For a curve defined implicitly by an equation, students derive an expression for dy/dx, locate points where the tangent line is horizontal and where it is vertical, and use related rates to find the rate of change of a particle's vertical position as it moves along the curve.
For a curve defined implicitly by an equation, students derive an expression for dy/dx, locate points where the tangent line is horizontal and where it is vertical, and use related rates to find the rate of change of a particle's vertical position as it moves along the curve.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
Differentiate $6xy=2+y^3$ implicitly with respect to $x$, using the **product rule** on the left and the **chain rule** on $y^3$: $$\frac{d}{dx}(6xy)=\frac{d}{dx}(2+y^3)\;\Rightarrow\; 6y+6x\frac{dy}{dx}=3y^2\frac{dy}{dx}.$$ Collect the $\frac{dy}{dx}$ terms: $$6y=3y^2\frac{dy}{dx}-6x\frac{dy}{dx}=\big(3y^2-6x\big)\frac{dy}{dx}.$$ Divide by $3$ and solve: $$\frac{dy}{dx}=\frac{6y}{3y^2-6x}=\frac{2y}{y^2-2x}.\qquad\blacksquare$$
2 points. P1 (implicit differentiation): correct differentiation of 6xy=2+y^3, i.e. 6y+6x·y'=3y^2·y' (y' notation acceptable). P2 (verification): solving to reach dy/dx=2y/(y^2-2x); cannot be earned without P1. Presenting 2y=(y^2-2x)·dy/dx is sufficient for P2 if there are no later errors.
The tangent line is **horizontal** where $\frac{dy}{dx}=\frac{2y}{y^2-2x}=0$, which requires the numerator $2y=0$, i.e. $y=0$ (with denominator $y^2-2x\ne0$). Substitute $y=0$ into the curve $6xy=2+y^3$: $$6x(0)=2+0^3\;\Rightarrow\; 0=2,$$ which is impossible. So there is **no point on the curve where $y=0$**, and therefore **no point at which the tangent line is horizontal.**
2 points. P1 (sets 2y=0): any of 2y=0, y=0, dy/dx=0, or 2y/(y^2-2x)=0. P2 (answer with reason): substitute y=0 into the curve, obtain 6x·0=2 (no solution), and conclude no horizontal tangent. A response need not state y^2-2x≠0.
The tangent line is **vertical** where the denominator of $\frac{dy}{dx}$ is zero (and the numerator is not): $y^2-2x=0$, i.e. $x=\frac{y^2}{2}$ (with $2y\ne0$). Substitute $x=\frac{y^2}{2}$ into the curve $6xy=2+y^3$: $$6\left(\frac{y^2}{2}\right)y=2+y^3\;\Rightarrow\; 3y^3=2+y^3\;\Rightarrow\; 2y^3=2\;\Rightarrow\; y=1.$$ Then $x=\frac{1^2}{2}=\frac{1}{2}$. Check on the curve: $6\cdot\frac{1}{2}\cdot 1=3$ and $2+1^3=3$. ✓ The tangent line is vertical at the point $\boxed{\left(\tfrac{1}{2},\,1\right)}$.
3 points. P1 (sets y^2-2x=0): also earned by y^2=2x or y=±sqrt(2x). P2 (substitutes x=y^2/2 into 6xy=2+y^3): equivalently substitute x=(2+y^3)/(6y) into y^2-2x=0. P3 (answer): both coordinates of (1/2,1), labeled. A response that states (1/2,1) but does not verify it lies on the curve does not earn P2 or P3.
A particle moves along the curve, so $x$ and $y$ are functions of time $t$. **Differentiate $6xy=2+y^3$ with respect to $t$** (product rule on $6xy$, chain rule on $y^3$): $$6y\frac{dx}{dt}+6x\frac{dy}{dt}=3y^2\frac{dy}{dt}.$$ At the instant the particle is at $\left(\tfrac12,-2\right)$ with $\frac{dx}{dt}=\frac{2}{3}$: $$6(-2)\!\left(\tfrac{2}{3}\right)+6\!\left(\tfrac12\right)\frac{dy}{dt}=3(-2)^2\frac{dy}{dt}.$$ Simplify: $$-8+3\frac{dy}{dt}=12\frac{dy}{dt}\;\Rightarrow\; -8=9\frac{dy}{dt}\;\Rightarrow\; \frac{dy}{dt}=-\frac{8}{9}.$$ The particle's vertical position is changing at $\boxed{\dfrac{dy}{dt}=-\dfrac{8}{9}\text{ unit per second}}$.
2 points. P1 (implicit differentiation w.r.t. t): present one or more of the terms 6y·dx/dt, 6x·dy/dt, or 3y^2·dy/dt. P2 (answer): dy/dt=-8/9. Units do not affect scoring. An unsupported -8/9 earns no points. Alternate path dy/dt=(dy/dx)·(dx/dt)=(2(-2)/((-2)^2-2(1/2)))·(2/3)=(-4/3)(2/3)=-8/9 is equally valid.
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