semi circle turning point: why (0,1) rather than anywhere else on the semi-circle?

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apple2357

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If you take the equation of a semi circle y= root(1-x^2) and try to find its turning point you get (0,1). I don't understand why it is turning at (0,1) rather than anywhere else on the semi-circle? Can we talk about a 'turning-point' on a semi circle?
 
apple2357 said:
If you take the equation of a semi circle y= root(1-x^2) and try to find its turning point you get (0,1). I don't understand why it is turning at (0,1) rather than anywhere else on the semi-circle? Can we talk about a 'turning-point' on a semi circle?
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What is the mathematical definition of turning point of a function?
 
A turning point is where the derivative changes sign? Or where the derivative is zero?
 
apple2357 said:
A turning point is where the derivative changes sign? Or where the derivative is zero?
Click to expand...
\(\displaystyle D_x(\sqrt{1-x^2})=\dfrac{-x}{\sqrt{1-x^2}}\) now apply that definition.
 
apple2357 said:
If you take the equation of a semi circle y= root(1-x^2) and try to find its turning point you get (0,1). I don't understand why it is turning at (0,1) rather than anywhere else on the semi-circle? Can we talk about a 'turning-point' on a semi circle?
Click to expand...

As I understand it, you correctly found the turning point of this function to be (0,1), but are wondering why any place on the curve isn't equally a "turning point", since it has the same curvature everywhere, always "turning". Is that what you have in mind?

As has been pointed out, the answer lies in the definition:

A turning point is a point at which the derivative changes sign. A turning point may be either a relative maximum or a relative minimum (also known as local minimum and maximum). If the function is differentiable, then a turning point is a stationary point; however not all stationary points are turning points.
​

This definition clearly fits only the one point you found. The important thing is that it is a property of a function (that is, of how y relates to x along the graph), not of the curve itself (thought of without reference to the axes). Curvature is an intrinsic property of a curve; turning points are relative to the axes. That's the difference. A turning point is not just "any point where the curve is turning", but specifically where it turns around, from moving up to moving down or vice versa.
 
Dr.Peterson said:
As I understand it, you correctly found the turning point of this function to be (0,1), but are wondering why any place on the curve isn't equally a "turning point", since it has the same curvature everywhere, always "turning". Is that what you have in mind?

As has been pointed out, the answer lies in the definition:
A turning point is a point at which the derivative changes sign. A turning point may be either a relative maximum or a relative minimum (also known as local minimum and maximum). If the function is differentiable, then a turning point is a stationary point; however not all stationary points are turning points.
​

This definition clearly fits only the one point you found. The important thing is that it is a property of a function (that is, of how y relates to x along the graph), not of the curve itself (thought of without reference to the axes). Curvature is an intrinsic property of a curve; turning points are relative to the axes. That's the difference. A turning point is not just "any point where the curve is turning", but specifically where it turns around, from moving up to moving down or vice versa.
Click to expand...

That's exactly my question and you have given me good explanation! Thank you!
 
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