What Are the Inner and Outer Radii of y=arccos(x)?

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The inner and outer radii of the function y=arccos(x) depend on the context of its application. In Cartesian coordinates, the inner radius is 0 and the outer radius is 1, as the graph represents a semi-circle with a radius of 1. In polar coordinates, the equation transforms to r=arccos(θ), where the inner radius remains 0, but the outer radius varies with θ, approaching 1 as θ nears π/2 and extending to infinity as θ approaches 0. Thus, the interpretation of radii is context-dependent, but generally, the inner radius is 0 and the outer radius is 1 in standard scenarios. Understanding these radii is crucial for applications involving volume of rotation.
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What would the inner and outter radius of y=arccos(x)
 
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cmab said:
What would the inner and outter radius of y=arccos(x)

You need to state the whole problem. I assume this function is one part of specifying a volume of rotation. How are the boundaries defined and where is the axis of rotation?
 
be

The inner and outer radius of y=arccos(x) would depend on the specific context in which the equation is being used. If we are considering the graph of this equation in the Cartesian coordinate system, then the inner radius would be 0 and the outer radius would be 1. This is because the range of the arccosine function is limited to the interval [0, π], and the graph of this function forms a semi-circle with a radius of 1 centered at the origin.

However, if we are looking at this equation in the context of polar coordinates, then the inner and outer radius would be different. In polar coordinates, the equation y=arccos(x) can be written as r=arccos(θ), where r is the distance from the origin and θ is the angle measured from the positive x-axis. In this case, the inner radius would be 0, as the point (0,0) is the origin. The outer radius would depend on the value of θ. As θ approaches π/2, the outer radius would approach 1, and as θ approaches 0, the outer radius would approach infinity.

Overall, the inner and outer radius of y=arccos(x) can vary depending on the context in which it is being used, but in general, the inner radius would be 0 and the outer radius would be 1.
 
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