# Why does this concavity function not work for this polar fun

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In summary, the conversation discusses the polar equation 1/[√(sinθcosθ)] and the derivatives dy/dx and d2y/dx2. There is confusion about the concavity of the function and how it appears when graphed. The use of the chain rule is also questioned. Further clarification and explanation are needed to understand the behavior of the function.
For the polar equation 1/[√(sinθcosθ)]

I found the slope of the graph by using the chain rule and found that dy/dx=−tan(θ)

and the concavity d2y/dx2=2(tanθ)^3/2

This is a pretty messy derivative so I checked it with wolfram alpha and both functions are correct (but feel free to check in case I made a mistake that would explain this all).

My problem is that the concavity function indicates that it will never be concave down and always concave up but if you graph the function it looks like a typical 1/x function, which means IT IS concave down in Q3.

I found this extremely interesting and it's killing to figure out why, is it because my concavity function is in terms of theta? or perhaps was there a +/- or something?

If someone can shine some light as to why it's like this or push me in the right direction, it'd be greatly appreciated.

For the polar equation 1/[√(sinθcosθ)]
This is not an equation. An equation always has "=" in it somewhere.
I found the slope of the graph by using the chain rule and found that dy/dx=−tan(θ)
I don't see how you get this. The Cartesian form of your equation is y = 1/x (which you state below), so dy/dx = -1/x^2. If we convert the right side to polar form, we get dy/dx = -1/(r^2 cos^2(θ).
and the concavity d2y/dx2=2(tanθ)^3/2

This is a pretty messy derivative so I checked it with wolfram alpha and both functions are correct (but feel free to check in case I made a mistake that would explain this all).

My problem is that the concavity function indicates that it will never be concave down and always concave up but if you graph the function it looks like a typical 1/x function, which means IT IS concave down in Q3.
I found this extremely interesting and it's killing to figure out why, is it because my concavity function is in terms of theta? or perhaps was there a +/- or something?

If someone can shine some light as to why it's like this or push me in the right direction, it'd be greatly appreciated.

Mark44 said:
This is not an equation. An equation always has "=" in it somewhere.
I don't see how you get this. The Cartesian form of your equation is y = 1/x (which you state below), so dy/dx = -1/x^2. If we convert the right side to polar form, we get dy/dx = -1/(r^2 cos^2(θ).
Sorry about being unclear, I meant the equation as r(theta) =

And while that is true, why would the chain rule yield that answer. I am asking that you use the chain rule and find the slope and concavity functions and explain why they won't correspond with the concavity of 1/x.

And while that is true, why would the chain rule yield that answer.
I suspect that you used the chain rule incorrectly. Since x and y are functions of both r and ##\theta##, there are partial derivatives involved.

Please show your work for how you got dy/dx = -tan(##\theta##).

but if you graph the function

How are you graphing the function? The calculations you are doing for concavity are relevant to the shape of a graph that plots ##\theta## on the x-axis and ##\frac{1}{\sqrt{\sin(\theta) \cos(\theta)}}## on the y-axis. They aren't relevant to a plot in polar coordinates.

For example, if ##f(x)## is a function such that ##f'(x)## is zero on an interval then we can say the graph of ##f(x)## is a horizontal line - provided we are talking about the normal sort of graph in cartesian coordinates. If we have the equation ##r(\theta) = 15## then ##r'(\theta) = 0##, but this does not show the graph of ##r(\theta)## is a horizontal line when we plot points ##(15\cos(\theta), 15\sin(\theta))## on x-y graph paper.

## 1. Why is the concavity function not working for this polar function?

The concavity function is designed to work with Cartesian coordinates, where the x and y values are independent of each other. Polar coordinates, on the other hand, use a different system of coordinates where the distance from the origin and the angle are used to locate a point. Therefore, the concavity function will not work for polar functions because it is not compatible with the polar coordinate system.

## 2. Can I convert a polar function to a Cartesian one to use with the concavity function?

Yes, it is possible to convert a polar function to a Cartesian one by using the following formulas: x = r * cos(theta) and y = r * sin(theta). However, keep in mind that this conversion may not always be accurate, and it is better to use a concavity function specifically designed for polar coordinates.

## 3. Is there a specific concavity function for polar coordinates?

Yes, there are concavity functions that are specifically designed to work with polar coordinates. These functions take into account the unique properties of polar coordinates and can accurately calculate the concavity of a polar function.

## 4. How does the concavity function work for Cartesian coordinates?

The concavity function for Cartesian coordinates uses the first and second derivatives of a function to determine the direction and shape of the curve at a given point. It calculates the rate of change of the slope and determines whether it is increasing or decreasing, which helps to determine the concavity of the function.

## 5. Can the concavity function be used for any type of function?

The concavity function is designed to work with continuous functions that have a well-defined derivative. Therefore, it may not work for discontinuous or undefined functions. It is always best to check the properties of the function before using the concavity function to ensure accurate results.

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