For a curve defined implicitly by a cubic equation, students derive dy/dx, use the tangent line to approximate the y-coordinate of a nearby point, find a point where the tangent line is vertical, and apply related rates to a second implicit curve to find a rate of change.
For a curve defined implicitly by a cubic equation, students derive dy/dx, use the tangent line to approximate the y-coordinate of a nearby point, find a point where the tangent line is vertical, and apply related rates to a second implicit curve to find a rate of change.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
Differentiate $y^3-y^2-y+\tfrac{1}{4}x^2=0$ implicitly with respect to $x$, treating $y$ as a function of $x$: $$3y^2\frac{dy}{dx}-2y\frac{dy}{dx}-\frac{dy}{dx}+\frac{x}{2}=0.$$ Group the $\frac{dy}{dx}$ terms: $$\big(3y^2-2y-1\big)\frac{dy}{dx}=-\frac{x}{2}.$$ Solve: $$\frac{dy}{dx}=\frac{-x}{2\big(3y^2-2y-1\big)}.$$
2 points. P1: the correct implicit differentiation of the whole equation (3y^2 y' - 2y y' - y' + x/2 = 0); any notation for dy/dx is fine. P2: completing the algebra to the requested form; presenting (3y^2-2y-1)(dy/dx) = -x/2 with no later error is sufficient. P2 requires P1 first.
At $(2,-1)$ the slope is $$\frac{dy}{dx}\Big|_{(2,-1)}=\frac{-2}{2\big(3(-1)^2-2(-1)-1\big)}=\frac{-2}{2(3+2-1)}=\frac{-2}{8}=-\frac{1}{4}.$$ The tangent line at $(2,-1)$ gives the linear approximation $$y\approx -1-\tfrac{1}{4}(x-2).$$ At $x=1.6$: $$y\approx -1-\tfrac{1}{4}(1.6-2)=-1-\tfrac{1}{4}(-0.4)=-1+0.1=-0.9.$$ So the $y$-coordinate of $P$ is approximately $-0.9$.
2 points. P3: the slope -1/4 at (2,-1). P4: the tangent-line approximation evaluated at x=1.6, i.e. -1 - (1/4)(1.6-2) = -0.9. Writing -1 - (1/4)(1.6-2) banks P4 (later simplification errors ignored); a slope other than -1/4 that has been declared as dy/dx still earns P4 only for the value -1 + k(-0.4) consistent with it.
The tangent line is vertical where $\frac{dy}{dx}$ is undefined, i.e. where the denominator is zero (and the numerator is not): $$2\big(3y^2-2y-1\big)=0 \;\Rightarrow\; 3y^2-2y-1=0.$$ Factor: $3y^2-2y-1=(3y+1)(y-1)=0$, so $y=-\tfrac{1}{3}$ or $y=1$. Since the point $S$ has $y>0$, we take $$y=1.$$ The vertical tangent (for $x>0,\,y>0$) occurs at the point on $G$ where $y=1$.
2 points. P5: sets the denominator equal to zero — any of 2(3y^2-2y-1)=0, 3y^2-2y-1=0, or the factored (3y+-1)(y+-1)=0. P6: the correct y-coordinate y=1 (must clearly select y=1 as the only valid solution given y>0; merely listing y=-1/3 and y=1 without choosing does not earn P6). P6 requires P5.
On curve $H$: $2xy+\ln y=8$, with $x$ and $y$ both functions of $t$. Differentiate with respect to $t$ (product rule on $2xy$, chain rule on $\ln y$): $$2\frac{dx}{dt}\,y+2x\frac{dy}{dt}+\frac{1}{y}\frac{dy}{dt}=0.$$ Substitute the instant $(x,y)=(4,1)$ with $\frac{dx}{dt}=3$: $$2(3)(1)+2(4)\frac{dy}{dt}+\frac{1}{1}\frac{dy}{dt}=0 \;\Rightarrow\; 6+8\frac{dy}{dt}+\frac{dy}{dt}=0.$$ So $6+9\frac{dy}{dt}=0$, giving $$\frac{dy}{dt}=-\frac{6}{9}=-\frac{2}{3}.$$
3 points. P7: implicitly differentiates 2xy+ln y=8 with respect to t with at most one error. P8: the correct related-rates equation 2(dx/dt)y + 2x(dy/dt) + (1/y)(dy/dt) = 0. P9: the value dy/dt = -2/3; eligibility requires P7 and P8 with no differentiation errors. (Alternate path: compute dy/dx = -2/9 at (4,1), then dy/dt = (dy/dx)(dx/dt) = (-2/9)(3) = -2/3.)
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