Thursday, November 09, 2006

Chapter 3 Test #11

11. (25 points) Use implicit differentiation to determine all points on the ellipse where the tangent line is undefined.

x2 + xy + y2 = 12

First, we have to differentiate the given equation (since what we’re going to observe is the slope of the tangent line, which is the derivative of the equation):

2x + (xy’ + y (1)) + 2yy’ = 0

(Remember to use the PRODUCT RULE for xy)

Get all y’ terms on one side:

xy’ + 2yy’ = -2x – y

Solve for y’:

y’ = -2x – y

x + 2y

We’re looking for when the slope of the tangent line is undefined. That is, when is y’ undefined? Since y’ is equal to a fraction, it would be undefined when the denominator is equal to 0:

x + 2y = 0

Solving for x: x = -2y

Now we can substitute –2y for all x values in the original equation:

(-2y)2 + (-2y)(y) + y2 = 12

Solving for y:

4y2 + (-2y2) + y2 = 12

3y2 = 12

y2 = 4

y = + or – 2 (don’t forget that square roots are going to be + and –)

These are the values of y of the points where the slope of the tangent line is undefined. To find the x values at these points, we can plug each y value into the x = -2y equation.

x = -2(2) = -4 Point #1: (-4, 2)

x = -2(-2) = 4 Point #2: (4, 2)

These are the points where the slope of the tangent line (i.e. the derivative of the original equation) is undefined.

Hope that helps everybody. Tomorrow's Friday, party up the wazoo! ~8)

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