anonymous
  • anonymous
x^2+y^3+4y=4 ,what is the equation of the tangent to the curve of the above equation at -1,1
MIT 18.01 Single Variable Calculus (OCW)
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
implicit derivation: 2x + 3(y^2)(y') + 4(y') = 0 y' = (-2x) / (3y^2 + 4) replace: y'(-1, 1) = 2/7
TuringTest
  • TuringTest
The comment was withdrawn, but it was stated earlier that (-1,1) is not on the graph of x^2+y^3+4y=4. That is true.
anonymous
  • anonymous
Although the point is not a point of tangency, doesn't necessarily mean that there does not exist a line tangent to the curve. In fact if one takes the derivative using implicit differentiation and substitute it into the point slope equation for the slope one can find the point of tangency. One can then use this point of tangency to find the equation of the line. I used winplot to find the point.... here is the equation of the line tangent to the graph passing through (-1,1). (y-0.82342576161)=((-2*0.38469416012450)/(3*(0.82342576161)^2+4))(x-0.38469416012450 )

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anonymous
  • anonymous
One could also use the following equation: (y-1)=((-2*0.38469416012450)/(3*(0.82342576161)^2+4))(x+1)
anonymous
  • anonymous
abermejo was correct if the point had been a point of tangency. And it is his equation for the slope that is used in my equation of the line.
anonymous
  • anonymous
thax loads all of you

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