AP Calculus AB 5.12 Exploring Behaviors of Implicit Relations Questions

Practice AP Calculus Unit 5.12 questions on finding critical points of implicit relations where the derivative is zero or undefined.

Covered syllabus points

FUN-4.D—Determine critical points of implicit relations FUN-4.E—Justify conclusions about the behavior of an implicitly defined function based on evidence from its derivatives

[Maximum number: 2]

Consider the curve G defined by the equation $y^{3}-y^{2}-y+\frac{1}{4} x^{2}=0$ .

For x>0 and y>0, there is a point S on the curve G at which the line tangent to the curve at that point is vertical. Find the y-coordinate of point S. Show the work that leads to your answer.

C For x>0 and y>0, there is a point S on the curve G at which the line tangent to the curve at that

point is vertical. Find the y-coordinate of point S. Show the work that leads to your answer.

For x>0, the curve G has a vertical tangent line when $2\left(3 y^{2}-2 y-1\right)=0$ .	Sets denominator equal to 0	Point 5 (P5)
$2\left(3 y^{2}-2 y-1\right)=0 \Rightarrow 2(3 y+1)(y-1)=0$	Answer	Point 6 (P6)
Because y>0, it follows that y=1.
The line tangent to the curve is vertical at the point on the curve where y=1.

Scoring Notes for Part C

- P5 is earned with any of $2\left(3 y^{2}-2 y-1\right)=0,3 y^{2}-2 y-1=0,2(3 y \pm 1)(y \pm 1)=0$ , or

$(3 y \pm 1)(y \pm 1)=0$ .

- To be eligible for P6, a response must have earned P5.

- A response does not need to consider the numerator of $\frac{d y}{d x}=\frac{-x}{2\left(3 y^{2}-2 y-1\right)}$ to earn P5 or P6;

considering the denominator is sufficient.

- A response that states solutions of $y=-\frac{1}{3}$ and y=1, but does not clearly identify that y=1 is the

only solution to the prompt, does not earn P6.

- In the presence of algebraic work to find the value of y, P6 is earned only if the algebraic work is

correct.

- A response of " $2\left(3 y^{2}-2 y-1\right)=0, y=1$ " earns both P5 and P6.

Given the following differential equation tart fraction numerator d y over denominator d x end fraction equals x minus 2 y plus 1

Find the second derivative, et fraction numerator d squared y over denominator d x squared end fraction , in terms of x and y. The region in the x y-plane where all the solution curves to the differential equation are concave down can be expressed as a linear inequality. Find this region.

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Covered syllabus points

Question 1

Question 2