Seven of nine. This week we continue our look at the 2021 free-response questions with an eye to ways to adapt and expand the questions. Hopefully, you will find ways to use this and other free-response questions to help your students learn more and be better prepared for the exams.

## 2021 BC 2

This is a Parametric and Vector Equation (Type 8) question and contains topic from Unit 9 of the current Course and Exam Description. The vector equation of the velocity of a particle moving in the xy-plane is given along with the position of the particle at t = 0. No units were given.

The stem for 2021 BC 2 is next. (Note the $\displaystyle \left\langle \text{ } \right\rangle$ notation for vectors. Any of the usual notations may be used by students, but be sure to show them the others in case the one their book usage is different than the exam’s.)

Part (a): Students were asked to find the speed and acceleration of the particle at t = 1.2. This is a calculator active questions and the students were expected, but not actually required, to use their calculator. With their calculator in parametric mode, students should begin by entering the velocity as xt1(t) and yt1(t).

Discussion and ideas for adapting this question:

• There is little I can suggest here other than changing the time.
• At the given time and other times, you can ask in what direction is the particle moving and which way the acceleration is pulling the velocity.
• Ask student to do this without using their calculator. The answer need not be simplified or expressed as a decimal.

Part (b): Asked the students to find the total distance traveled by the particle over a given the time interval. This must be done on a calculator. Be sure your students know how to enter the expression using the already entered values for xt1(t) and yt1(t). The calculator entry should look like this.

$\displaystyle \int_{0}^{{1.2}}{{\sqrt{{{{{\left( {\text{xt}1(t)} \right)}}^{2}}+{{{\left( {\text{yt1}(t)} \right)}}^{2}}}}}}dt$

Discussion and ideas for adapting this question:

• Use different intervals.
• Discuss the similarities with the number line distance formula. In linear motion, the distance is simply the integral of the absolute value of the velocity. Since $\displaystyle \int_{a}^{b}{{\left| {v\left( t \right)} \right|}}dt=\int_{a}^{b}{{\sqrt{{{{{\left( {v(t)} \right)}}^{2}}}}}}dt$, this is the same formula reduced to one dimension.

Part (c): The situation is reduced to a one-dimensional problem: students were asked to find the coordinates of the point at which the particle is farthest left and explain why there is no point farthest to the right.

Discussion and ideas for adapting this question:

• Discuss how to do this and how students should present their answer and explanation.
• Show that this is the same as an extreme value problem and done the same way (i.e., find where the derivative is zero, and show that this is a minimum (farthest left), etc.).
• Discuss how you know there is no maximum and interpret this in the context of the equation.

For further exploration. Try graphing the path of the particle. Discuss how to do that with your class. See what they suggest. Here a few approaches.

• The first thought may be to integrate the velocity vector as an initial value problem. Unfortunately, this cannot be done. Neither the x-component nor the y-component can be integrated in terms of Elementary Functions. Even WolframAlpha.com is no help.
• Having entered the velocity vector as xt1(t) and yt1(t), as suggested above, enter something like this depending on your calculator’s syntax and then graph in a suitable window. Compare the graph with the previous analysis in part (c)?

$\displaystyle \text{x2t}(t)=-2+\int_{0}^{t}{{\text{x1t}(t)dt}}$

$\displaystyle \text{y2t}(t)=5+\int_{0}^{t}{{\text{y2t}(t)dt}}$

• You may also try expressing the components of velocity as a Taylor series centered at some positive number, a, not at zero. Integrate that to get an approximation to graph. Be sure to adjust things so the initial point is on the graph. WolframAlpha will help here. The one problem here is that the y-component is not defined for negative numbers. Therefore, zero cannt be then center and the largest the interval of convergence can be is [0, 2a] (Why?) and may not even by that large. This is an interesting approach mathematically but will not help with most of the graph.

Personal opinion: I do not think much of this question because all the first two parts require is entering the formula in your calculator and computing the answer, and the third part is really an AB level question. Just my opinion.

Next week 2021 BC 5

I would be happy to hear your ideas for other ways to use this question. Please use the reply box below to share your ideas.