Consider the parametric equations 𝑥 of 𝑡 is equal to 𝑡 cubed minus one and 𝑦 of 𝑡 is equal to 𝑡, where 𝑡 is greater than negative two and 𝑡 is less than one. Which of the following is the sketch of the given equations? Option (A), option (B), option (C), option (D), or option (E).
We’re given a pair of parametric equations, 𝑥 of 𝑡 is equal to 𝑡 cubed minus one and 𝑦 of 𝑡 is equal to 𝑡, where our values of 𝑡 range from negative two to one. We need to decide which of our graphs is the sketch of the curve of these parametric equations. To do this, we need to recall what we mean by parametric equations. We’re given functions 𝑥 of 𝑡 and 𝑦 of 𝑡. These take inputs of 𝑡 and tell us the 𝑥-coordinate and 𝑦-coordinate of our curve. So we need to substitute values of 𝑡 into both of these functions to find coordinates of our curve. We can then plot these on our graphs and see which of our curves matches with these values.
We can do this using a table. However, the first question we need to ask is, which values of 𝑡 should we use? Remember, the question tells us that our curve will only take inputs of 𝑡 greater than negative two and less than one. But remember, this means we can take values of 𝑡 very close to negative two and very close to one. So the endpoints of our curve will approach the values where 𝑡 is equal to negative two and 𝑡 is equal to one. So we’ll use these values in our table to help us find the endpoints of our curve. It then makes sense to just try using the integers values between these two values of 𝑡.
We now need to substitute these values of 𝑡 into our functions. Let’s start with negative two in 𝑥 of 𝑡. This gives us 𝑥 of negative two is equal to negative two cubed minus one. And if we evaluate this expression, we get negative nine. So when 𝑡 is negative two, 𝑥 of 𝑡 is negative nine. We can do the same with 𝑦 of 𝑡, but the function 𝑦 of 𝑡 is just equal to 𝑡. So we just get that 𝑦 of negative two is equal to negative two. So we’ve found the first endpoint of our parametric curve. It’s the point negative nine, negative two.
In fact, we can plot this point on all five of our graphs. We can see that only options (C) and (E) pass through this point. Since this point should be one of the endpoints of our curve, we can conclude that it can’t be options (A), (B), or (D). Let’s now find another coordinate. We’ll substitute 𝑡 is equal to negative one. We get 𝑥 of negative one is negative one cubed minus one and 𝑦 of negative one is equal to negative one. Evaluating these expressions, we get negative two, negative one. So we know that our parametric curve must pass through the point negative two, negative one.
Once again, we can plot this point on our curve. We see that options (A), (B), and (D) don’t pass through this point. However, both options (C) and (E) do pass through this point, so we’ll find another coordinate. This time, we’ll substitute 𝑡 is equal to zero. This gives us 𝑥 of zero is zero cubed minus one and 𝑦 of zero is equal to zero. Evaluating these, we get negative one, zero. So we know our parametric curve passes through the point negative one, zero. And plotting these on our curves, we can see both graphs (C) and (E) pass through this point.
We’ll do the same when 𝑡 is equal to one. We get 𝑥 of one is one cubed minus one and 𝑦 of one is one. Plotting this point on all of our graphs, we can see only the curves in options (C) and (E) pass through this point. In fact, we can see the curves in options (C) and (E) have exactly the same shape. So no matter how many values of 𝑡 we take, both curves will pass through the coordinates we get using this method. So to decide which of these two sketches is correct, we need to use another fact about how we sketch parametric curves.
When we sketch a parametric curve, we always do this by increasing our value of 𝑡. So when we were sketching this parametric curve, we would’ve started at the point negative nine, negative two and we would’ve increased our value of 𝑡 up to our endpoint zero, one. We represent this on a sketch by using arrows. We want to start at the point negative nine, two and end at the point zero, one. So our arrows should be pointing from negative nine, two to zero, one. And the arrows are only pointing in this direction in option (E). So the correct sketch must be option (E).
Therefore, we were able to show of the five possible sketches of the parametric curve defined by the equations 𝑥 of 𝑡 is equal to 𝑡 cubed minus one and 𝑦 of 𝑡 is equal to 𝑡, where 𝑡 is greater than negative two and 𝑡 is less than one, must be the curve given in option (E).