2002 ap® calculus ab free - response questions (form b)\n5. consider the differential equation…

Explanation:

Step1: Recall slope of tangent line

The slope of the tangent line $y = - 2$ is $m = 0$. Given $\frac{dy}{dx}=\frac{3 - x}{y}$, when $y=-2$ and $\frac{dy}{dx}=0$, we set $\frac{3 - x}{y}=0$. $$\frac{3 - x}{-2}=0$$

Step2: Solve for $x$

Cross - multiply to get $3 - x=0$, so $x = 3$.

Step3: Use the second - derivative test

Differentiate $\frac{dy}{dx}=\frac{3 - x}{y}$ with respect to $x$ using the quotient rule. First, rewrite it as $y\frac{dy}{dx}=3 - x$. Differentiating both sides with respect to $x$ gives $y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=-1$. When $x = 3$ and $y=-2$, $\frac{dy}{dx}=0$. Then $-2\frac{d^{2}y}{dx^{2}}+0=-1$, so $\frac{d^{2}y}{dx^{2}}=\frac{1}{2}>0$. So $f$ has a local minimum at $x = 3$.

Step4: Solve the differential equation for part (b)

Separate variables in $\frac{dy}{dx}=\frac{3 - x}{y}$, we get $y;dy=(3 - x)dx$. Integrate both sides: $\int y;dy=\int(3 - x)dx$. $$\frac{y^{2}}{2}=3x-\frac{x^{2}}{2}+C$$

Step5: Use the initial condition

Given $g(6)=-4$, substitute $x = 6$ and $y=-4$ into $\frac{y^{2}}{2}=3x-\frac{x^{2}}{2}+C$. $$\frac{(-4)^{2}}{2}=3\times6-\frac{6^{2}}{2}+C$$ $$8 = 18 - 18+C$$ $C = 8$.

Step6: Find $y = g(x)$

So $\frac{y^{2}}{2}=3x-\frac{x^{2}}{2}+8$, and $y^{2}=6x - x^{2}+16$, then $y=-\sqrt{6x - x^{2}+16}$ (since $y=-4$ when $x = 6$).

Answer:

(a) The $x$-coordinate of the point of tangency is $x = 3$. $f$ has a local minimum at this point. (b) $y=-\sqrt{6x - x^{2}+16}$