# Why is the derivative of a polar function dy/dx?

Tags:
1. Nov 20, 2014

1. The problem statement, all variables and given/known data
$r = 2\cos(\theta)$

2. Relevant equations

3. The attempt at a solution

Why do textbook state that the derivative of the polar function (symbolic) is dy/dx and not $dr/d\theta$? It is a function of theta, then why is the derivative dy/dx?

Idea: Even though the equation is polar, you are defining the function in the xy-plane, so therefore, the function really is parametric y = f(x), but the parameter is theta?

2. Nov 20, 2014

### Staff: Mentor

I don't believe that they do this. If you're given a polar equation such as the one above, you have r as a function of $\theta$, so $dr/d\theta$ would be the derivative of this function. The textbook could ask that you calculate dy/dx by converting the equation into Cartesian form and then differentiating, but I don't believe that they would say that dy/dx is the derivative of the polar function.

3. Nov 20, 2014

### Staff: Mentor

Is that exactly what the book asked for, or did it say something like "what is the slope of the tangent to the locus of points defined by the function r = 2 cosΘ?"

Chet

4. Dec 20, 2014

@chet, sorry, I didnt see this. The book asked for "slope of the tangent line at Θ = pi/6" for example.

But the question is:

In polar derivatives, are you parametrizing an equation of xy in terms of theta?

5. Dec 20, 2014

### Staff: Mentor

They give you a function of r and theta, and they seem to want the slope on an x-y plot. I don't think the problem statement is very clear.

Chet

6. Dec 21, 2014

Yes, they give a polar function, in circular coordinates. Obviously, the derivative is not defined in the circular coordinates.

So I think what they are saying is that:

In an xy plane, you parametrize x, and y with respect to theta. Then dy/dx tells the slope at some point theta. It is really a parametric equation I suppose?

7. Dec 21, 2014

### ehild

You can write x and y in terms of the polar coordinates, and also the differentials dx and dy. - how?
$$\frac{dy}{dx}=\frac {\frac{\partial y} {\partial r} dr + \frac {\partial y} {\partial \theta } d \theta } {\frac {\partial x} {\partial r} dr + \frac {\partial x} {\partial \theta }d \theta }$$

From the function, r= 2 cos(θ) and dr = -2sin(θ)dθ. Substitute into the expression for dy/dx.

8. Dec 21, 2014

Exactly, it is still dy/dx. It is describing a cartesian plane slope still, horizontally.

9. Dec 21, 2014

### ehild

The derivative of a polar function r(θ) is dr/dθ. In this case, it is dr/dθ = -2sin(θ). If you plot r(θ) on the way that θ is on the horizontal axis and r is on the vertical axis, you get a simple cosine plot. But you can plot the r(θ) function on the (x,y) plane, in polar coordinates (r is the distance from origin and theta is the angle measured from the positive x axis).
See the figure. The plot of the given function looks similar. The red line is the tangent of the parametric curve (x(r,θ) , y(r,θ)) and dy/dx is the slope of that tangent line, you can find it by implicit derivation.

10. Dec 21, 2014

### Staff: Mentor

For this problem, it seems simpler the parametric way Amad27 suggested:
y = rsinθ=2sinθcosθ=sin(2θ)
x = rcosθ=2cos2θ=(1+cos(2θ))/2
Then get dy/dθ and dx/dθ in terms of θ, and divide to get dy/dx in terms of θ.

Chet

11. Dec 21, 2014

### vela

Staff Emeritus
The slope of the tangent line in the context of this problem is dy/dx. It's not just any derivative; it's the specific derivative dy/dx. Regardless of the representation of the function, whether it's in cartesian or polar coordinates, the book is asking you to find dy/dx at the point in question.

12. Dec 22, 2014

It does not specifically say find "dy/dx" it says find the slope of the tangent line. Which could easily fool one person to be dr/dtheta

13. Dec 22, 2014

### vela

Staff Emeritus
Part of learning calculus is understanding what the derivatives represent. You are supposed to recognize that "the slope of the tangent line" corresponds to the derivative dy/dx. No one's being fooled into calculating $dr/d\theta$. If someone does that, it means they didn't read the problem carefully or they don't understand the geometric meaning of the various derivatives.

14. Dec 22, 2014

### Staff: Mentor

I understand your confusion, and think they should have worded the problem better.

Chet

15. Dec 23, 2014

Thank you @Chestermiller. I appreciate everyone's help here.

16. Dec 23, 2014

@vela, this is a terrible point.

Consider the function.

$$h(a) = a^2 + 2a - 3$$

what is the slope of the tangent line at a=1 for example?

There is no dy/dx here, but rather dh/da. So your point that the derivative is dy/dx, is absolutely false.

17. Dec 23, 2014

### vela

Staff Emeritus
That's a straw man. In the context of this problem about polar coordinates, there is no ambiguity the way the problem is stated in the book. It asked for the slope of the tangent line at a point; it didn't ask for, as you initially claimed, the derivative of a polar function. Sure, if you want to completely ignore the conventions relating polar and cartesian coordinates and take some contorted interpretation of "slope of the tangent line," you can argue that the problem is unclear, but I doubt your professor will have much sympathy when you're marked wrong on the exam for calculating $dr/d\theta$ instead of $dy/dx$.

18. Dec 23, 2014

### ehild

What is the whole original text of the problem? If it asks to find the derivative of the function r(θ) = 2 cos(θ), it is dr/dθ= -2sin(θ).
If the question is "What is the slope of the tangent line drawn to the polar curve r=2 cos(θ) (θ is measured from the positive horizontal axis) at θ=pi/6 " then it is dy/dx at θ=pi/6.
A curve is not the same as a function.

19. Dec 23, 2014