I would just like to check that I have done this correctly. I have to find the equation of the normal to the curve 3x^2+2xy^2+6=2y^3 at the point (2,3).
First I have differentiated wrt x giving: dy/dx = 3x+y^2/3y^2-2xy.
Then using x=2 and y =3: dy/dx = 1.
So the equation of the tangent is (using y=mx+c): 3=(1)(2)+c.
So c=1. Therefore y=x+1.
Now the normal to this is: y=(-1)x+c, passing through (2,3).
So 3=(-1)(2)+c, so c=5. Therefore the equation of the normal is y=-x+5.
Is this correct.
2007-03-24
01:22:17
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4 answers
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asked by
eazylee369
4
in
Mathematics