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2024 AP Calculus AB FRQ Question 5: Implicit Curve — Tangent Lines & Related Rates

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2024.

Question 5

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Implicit Curve — Tangent Lines & Related Rates

Implicit Differentiation

Hard
Consider the curve defined by the equation x2+3y+2y2=48x^2+3y+2y^2=48. It can be shown that dydx=−2x3+4y\dfrac{dy}{dx}=\dfrac{-2x}{3+4y}.
Part AMedium2 points
There is a point on the curve near (2,4)(2,4) with xx-coordinate 3. Use the line tangent to the curve at (2,4)(2,4) to approximate the yy-coordinate of this point.

Answer

y≈4−419(3−2)=7219y \approx 4-\dfrac{4}{19}(3-2)=\dfrac{72}{19}
Full Solution & Work

Find the slope at (2,4)

dydx∣(2,4)=−2(2)3+4(4)=−419\left.\frac{dy}{dx}\right|_{(2,4)} = \frac{-2(2)}{3+4(4)} = \frac{-4}{19}

Apply the tangent line approximation

y≈4−419(3−2)=7219y \approx 4-\frac{4}{19}(3-2) = \frac{72}{19}

AP Scoring — 2 Points

**P1**: Correctly finding dydx∣(2,4)=−419\left.\frac{dy}{dx}\right|_{(2,4)}=-\frac{4}{19}, even if not labeled or used as a tangent-line slope.
**P2**: Correct approximation, evaluated at x=3x=3 using a linear approximation through (2,4)(2,4) with slope −419-\frac{4}{19}.
Part BHard2 points
Is the horizontal line y=1y=1 tangent to the curve? Give a reason for your answer.

Answer

No — the only candidate point of tangency, (0,1)(0,1), is not on the curve.
Full Solution & Work

Find where dy/dx = 0

dydx=−2x3+4y=0  ⟹  x=0\frac{dy}{dx} = \frac{-2x}{3+4y}=0 \implies x=0

So if y=1y=1 is tangent to the curve, the point of tangency must be (0,1)(0,1).

Check whether (0,1) lies on the curve

02+3(1)+2(1)2=5≠480^2+3(1)+2(1)^2 = 5 \neq 48

The point (0,1)(0,1) is not on the curve. Therefore, the horizontal line y=1y=1 is **not** tangent to the curve.

AP Scoring — 2 Points

**P1**: Considering dydx=0\frac{dy}{dx}=0, −2x=0-2x=0, or x=0x=0 — or, alternatively, identifying the point (0,1)(0,1) directly.
**P2**: A reason that y=1y=1 is not tangent to the curve; merely stating "(0,1)(0,1) is not on the curve" without checking it against the equation is insufficient.
Part CMedium1 point
The curve intersects the positive xx-axis at the point (48,0)(\sqrt{48},0). Is the line tangent to the curve at this point vertical? Give a reason for your answer.

Answer

No — the denominator of dydx\dfrac{dy}{dx} at that point is 3+4(0)=3≠03+4(0)=3\neq0.
Full Solution & Work

Evaluate the denominator at (√48, 0)

dydx∣(48,0)=−2483+4(0)\left.\frac{dy}{dx}\right|_{(\sqrt{48},0)} = \frac{-2\sqrt{48}}{3+4(0)}

The denominator, 3+4(0)3+4(0), does not equal 0.

Conclude

Since the slope of the tangent line is defined (finite) at (48,0)(\sqrt{48},0), the tangent line at this point is **not** vertical.

AP Scoring — 1 Point

**P1**: A response does not need to consider the numerator; showing the denominator is nonzero at (48,0)(\sqrt{48},0) is sufficient. Must clearly demonstrate the slope is defined and conclude "no."
Part DHard4 points
For time t≥0t \geq 0, a particle is moving along another curve defined by the equation y3+2xy=24y^3+2xy=24. At the instant the particle is at the point (4,2)(4,2), the yy-coordinate of the particle's position is decreasing at a rate of 2 units per second. At that instant, what is the rate of change of the xx-coordinate of the particle's position with respect to time?

Answer

dxdt=10\dfrac{dx}{dt}=10 units per second
Full Solution & Work

Differentiate implicitly with respect to t

3y2dydt+2xdydt+2ydxdt=03y^2\frac{dy}{dt}+2x\frac{dy}{dt}+2y\frac{dx}{dt}=0

Substitute known values

At (x,y)=(4,2)(x,y)=(4,2) with dydt=−2\dfrac{dy}{dt}=-2:
3(2)2(−2)+2(4)(−2)+2(2)dxdt=03(2)^2(-2)+2(4)(-2)+2(2)\frac{dx}{dt}=0

−24−16+4dxdt=0-24-16+4\frac{dx}{dt}=0

Solve for dx/dt

4dxdt=40  ⟹  dxdt=10 units per second4\frac{dx}{dt}=40 \implies \frac{dx}{dt}=10 \text{ units per second}

AP Scoring — 4 Points

**P1**: Attempts implicit differentiation of y3+2xy=24y^3+2xy=24 with respect to tt, with at most one error.
**P2**: A correct equation equivalent to 3y2dydt+2xdydt+2ydxdt=03y^2\frac{dy}{dt}+2x\frac{dy}{dt}+2y\frac{dx}{dt}=0.
**P3**: Correctly uses dydt=−2\frac{dy}{dt}=-2 (a response does not need to explicitly declare this — substituting it directly into the implicit equation is sufficient). Using both +2+2 and −2-2 anywhere in the same response does not earn this point.
**P4**: Cannot be earned without the first 3 points. Earned only for the value 10, with no mistakes in the supporting work.

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