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

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.

#### Mark44

Mentor
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.

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.

#### Mark44

Mentor
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($\theta$).

#### Stephen Tashi

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.