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2021 AP Calculus AB FRQ Question 5: Implicit Curve — Tangent Lines & Extrema

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2021.

Question 5

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Implicit Curve — Tangent Lines & Extrema

Implicit Differentiation

Hard
Consider the function y=f(x)y=f(x) whose curve is given by the equation 2y2−6=ysin⁡x2y^2-6=y\sin x, for y>0y>0.
Part AMedium2 points
Show that dydx=ycos⁡x4y−sin⁡x\dfrac{dy}{dx}=\dfrac{y\cos x}{4y-\sin x}.

Answer

dydx=ycos⁡x4y−sin⁡x\dfrac{dy}{dx}=\dfrac{y\cos x}{4y-\sin x} (shown by implicit differentiation)
Full Solution & Work

Differentiate both sides implicitly

ddx(2y2−6)=ddx(ysin⁡x)  ⟹  4ydydx=dydxsin⁡x+ycos⁡x\frac{d}{dx}(2y^2-6) = \frac{d}{dx}(y\sin x) \implies 4y\frac{dy}{dx} = \frac{dy}{dx}\sin x+y\cos x

Solve for dy/dx

dydx(4y−sin⁡x)=ycos⁡x  ⟹  dydx=ycos⁡x4y−sin⁡x\frac{dy}{dx}(4y-\sin x) = y\cos x \implies \frac{dy}{dx}=\frac{y\cos x}{4y-\sin x}

AP Scoring — 2 Points

**P1**: Correct implicit differentiation, including the Product Rule on ysin⁡xy\sin x.
**P2**: Requires P1. Correctly isolates dydx\frac{dy}{dx} to reach the given expression.
Part BMedium2 points
Write an equation for the line tangent to the curve at the point (0,3)(0,\sqrt3).

Answer

y=14x+3y=\dfrac14 x+\sqrt3
Full Solution & Work

Find the slope at (0, √3)

dydx∣(0,3)=3cos⁡(0)43−sin⁡(0)=343=14\left.\frac{dy}{dx}\right|_{(0,\sqrt3)} = \frac{\sqrt3\cos(0)}{4\sqrt3-\sin(0)} = \frac{\sqrt3}{4\sqrt3}=\frac14

Write the tangent line

y−3=14(x−0)  ⟹  y=14x+3y-\sqrt3 = \frac14(x-0) \implies y=\frac14 x+\sqrt3

AP Scoring — 2 Points

**P1**: Correct slope, 14\frac14.
**P2**: Correct tangent-line equation.
Part CMedium2 points
For 0≤x≤π0\leq x\leq\pi and y>0y>0, find the coordinates of the point where the line tangent to the curve is horizontal.

Answer

(π2,2)\left(\dfrac{\pi}{2},2\right)
Full Solution & Work

Set dy/dx = 0

Since y>0y>0, dydx=0\frac{dy}{dx}=0 requires cos⁡x=0\cos x=0, so x=π2x=\frac{\pi}{2} within [0,π][0,\pi].

Solve for y at x = π/2

Substitute into the curve equation: 2y2−6=ysin⁡(π/2)=y2y^2-6=y\sin(\pi/2)=y.
2y2−y−6=0  ⟹  y=1±74  ⟹  y=2 or y=−1.52y^2-y-6=0 \implies y=\frac{1\pm7}{4} \implies y=2 \text{ or } y=-1.5

Since y>0y>0, y=2y=2.

State the point

The tangent line is horizontal at (π2,2)\left(\dfrac{\pi}{2},2\right).

AP Scoring — 2 Points

**P1**: Correctly sets cos⁡x=0\cos x=0 to find x=π/2x=\pi/2.
**P2**: Correctly solves the resulting quadratic for yy and selects the positive root, y=2y=2.
Part DHard2 points
Determine whether ff has a relative minimum, a relative maximum, or neither at the point found in part (c). Justify your answer.

Answer

Relative maximum.
Full Solution & Work

Analyze the sign of dy/dx near x = π/2

Near x=π2x=\frac{\pi}{2}, y≈2>0y\approx2>0 and 4y−sin⁡x≈8−1=7>04y-\sin x\approx8-1=7>0 stay positive, so the sign of dydx=ycos⁡x4y−sin⁡x\frac{dy}{dx}=\frac{y\cos x}{4y-\sin x} is governed by cos⁡x\cos x.

Determine the sign change

cos⁡x>0\cos x>0 for xx slightly less than π2\frac{\pi}{2}, and cos⁡x<0\cos x<0 for xx slightly greater than π2\frac\pi2. So dydx\frac{dy}{dx} changes from **positive to negative** as xx increases through π2\frac{\pi}{2}.

Conclude

Since y′y' changes from positive to negative at x=π2x=\frac\pi2, ff has a **relative maximum** there.

AP Scoring — 2 Points

**P1**: Correct sign analysis of dydx\frac{dy}{dx} on either side of x=π/2x=\pi/2 (via the sign of cos⁡x\cos x, with yy and 4y−sin⁡x4y-\sin x remaining positive).
**P2**: Correct conclusion, "relative maximum," tied to that sign change.

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