AP Calculus AB Unit 5.12: Implicit Critical Points

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

(b) 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

5.12 Exploring Behaviors of Implicit Relations question 1

5.12 Exploring Behaviors of Implicit Relations question 2