AP Calc FR Help, Please!

Test Preparation AP Test Preparation
1

<p>My teacher assigned us FR questions from old exams and I'm having trouble with Part C on this one. Help would be appreciated!</p>

<p>Question: </p>

<p>**Given the curve x^2 - xy + y^2 = 9.</p>

<p>a) Write a general expression for the slope of the curve.</p>

<p>b) Find the coordinates of the points on the curve where the tangents are vertical.</p>

<p>c) At the point (0,3) find the rate of the change in the slope of the curve with respect to x. **</p>

<p>What I've solved so far:</p>

<p>a) by using implicit differentiation, I got the derivative as: (dy/dx) = (-2x +y) / (-x +2y)</p>

<p>b) the tangents are vertical when the derivative is undefined, so I set the denominator in the derivative, (-x +2y), equal to zero. Then I solved for y, which = +/- sqrt(3). Then I plugged y into -x +2y = 0 to solve for y, and I got +/- 2(sqrt3). So I know the coordinates.</p>

<p>**c) I know that finding the second derivative will give me the rate of change in the slope, but I can't seem to get a 2nd derivative that makes sense. I either lose or gain an extra (dy/dx). I've been using the quotient rule.</p>

<p>This is where I need help. Please show/explain how you did the work, b/c I'm a visual learner and just an answer won't really teach me how to find a 2nd derivative with implicit differentiation. **</p>

<p>Thanks!</p>

2

<p>x^2 - xy + y^2 = 9
2x-x(dy/dx)-y+2y(dy/dx)=0
2-x(d^2y/dx^2)-(dy/dx)+2y(d^2y/dx^2)+2(dy/dx)^2=0
2-0-.5+6(d^2y/dx^2)+2(1/4)=0
2+6(d^2y/dx^2)=0
d^2y/dx^2=-2/6
Quickly done; could've made a careless error.
(dy/dx)=first derivative of y with respect to x
(d^2y/dx^2)=second derivative of y with respect to x</p>

3

<p>Thanks, aimee.</p>

<p>I got the same answer when I did the part C, earlier. One of the many times I did part C. lol, thanks again!</p>

4

<p>I disagree with the above process slightly, and that changed the answer just a smidge.</p>

<p>x^2 - xy + y^2
2x - x(dy/dx) - y + 2y(dy/dx) = 0
2 - [x(d^2y/dx^2) + (dy/dx)] - (dy/dx) + [2y(d^2y/dx^2) + 2(dy/dx)(dy/dx)] = 0
[terms inside the brackets represent use of the Product Rule]</p>

<p>The above answer seemed to be missing one of the -dy/dx terms, one created by the Product Rule for the derivative of x(dy/dx) and the other created by the derivative of y.</p>

<p>At (0,3), dy/dx = 1/2</p>

<p>2 - 0 - 1/2 - 1/2 + 6(d^2y/dx^2) + 1/2 = 0
3/2 + 6(d^2y/dx^2) = 0
d^2y/dx^2 = (-3/2)/6 = -3/12 = -1/4</p>

<p>--</p>

<p>An alternative method:</p>

<p>dy/dx = (-2x + y)/(-x + 2y)
d^2y/dx^2 = [(-x + 2y)(-2 + dy/dx) - (-2x + y)(-1 + 2 dy/dx)] / (-x + 2y)^2
d^2y/dx^2 = [(6)(-2 + 1/2) - (3)(-1 + 1)] / (6^2)
d^2y/dx^2 = -9/36 = -1/4</p>