cos of 𝑡 is defined for all real values of 𝑡. So we need to ask the question, when can’t we divide by the cos of 𝑡? Well, we can’t do this when the cos of 𝑡 is equal to zero.
There’s a few different ways of finding the values of 𝑡 where the cos of 𝑡 is equal to zero. We’ll do this by plotting a graph of the cos of 𝑡. We know the coordinates of these four 𝑥-intercepts. We know they’re at negative three 𝜋 by two, negative 𝜋 by two, 𝜋 by two, and three 𝜋 by two. But we need to find all of the 𝑥-intercepts. To do this, we need to remember the graph of the cos of 𝑡 will repeat every two 𝜋. This means we can add or subtract integer multiples of two 𝜋 from these 𝑥-intercepts to find all the rest of the 𝑥-intercepts.
If we add or subtract integer multiples of two 𝜋 from 𝜋 by two, three 𝜋 by two, negative 𝜋 by two, and negative three 𝜋 by two, we’ll just end up with an odd integer multiple of 𝜋 by two. In other words, we’ve shown the cos of 𝑡 will be equal to zero whenever 𝑡 is equal to two 𝑛 plus one times 𝜋 by two for any integer 𝑛. In fact, this is all of the values of 𝑡 where the cos of 𝑡 is equal to zero. And remember, these values of 𝑡 will be where we’re dividing by zero in the sec of 𝑡. So we can equivalently write this as the sec of 𝑡 will not be defined for any value of 𝑡 equal to two 𝑛 plus one times 𝜋 by two for any integer value of 𝑛. And of course, it will be defined for any other value of 𝑡. So we’ve found the domain of the sec of 𝑡.
Let’s now move on to finding the domain of our second component function. This means we want to find the domain of the function the square root of three 𝑡 plus two. This is much easier. We know we can take the square root of any nonnegative number. In other words, the square root of three 𝑡 plus two won’t be defined when three 𝑡 plus two is negative. All the other values of 𝑡 will be in the domain of this function. So we want to find the values of 𝑡 where three 𝑡 plus two is less than zero. And to do this, we just rearrange our inequality. We’ll subtract two from both sides of the inequality and divide through by three. This gives us that 𝑡 will be less than negative two over three.
So our second component function is not defined for all values of 𝑡 less than negative two over three. And it is defined for all other values of 𝑡. Finally, we want to find the domain of our third component function, one divided by five minus 𝑡. We can see that our function is one divided by five minus 𝑡. It’s just the quotient of two numbers. The quotient of two numbers is defined as long as we’re not dividing by zero. So this function is defined for all values of 𝑡 except when the denominator five minus 𝑡 is equal to zero. And of course, we can solve this for 𝑡 by adding 𝑡 to both sides of the equation. This gives us that this function is not defined when 𝑡 is equal to five.
So we’ve now found every single value of 𝑡 where one of our component functions is not defined. For all other values of 𝑡, all three of our component functions are defined. This means the domain of our function 𝐫 of 𝑡 will be when 𝑡 is not in the form two 𝑛 plus one times 𝜋 by two for any integer 𝑛, 𝑡 is not allowed to be less than negative two over three, and 𝑡 is not allowed to be equal to five.
We can write this more compactly, saying that 𝑡 is not allowed to be less than negative two over three is the same as saying 𝑡 is greater than or equal to negative two over three. Of course, we can write in shorthand notation 𝑡 is not equal to five. Finally, we showed that 𝑡 is not allowed to be an odd integer multiple of 𝜋 by two. We’ll write this as two 𝑛 plus one all divided by two multiplied by 𝜋.
Therefore, we were able to show the vector-valued function 𝐫 of 𝑡 is equal to the sec of 𝑡 𝐢 plus the square root of three 𝑡 plus two 𝐣 plus one over five minus 𝑡 𝐤 is defined for all real values of 𝑡, where 𝑡 is greater than or equal to negative two over three. 𝑡 is not allowed to be equal to five. And 𝑡 is not allowed to be equal to two 𝑛 plus one all divided by two times 𝜋 where 𝑛 is an integer.
