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2023 AP Calculus AB FRQ Question 6: Implicit Curve — Tangent Lines & Related Rates

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2023.

Question 6

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Implicit Curve — Tangent Lines & Related Rates

Implicit Differentiation

Hard
Consider the curve given by the equation 6xy=2+y36xy = 2+y^3.
Part AMedium2 points
Show that dydx=2yy2−2x\dfrac{dy}{dx}=\dfrac{2y}{y^2-2x}.

Answer

dydx=2yy2−2x\dfrac{dy}{dx}=\dfrac{2y}{y^2-2x} (shown by implicit differentiation)
Full Solution & Work

Differentiate both sides implicitly

ddx(6xy)=ddx(2+y3)  ⟹  6y+6xdydx=3y2dydx\frac{d}{dx}(6xy) = \frac{d}{dx}(2+y^3) \implies 6y+6x\frac{dy}{dx} = 3y^2\frac{dy}{dx}

Solve for dy/dx

2y=dydx(y2−2x)  ⟹  dydx=2yy2−2x2y = \frac{dy}{dx}\big(y^2-2x\big) \implies \frac{dy}{dx} = \frac{2y}{y^2-2x}

AP Scoring — 2 Points

**P1**: Correct implicit differentiation of 6xy=2+y36xy=2+y^3 (alternate notations for dydx\frac{dy}{dx}, such as y′y', are accepted).
**P2**: Requires P1. It is sufficient to present 2y=dydx(y2−2x)2y=\frac{dy}{dx}(y^2-2x), provided there are no subsequent errors.
Part BMedium2 points
Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.

Answer

No such point exists.
Full Solution & Work

Set dy/dx = 0

For the tangent line to be horizontal, it is necessary that 2y=02y=0 (so y=0y=0) and that y2−2x≠0y^2-2x\neq0.

Test y = 0 in the original curve equation

Substituting y=0y=0 into 6xy=2+y36xy=2+y^3 yields 6x⋅0=26x\cdot0=2, i.e. 0=20=2, which has **no solution**.

Conclude

Therefore, there is **no point** on the curve at which the tangent line is horizontal.

AP Scoring — 2 Points

**P1**: Earned with any of 2y=02y=0, y=0y=0, dydx=0\frac{dy}{dx}=0, dy=0dy=0, y′=0y'=0, or 2yy2−2x=0\frac{2y}{y^2-2x}=0.
**P2**: The answer with reason — a response does not need to also state that y2−2x≠0y^2-2x\neq0 at that point.
Part CHard3 points
Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.

Answer

(12,1)\left(\dfrac{1}{2},1\right)
Full Solution & Work

Set the denominator equal to 0

For a vertical tangent it is necessary that 2y≠02y\neq0 and that y2−2x=0y^2-2x=0 (so x=y22x=\frac{y^2}{2}).

Substitute into the curve equation

Substituting x=y22x=\frac{y^2}{2} into 6xy=2+y36xy=2+y^3:
3y2⋅y=2+y3  ⟹  3y3=2+y3  ⟹  2y3=2  ⟹  y=13y^2\cdot y = 2+y^3 \implies 3y^3=2+y^3 \implies 2y^3=2 \implies y=1

Find x

Substituting y=1y=1 into 6xy=2+y36xy=2+y^3: 6x=2+1=3  ⟹  x=126x=2+1=3 \implies x=\frac{1}{2}.
The tangent line to the curve is vertical at the point (12,1)\left(\frac12,1\right).

AP Scoring — 3 Points

**P1**: Presenting y2=2xy^2=2x or y=2xy=\sqrt{2x}.
**P2**: Substituting y=2xy=\sqrt{2x} (or x=y22x=\frac{y^2}{2} from y2−2x=0y^2-2x=0) into the original curve equation.
**P3**: Both coordinates of the point (12,1)\left(\frac12,1\right), labeled — a response that identifies the point without verifying it satisfies the curve equation does not earn P2 or P3.
Part DHard2 points
A particle is moving along the curve. At the instant when the particle is at the point (12,−2)\left(\dfrac{1}{2},-2\right), its horizontal position is increasing at a rate of dxdt=23\dfrac{dx}{dt}=\dfrac{2}{3} unit per second. What is the value of dydt\dfrac{dy}{dt}, the rate of change of the particle's vertical position, at that instant?

Answer

dydt=−89\dfrac{dy}{dt}=-\dfrac{8}{9} unit per second
Full Solution & Work

Differentiate 6xy = 2+y³ implicitly with respect to t

6ydxdt+6xdydt=3y2dydt6y\frac{dx}{dt}+6x\frac{dy}{dt} = 3y^2\frac{dy}{dt}

Substitute the known values

At (x,y)=(12,−2)(x,y)=\left(\frac12,-2\right) with dxdt=23\frac{dx}{dt}=\frac23:
6(−2)(23)+6(12)dydt=3(−2)2dydt6(-2)\left(\frac23\right)+6\left(\frac12\right)\frac{dy}{dt} = 3(-2)^2\frac{dy}{dt}

−8+3dydt=12dydt-8+3\frac{dy}{dt}=12\frac{dy}{dt}

Solve for dy/dt

−8=9dydt  ⟹  dydt=−89 unit per second-8 = 9\frac{dy}{dt} \implies \frac{dy}{dt} = -\frac{8}{9} \text{ unit per second}

AP Scoring — 2 Points

**P1**: Presenting one or more of the terms 6ydxdt6y\frac{dx}{dt}, 6xdydt6x\frac{dy}{dt}, or 3y2dydt3y^2\frac{dy}{dt}.
**P2**: The correct answer −89-\frac89 — an unsupported response of −89-\frac89 alone earns no points.

Common mistake: Alternate route: use $\frac{dy}{dx}\big|_{(1/2,-2)}=\frac{2(-2)}{(-2)^2-2(1/2)}=-\frac43$ from part (a)'s formula, then $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}=-\frac43\cdot\frac23=-\frac89$.

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