What does "with respect to.." actually mean in differentiation?

What does "with respect to d-thetha" actually mean?

I'm trying to integrating 1/2r^2 dtheta and the dtheta is putting me off :P

Just like if you were to differentiate e^udy/dx it would be e^udy/dx + e^ud^2y/dx^2 dx/du I understand the use of the product rule but I'm confused by why they wrote dx/du

What does "with respect to d-thetha" actually mean?

I'm trying to integrating 1/2r^2 dtheta and the dtheta is putting me off :P

Just like if you were to differentiate e^udy/dx it would be e^udy/dx + e^ud^2y/dx^2 dx/du I understand the use of the product rule but I'm confused by why they wrote dx/du

How can a $d\theta$ "put you off"? Do you get "put off" by a dx when you're asked to integrate f(x) with respect to x?

You need to write r as a function of theta and then integrate 'normally' i.e. exactly the same as if r were a function of x and the dtheta just said dx

is it polar coordinates that you are doing?

you jsut write r as f(theta) and integrate

you cannot integrate r d theta the same way as you cannot integrate f(x) dy

The 'with respect to' tells you which letter is the variable - all other letters can be considered as constants.

So r is a constant in this case?

dx doesn't put me off because we did it so many times that it became the norm but even then I didn't really understand why the dx was there!

Just like I don't really know what you mean when you say "r as a function of theta"

So r is a constant in this case?

Unless you post the question I can't tell - it could be a function of theta.

Come on, this is GCSE stuff now! "r is a function of theta" means that theta is the independent variable and r is the dependent variable, i.e. r can be written in terms of theta (or "as a function of theta", as we usually say).

No (not usually!) - r should be given to you as a function of theta! Can you upload a picture of the actual question please?

Oh I see! So theta is basically "y" and r as a function of theta can be shown from theta = r^2 + 5 etc etc


Find the integral of 1/2r^2dtheta

Well in your terms, theta is more like "x" and r is more like "y", but it depends on the question!


Are you missing something out here??? That does not look like a complete question to me!

Apologies! There's also "given r=3sin2theta"

Apologies! There's also "given r=3sin2theta"

What would you do if the question said

Evaluate $\displaystyle \int \frac{1}{2}y^2 \, \mathrm{d}x$ given that $y = 3 \sin 2x$?

Can't remember it exactly but I would differentiate 9sin^2x to give 56sin2xcos2x which is also 14sin4x

But you're integrating $y^2$ not differentiating it.