Ap Calculus BC 2002 Form B Question 5?

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1 Answer
Noah G
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=3−xy

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

0=3−x−2

x=3

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

d2ydx2=−1(y)−(3−x)(dydx)y2

d2ydx2=−y−(3−x)3−xyy2

d2ydx2=−y−(3−x)2yy2

d2ydx2=−y2−(3−x)2y3

Since this d2ydx2 is positive at the point (3,−2), 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.

dydx(y)=3−x

dy(y)=3−xdx

Integrate both sides.

∫ydy=∫3−xdx

12y2=−12x2+3x+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 36 marks! It's ok to rewrite as a different letter before you take the square root.

12y2=−12x2+3x+C

Now solve for C.

12(−4)2=−12(62)+3(6)+C

8+18−18=C

C=8

It now follows that

12y2=−12x2+3x+8

y2=−x2+6x+16

y=±√−x2+6x+16

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

y=−√6x−x2+16

Hopefully this helps!

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