If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Differential Calculus

### Course: Differential Calculus>Unit 6

Lesson 5: Polar functions

# Polar functions derivatives

Finding derivatives of 𝑟, 𝘹, and 𝘺 of a function given in polar coordinates.

## Want to join the conversation?

• If anyone is struggling to find the intro to polar coordinates like I did, here's a link https://www.khanacademy.org/deprecated/dep-math/dep-trig/v/polar-coordinates-1 • This is what I learned on this video and just want to verify if they're correct.

1) Calculating y' in terms of theta will give you the rate of change of the y-value as theta changes,
2) Calculating x' in terms of theta will give you the rate of change of the x-value as theta changes, and
3) The rate of change of y with respect to x will give you the slope of the graph defined by the polar function at any point. • Why don't you get the same result for x' and y' if you differentiate r, and then convert the result to Cartesian coordinates? • Because at that point you aren't calculating x' and y' but rather the x and y values of r' which isn't the change that x and y go through as theta changes, but rather a cartesian representation of the change r goes through. The simplest way to get x' and y' is to follow this method Sal used as it shows clearly that first you are obtaining functions x(theta) and y(theta) and from there you are taking their change for the change in theta (i.e.: taking their derivatives).
• So, to differentiate a polar function, we first express x and y as functions of the third parameter theta, which is in fact "parametrization", and then differentiate the x- and y-components separately. Do I got this right? • I am confused why evaluating the derivative of the polar expression--r'(theta) = 2 cos(2 theta)) -- at pi/4 equals zero, while the dy/dt / dx/dt evaluation of r(theta)=sin(2theta) equals negative 1. Its seems like the dy/dt / dx/dt derivative of the initial expression should equal xy expression of the polar derivative! • The dy/dt/dx/dt evaluation is describing the change in y of the function with respect to x. The evaluation of r'(theta) is describing the change in the radius of the function, the distance from the point on the function the the origin, with respect to theta. These two evaluations describe change in the function at the same point, just using different variables.
• What does the derivative of r'(theta)= 2cos(2theta) actually mean in itself (without respect to x or y)? I thought it would refer to the slope of the r-vector, but clearly that isn't the case (or at pi/4 the slope would be zero). • i dont know how to calculate horizontal distances • In polar? If you have two polar coordinates you can simply convert them to rectangular (Cartesian) form and then subtract the 𝑥-coordinates to find the horizontal distance between the two coordinates. If you're not sure how to do this, I suggest watching the videos on the polar coordinate system on KA.
• If you were to find the magnitude of the x'(theta) and y'(theta) components, would it be equal to r'(theta)? • I tried using sin(2x)=2*sinx*cosx to turn r=sin(2θ) into x,y form. I got (x^2 + y^2)^3/2 but then I graphed to check if I did correct, and it did not have the same graph as r=sin(2θ). Where did I go wrong? Any ideas why what I did has error? • I am not sure what steps you took here, but whatever you did, you started with an equation (r=sin(2θ)) and ended up with an expression (there's no = sign in (x²+y²)^(3/2)).

If you apply the identity, we get r=2sin(θ)cos(θ). Multiplying by r² gives
r³=2(rsinθ)(rcosθ)
r³=2(y)(x)

Now from the identity r²=x²+y², we can raise both sides to the 3/2 to get r³=(x²+y²)^(3/2).

If we back-substitute this expression for r³ into our equation, we get (x²+y²)^(3/2)=2xy.

If you graph r=sin(2θ), you get a four-petaled rose. If you graph the equation above, you get exactly half of this graph, the petals in the first and third quadrants. This is because we took a square root when we raised the equation to the 3/2 power, so we are missing a ± sign.

Indeed, if we replace 2xy by -2xy, we get the other two petals of the graph instead, the second- and fourth-quadrant petals. So if we took the union of these graphs (algebraically, if we include the ± sign), we get exactly the same curve in both cases.
• 1. at , what makes the line from the origin to a point on the graph of r=sin(2θ) be r? Is it because it's defined that way?

2. sort of related to the above question. at , for r=sin(2θ), how do we know it's y=rsinθ? I know y=rsinθ works for the unit circle, but how do we know that it also works for the graph of r=sin(2θ)? thanks. 