# Ap Calculus BC 2002 Form B Question 5?

## Apr 28, 2018

a) If $y = - 2$ is tangent to the graph, it means that the slope of that particular tangent is $0$. Thus:

$0 = \frac{3 - x}{y}$

We also know that $y = - 2$. Now all we must do is solve for $x$ which is pretty simple.

$0 = \frac{3 - x}{- 2}$

$x = 3$

To determine the nature of this critical point, we must determine the second derivative.

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 1 \left(y\right) - \left(3 - x\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{y} ^ 2$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- y - \left(3 - x\right) \frac{3 - x}{y}}{y} ^ 2$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- y - {\left(3 - x\right)}^{2} / y}{{y}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- {y}^{2} - {\left(3 - x\right)}^{2}}{y} ^ 3$

Since this $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ is positive at the point $\left(3 , - 2\right)$, this is a minimum (remember a minimum is always concave up and a maximum always concave down).

b) This is a classic differential equation solving problem. Start by separating the x and the y.

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(y\right) = 3 - x$

$\mathrm{dy} \left(y\right) = 3 - x \mathrm{dx}$

Integrate both sides.

$\int y \mathrm{dy} = \int 3 - x \mathrm{dx}$

$\frac{1}{2} {y}^{2} = - \frac{1}{2} {x}^{2} + 3 x + C$

Never forget the constant of integration has to be included before you solve for $y$. If you forget $C$ when solving this type of problem on the exam, they will only be able to give you a maximum of $\frac{3}{6}$ marks! It's ok to rewrite as a different letter before you take the square root.

$\frac{1}{2} {y}^{2} = - \frac{1}{2} {x}^{2} + 3 x + C$

Now solve for $C$.

$\frac{1}{2} {\left(- 4\right)}^{2} = - \frac{1}{2} \left({6}^{2}\right) + 3 \left(6\right) + C$

$8 + 18 - 18 = C$

$C = 8$

It now follows that

$\frac{1}{2} {y}^{2} = - \frac{1}{2} {x}^{2} + 3 x + 8$

${y}^{2} = - {x}^{2} + 6 x + 16$

$y = \pm \sqrt{- {x}^{2} + 6 x + 16}$

But since the initial condition has $y$ negative, we only keep the negative sign.

$y = - \sqrt{6 x - {x}^{2} + 16}$

Hopefully this helps!