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2025 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 2025.

Question 6

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Implicit Curve — Tangent Lines & Related Rates

Implicit Differentiation

Hard
Consider the curve GG defined by the equation y3−y2−y+14x2=0y^3 - y^2 - y + \dfrac{1}{4}x^2 = 0.
Part AMedium2 points
Show that dydx=−x2(3y2−2y−1)\dfrac{dy}{dx} = \dfrac{-x}{2(3y^2-2y-1)}.

Answer

dydx=−x2(3y2−2y−1)\dfrac{dy}{dx} = \dfrac{-x}{2(3y^2-2y-1)} (shown by implicit differentiation)
Full Solution & Work

Differentiate both sides implicitly

ddx(y3−y2−y+14x2)=ddx(0)\frac{d}{dx}\left(y^3-y^2-y+\frac{1}{4}x^2\right) = \frac{d}{dx}(0)

3y2dydx−2ydydx−dydx+x2=03y^2\frac{dy}{dx} - 2y\frac{dy}{dx} - \frac{dy}{dx} + \frac{x}{2} = 0

Solve for dy/dx

(3y2−2y−1)dydx=−x2  ⟹  dydx=−x2(3y2−2y−1)(3y^2-2y-1)\frac{dy}{dx} = -\frac{x}{2} \implies \frac{dy}{dx} = \frac{-x}{2(3y^2-2y-1)}

AP Scoring — 2 Points

**P1**: Correct implicit differentiation of y3−y2−y+14x2=0y^3-y^2-y+\frac14 x^2=0 (alternate notations for dydx\frac{dy}{dx}, such as y′y', are accepted).
**P2**: Algebraic verification that isolates dydx\frac{dy}{dx} to reach the given expression — earned only if P1 was earned. Presenting (3y2−2y−1)dydx=−x2(3y^2-2y-1)\frac{dy}{dx}=-\frac{x}{2} is sufficient, provided no later error appears.
Part BMedium2 points
There is a point PP on the curve GG near (2,−1)(2,-1) with xx-coordinate 1.6. Use the line tangent to the curve at (2,−1)(2,-1) to approximate the yy-coordinate of point PP.

Answer

y≈−1−14(1.6−2)=−0.9y \approx -1 - \dfrac{1}{4}(1.6-2) = -0.9
Full Solution & Work

Find the slope at (2, −1)

dydx∣(2,−1)=−22(3(1)−2(−1)−1)=−22(3+2−1)=−28=−14\left.\frac{dy}{dx}\right|_{(2,-1)} = \frac{-2}{2(3(1)-2(-1)-1)} = \frac{-2}{2(3+2-1)} = \frac{-2}{8} = -\frac{1}{4}

Apply the tangent line approximation

y≈−1+(−14)(1.6−2)=−1−14(−0.4)=−1+0.1=−0.9y \approx -1 + \left(-\frac{1}{4}\right)(1.6-2) = -1 - \frac{1}{4}(-0.4) = -1+0.1 = -0.9

AP Scoring — 2 Points

**P3**: Correct slope of the tangent line, −14-\frac14, at (2,−1)(2,-1) — a response that declares any other nonzero slope cannot earn this point (though it may still earn P4 for consistent use of that value).
**P4**: Correct tangent-line approximation, clearly evaluated at x=1.6x=1.6: y≈−0.9y\approx -0.9.
Part CMedium2 points
For x>0x > 0 and y>0y > 0, there is a point SS on the curve GG at which the line tangent to the curve at that point is vertical. Find the yy-coordinate of point SS. Show the work that leads to your answer.

Answer

y=1y = 1
Full Solution & Work

A vertical tangent occurs where dy/dx is undefined

From part A, dydx=−x2(3y2−2y−1)\frac{dy}{dx}=\frac{-x}{2(3y^2-2y-1)} is undefined (vertical tangent, for x≠0x\neq0) when the denominator is 0:
2(3y2−2y−1)=02(3y^2-2y-1) = 0

Solve for y

3y2−2y−1=0  ⟹  (3y+1)(y−1)=0  ⟹  y=−13 or y=13y^2-2y-1 = 0 \implies (3y+1)(y-1) = 0 \implies y = -\frac{1}{3} \text{ or } y = 1

Apply the y > 0 restriction

Since the problem requires y>0y>0, only y=1y=1 is valid. So the yy-coordinate of point SS is 1\boxed{1}.

AP Scoring — 2 Points

**P5**: Sets the denominator equal to 0 — any of 2(3y2−2y−1)=02(3y^2-2y-1)=0, 3y2−2y−1=03y^2-2y-1=0, or its factored form.
**P6**: Correct answer y=1y=1, clearly identified as the only solution satisfying y>0y>0 (earned only if P5 was earned, and only if the algebra to isolate yy is correct).
Part DHard3 points
A particle moves along the curve HH defined by the equation 2xy+ln⁡y=82xy + \ln y = 8. At the instant when the particle is at the point (4,1)(4,1), dxdt=3\dfrac{dx}{dt} = 3. Find dydt\dfrac{dy}{dt} at that instant. Show the work that leads to your answer.

Answer

dydt=−23\dfrac{dy}{dt} = -\dfrac{2}{3}
Full Solution & Work

Differentiate implicitly with respect to t

ddt(2xy+ln⁡y)=ddt(8)\frac{d}{dt}\big(2xy+\ln y\big) = \frac{d}{dt}(8)

2dxdty+2xdydt+1ydydt=02\frac{dx}{dt}y + 2x\frac{dy}{dt} + \frac{1}{y}\frac{dy}{dt} = 0

Substitute the known values

At (x,y)=(4,1)(x,y)=(4,1) with dxdt=3\frac{dx}{dt}=3:
2(3)(1)+2(4)dydt+11dydt=02(3)(1) + 2(4)\frac{dy}{dt} + \frac{1}{1}\frac{dy}{dt} = 0

6+9dydt=06 + 9\frac{dy}{dt} = 0

Solve for dy/dt

dydt=−69=−23\frac{dy}{dt} = -\frac{6}{9} = -\frac{2}{3}

AP Scoring — 3 Points

**P7**: Attempts implicit differentiation of 2xy+ln⁡y=82xy+\ln y=8 with respect to tt, with at most one error.
**P8**: A correct equation equivalent to 2dxdty+2xdydt+1ydydt=02\frac{dx}{dt}y+2x\frac{dy}{dt}+\frac{1}{y}\frac{dy}{dt}=0 — requires P7 with no errors.
**P9**: Correct final value dydt=−23\frac{dy}{dt}=-\frac{2}{3}, requiring both P7 and P8.

Common mistake: An equally valid alternate route: solve $\frac{dy}{dx}$ implicitly with respect to $x$ first, evaluate it at $(4,1)$ to get $-\frac{2}{9}$, then multiply by $\frac{dx}{dt}=3$ using the chain rule $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$.

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