Question

If x=cos θ-r sin θ,y=sin θ+r cos θ
prove that dr/dx =x/r

Answer 1

I have
\( x^2 +y^2 =1+r^2 \)

Answer 2

could find
\(\dfrac{\partial r}{\partial x}\)
and
\(\dfrac{\partial r}{\partial y}\)

but dr/dx ?

Answer 3

I think it will help to use implicit differentiation for this problem.

Answer 4

then how will i get dy/dx ?

Answer 5

2x +2y dy/dx = 2r dr/dx

Answer 6

I don't think we need to worry about dy
We're just trying to find dr/dx, so we can treat y as a constant

Answer 7

If we do that, what would we get from the implicit differentiation?

Answer 8

if we are treating y as constant , then we get
\(\partial r/\partial x\)

not dr/dx ...

Answer 9

right ?

Answer 10

I think if we use implicit differentiation, we can use dr and dx

Answer 11

show some initial steps ?

Answer 12

We just have to take the derivative of each term separately
For the first term, what is the derivative of x^2?

Answer 13

2x dx

Answer 14

Right! Then since we're treating y as a constant, the derivative of y^2 will be 0

Answer 15

Then what is the derivative of 1+r^2?

Answer 16

"we're treating y as a constant"
then we are partially differentiating!

Answer 17

2x \(\partial x = 2r \partial r\)

Answer 18

Maybe, but I don't think it will matter here since y is in its own term
Another way to think about it is instead of having dx and dr, using dx/dt and dr/dt
Then you'll also have dy/dt, but if y is constant it will be 0

Answer 19

not sure about the whole "y is constant" thing....

Answer 20

it almost seems like they just used bad notation
I think they meant to use partial notation there, what they have with dr/dx makes no sense really... what did you get for dr/dx (partial)?

Answer 21

i was just reminding myself that x depends on r and theta...
but what you have here seems off to me...
isn't x^2 + y^2 = r^2 ?