Question Video: Identifying the Sketch of Two Parametric Equations Mathematics

Video Transcript

Consider the parametric equations 𝑥 of 𝑡 equals 𝑒 to the power of 𝑡 and 𝑦 of 𝑡 equals 𝑒 to the power of three 𝑡 plus one, where 𝑡 is strictly greater than negative four but strictly less than 0.5. Which of the following is the sketch of the given equations?

Here, we are given a pair of parametric equations in the parameter 𝑡 and asked to determine which sketch out of the five given correctly represents them when the values of the parameter 𝑡 lie in the open interval from negative four to 0.5. We will do this by finding coordinate pairs that satisfy the given parametric equations over the open interval of 𝑡-values from negative four to 0.5.

To find the coordinate pairs 𝑥, 𝑦 that satisfy the given parametric equations over the open interval of 𝑡-values from negative four to 0.5, we will substitute values of 𝑡 between negative four and 0.5 into the parametric equations. Note that even though the coordinate pairs associated to the values of 𝑡 equal to negative four and 0.5 are not to be included in our sketch, we still need to find them as our curve approaches those coordinate pairs. Let’s increase the values of 𝑡 by 0.5 each time, starting with negative four.

Substituting 𝑡 equals negative four into 𝑥 of 𝑡 equals 𝑒 to the power of 𝑡, we obtain 𝑒 to the power of negative four, which equals 0.02 to two decimal places. We will compute values for 𝑥 of 𝑡 and 𝑦 of 𝑡 accurate to two decimal places as the 𝑥- and 𝑦-axes on the graphs we are given increment in small steps, which happen to be of 0.5 units each. Substituting 𝑡 equals negative four into 𝑦 of 𝑡 equals 𝑒 to the power of three 𝑡 plus one, we obtain 𝑒 to the power of three times negative four plus one, which equals 1.00 to two decimal places, which is, of course, just one. So, the value 𝑡 equals negative four gives us the coordinate pair 0.02, one.

Substituting 𝑡 equals negative 3.5 into 𝑥 of 𝑡, we obtain 𝑒 to the power of negative 3.5, which equals 0.03 to two decimal places. Substituting 𝑡 equals negative 3.5 into 𝑦 of 𝑡, we obtain 𝑒 to the power of three times negative 3.5 plus one, which equals 1.00 or one to two decimal places. So, the value 𝑡 equals negative 3.5 gives us the coordinate pair 0.031. Continuing this process, we find that substituting the remaining values of 𝑡 into 𝑥 of 𝑡 equals 𝑒 to the power of 𝑡 gives us 0.05, 0.08, 0.14, 0.22, 0.37, 0.61, one, and 1.65, respectively, all computed to two decimal places.

Similarly, substituting the remaining values of 𝑡 into 𝑦 of 𝑡 equals 𝑒 to the power of three 𝑡 plus one gives us one for the 𝑡-values negative three, negative 2.5, and negative two. And 1.01, 1.05, 1.22, two, and 5.48 for the 𝑡-values negative 1.5, negative one, negative 0.5, zero, and 0.5, respectively, all computed to two decimal places. So, the remaining coordinate pairs are as shown.

Now that we have the coordinate pairs associated to each of our chosen values of 𝑡, we will see which, out of the five curves given to us, contains all of these coordinate pairs. We could also plot the coordinate pairs on a separate graph if we wanted to. And we would indeed do that if we were sketching the graph of the parametric equations from scratch. We see that only the curves in options (B) and (D) contain the coordinate pair one, two. So, the correct answer to the question cannot be one of options (A), (C), or (E).

Now, the scale of the 𝑥- and 𝑦-axes on the graphs given to us is too small to accurately plot most of the coordinate pairs that we have computed. However, we can roughly inspect that the coordinate pairs that we have computed will indeed follow the shape of the curve given to us in options (B) and (D), which differ only in the direction of the curve.

Note that when sketching parametric equations, we move in increasing values of 𝑡. And so, the direction arrows on the curve which correctly sketches the equations should move towards the coordinate pair 1.65, 5.48, which they indeed do in option (D). So, this sketch of the parametric equations 𝑥 of 𝑡 equals 𝑒 to the power of 𝑡 and 𝑦 of 𝑡 equals 𝑒 to the power of three 𝑡 plus one, where 𝑡 lies in the open interval from negative four to 0.5, is option (D).