implicit differentiaon/ sketching it (1 Viewer)
- Thread starter ..:''ooo
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At a point where dy/dx = 0 you have a stationary point and where dy/dx--->∞ you end up with vertical tangents. Taking an example from a textbook:
x<sup>2</sup> + y<sup>2</sup> = xy +3 (do the implicit thing)
(x-2y)dy/dx=(2x-y)
stationary points will exist where 2x-y=0 and these can either be found by inspection or by substituting 2x-y=0 into the original equation and solving simultaneously. Either way you end up with (1,2) and (-1,-2). These are your stationary points.
dy/dx --> ∞ i.e where x-2y-->0 so to find your vertical tangents you solve x-2y=0 which, once more, can be done via inspection or substitution (or whatever other method you see fit).
Using this information to draw the curve is something I hate but in this case you need to recognise either that the stationary points/vertical tangents imply an ellipsoid or simply that the equation itself x<sup>2</sup> + y<sup>2</sup> = xy +3 implies this. The critical points kind of give you bounds within which the curve then exists. It also helps to recognise that the curve has symmetry about y=x.
I hope that gives you an idea how to go about it.
x<sup>2</sup> + y<sup>2</sup> = xy +3 (do the implicit thing)
(x-2y)dy/dx=(2x-y)
stationary points will exist where 2x-y=0 and these can either be found by inspection or by substituting 2x-y=0 into the original equation and solving simultaneously. Either way you end up with (1,2) and (-1,-2). These are your stationary points.
dy/dx --> ∞ i.e where x-2y-->0 so to find your vertical tangents you solve x-2y=0 which, once more, can be done via inspection or substitution (or whatever other method you see fit).
Using this information to draw the curve is something I hate but in this case you need to recognise either that the stationary points/vertical tangents imply an ellipsoid or simply that the equation itself x<sup>2</sup> + y<sup>2</sup> = xy +3 implies this. The critical points kind of give you bounds within which the curve then exists. It also helps to recognise that the curve has symmetry about y=x.
I hope that gives you an idea how to go about it.
thunderdax
I AM JESUS LOL!
To implicitly differentiate, differentiate all x's normally, and with y's, differentiate like you would with x and then multiply by dy/dx(eg d(y^3)/dx=3y^2*dy/dx).
Also, you can find points of inflexion with implicit differentiation with the second derivative.
Also, you can find points of inflexion with implicit differentiation with the second derivative.