top of page
Search

# Early Solutions to the 2021 AP Calculus AB Free Response

Teaching AP Calculus is challenging but the 48 hours between the exam and the free response release are probably the hardest! We love seeing the new questions and working through them, always amazed at the original ways the College Board incorporates multiple concepts into a single question. You can see this year’s AB questions here. Although the official scoring guidelines don’t release until the summer, here’s a first attempt at solutions.

### Reflections:

The integrand in 1b was a product of the radius and the density, which added a new component to the traditional Riemann sum approximation. To find the height of each rectangle, students would have to multiply the radius by the density. The width of each rectangle is of course the length of the interval, or the change in the radius.

The spinning toy question (FRQ #3) will likely go down in history with the hot potato question of 2017 (2017AB4) in which students averaged only 1.54 out of 9 possible points! The introduction of a parameter “c” is somewhat unfamiliar in area/volume questions and the concept of a maximum radius in part b had many of my students stumped. Moving forward, we’ll want to continue to think about how we can help students “think on their feet” when it comes to tackling questions they’ve never seen before.

Question 5d required a bit more work than usual to justify the relative extrema. The second derivative test would work well, but requires the quotient rule, product rule, and implicit differentiation! I chose to go with the first derivative test and compare the behavior of dy/dx near the critical point by analyzing the signs of the numerator and denominator. We’ll have to see how the College Board allocates points on this one!

What did you think about this year's questions? How might they inform our teaching practice moving forward?

See All

## 6 commentaires

John Krieger
27 mai 2021

Can you explain question #4 in more detail, especially as to G being the integral of f. I don't understand how G(3) = -1/2(4 + 3)(1) or how G(2) = 0.

J'aime
Mariah Westerhausen
27 mai 2021
En réponse à

G is an accumulation function which means you find it by finding the amount of accumulation (area under the curve) of f. Therefore G(3) is the area under the curve from x=0 to x=3. Area under the x-axis is negative so the area under the curve from x=0 to x=2 has the same positive area as negative area and is equal to zero. The area under the curve from x=2 to x=3 is a trapezoid under the x-axis, so negative 1/2(base 1+base 2)(height).

J'aime

BRANDON KLINE
09 mai 2021

For 6d, do you think that students should also mention that dy/dt is positive, and then explain that the rate is decreasing because the signs for first and second derivatives are opposite?

If dy/dt were negative, then d^2y/dt^2 being negative would actually show that the rate is *increasing* (becoming more negative), no?

Thanks for posting these so fast! Planning to use these problems with my class this week.

J'aime
Keith Hatfield
10 mai 2021
En réponse à

The main reason this is true in particle motion is that they are asking if the object is speeding up/slowing down. Because speed is the absolute value of velocity, you need to look at the sign of both the velocity and acceleration to determine speeding up/slowing down.

J'aime
bottom of page