For a curve defined implicitly (with dy/dx supplied), students use the tangent line to approximate a nearby y-coordinate, test whether a given horizontal line and a point on the x-axis correspond to tangency, and apply related rates to a second implicit curve to find the rate of change of an x-coordinate.
For a curve defined implicitly (with dy/dx supplied), students use the tangent line to approximate a nearby y-coordinate, test whether a given horizontal line and a point on the x-axis correspond to tangency, and apply related rates to a second implicit curve to find the rate of change of an x-coordinate.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
The slope of the tangent at $(2,4)$ is $$\frac{dy}{dx}\Big|_{(2,4)}=\frac{-2(2)}{3+4(4)}=\frac{-4}{19}.$$ The tangent line at $(2,4)$ is $y-4=-\frac{4}{19}(x-2)$. Using it to approximate the $y$-coordinate of the nearby point with $x=3$: $$y\approx 4-\frac{4}{19}(3-2)=4-\frac{4}{19}=\frac{72}{19}\approx 3.789.$$
2 points: 1 for the tangent slope dy/dx|_{(2,4)}=-4/19; 1 for the linear approximation y approx 4-(4/19)(3-2)=72/19. An answer that omits the (3-2) step, or uses the point as if x already equals 3, does not earn the approximation point; the value need not be simplified.
A horizontal tangent requires $\frac{dy}{dx}=0$, i.e. $\frac{-2x}{3+4y}=0$, which forces $x=0$. If $y=1$ were tangent, the point of tangency would have to be $(0,1)$. Test whether $(0,1)$ lies on the curve: $$0^2+3(1)+2(1)^2=0+3+2=5\ne 48.$$ Since $(0,1)$ is **not** on the curve, the horizontal line $y=1$ is **not tangent** to the curve.
2 points: 1 for setting dy/dx=0 to get x=0 (so any tangency point on y=1 must be (0,1)); 1 for the conclusion 'not tangent' justified by showing (0,1) fails the curve equation (5 != 48). An alternative accepted method substitutes y=1 into the curve to get x=+/-sqrt(43), then shows dy/dx != 0 there; that method must consider both signs of x for the second point. Answering without a substitution check does not earn the second point.
A vertical tangent would require $\frac{dy}{dx}$ to be undefined, i.e. the denominator $3+4y=0$. At the point $(\sqrt{48},0)$, $y=0$, so $$3+4y=3+4(0)=3\ne 0.$$ The slope there is $\frac{dy}{dx}=\frac{-2\sqrt{48}}{3}$, which is **defined (finite)**. Therefore the tangent line at $(\sqrt{48},0)$ is **not vertical**.
1 point for clearly showing the slope at (sqrt48,0) is defined (denominator 3+4(0)=3 != 0) and answering 'no, not vertical.' A bare 'no' without evaluating the denominator/slope does not earn the point.
Differentiate the second curve $y^3+2xy=24$ implicitly with respect to time $t$: $$3y^2\frac{dy}{dt}+2\left(x\frac{dy}{dt}+y\frac{dx}{dt}\right)=0.$$ At $(4,2)$ with $\frac{dy}{dt}=-2$ (the $y$-coordinate is decreasing at $2$ units/sec): $$3(2)^2(-2)+2\left(4(-2)+2\frac{dx}{dt}\right)=0$$ $$3(4)(-2)+2(-8)+4\frac{dx}{dt}=0$$ $$-24-16+4\frac{dx}{dt}=0 \;\Rightarrow\; 4\frac{dx}{dt}=40 \;\Rightarrow\; \frac{dx}{dt}=10.$$ At that instant the $x$-coordinate is increasing at a rate of $10$ units per second.
4 points: 1 for correctly differentiating y^3 (3y^2 dy/dt); 1 for correctly differentiating 2xy with the product rule (2x dy/dt + 2y dx/dt) and setting the total to 0; 1 for substituting dy/dt=-2 at (4,2); 1 for the answer dx/dt=10. A sign error in dy/dt (using +2) or a missing product-rule term forfeits the affected points.
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