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December 18, 2014

December 18, 2014

Posted by **Lee** on Sunday, October 30, 2011 at 8:40pm.

- calculus -
**Steve**, Monday, October 31, 2011 at 11:50amjust plug and chug:

4(x^2 + y^2)(2x + 2yy') = 9(2x - 2yy')

Now, the tangent is horizontal when y' = 0

4(x^2 + y^2)(2x) = 9(2x)

8x^3 + 8xy^2 = 18x

2x(4x^2 + 4y^2 - 9) = 0

So, either x=0

substitute back into original equation:

2y^4 = -9y^2

y=0

or,

4x^2 + 4y^2 = 9

x^2 = (9 - 4y^2)/4

Substitute that back into the original equation

2((9 - 4y^2)/4 + y^2)^2 = 9((9 - 4y^2)/4 - y^2)

Expand the binomial and solve for y^2

- calculus -
**Kelvin**, Saturday, January 26, 2013 at 5:36pmTan^3(xy^2+y)=x

Use derivative

- calculus -
**Kelvin**, Saturday, January 26, 2013 at 5:40pmFind .... Tan^3(xy^2+y)=x

Use derivative

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