Estimate using the derivative: Am I making a sign error somewhere?

[ElijahRockers]

Homework Statement

Find the derivative of the transformation x=rcos(theta), y=rsin(theta).

Then estimate the cartesian coordinates for r=2.2 and theta=pi(1/6-1/60)

The Attempt at a Solution

I found the partials with respect to r and theta for both x and y.

I also wrote down the differentials (i think):

$dx=cos\theta dr-rsin\theta d\theta$

$dy=sin\theta dr+rcos\theta d\theta$

So I calculated x(r,theta) at (2,pi/6) then added the value of dx (r,theta,dr,dtheta) at (2,pi/6,.2,-pi/60) and got approx 1.95762. I calculated the x=2.2cos(pi(1/6-1/60)) and got a value very close to that, so I'm assuming I did that part right.

Where I am getting stuck is for the y coordinate.

I used the same exact procedure, but when I add dy to y(2,pi/6) the value increases to 1.00931, instead of decreasing to approx .99878.

My gut tells me I made a sign error somewhere or something, but I can't seem to find it. What am I doing wrong?

Thanks.

 

[SammyS]

Homework Statement

Find the derivative of the transformation x=rcos(theta), y=rsin(theta).

Then estimate the cartesian coordinates for r=2.2 and theta=pi(1/6-1/60)

The Attempt at a Solution

I found the partials with respect to r and theta for both x and y.

I also wrote down the differentials (i think):

$dx=cos\theta dr-rsin\theta d\theta$

$dy=sin\theta dr+rcos\theta d\theta$

So I calculated x(r,theta) at (2,pi/6) then added the value of dx (r,theta,dr,dtheta) at (2,pi/6,.2,-pi/60) and got approx 1.95762. I calculated the x=2.2cos(pi(1/6-1/60)) and got a value very close to that, so I'm assuming I did that part right.

Where I am getting stuck is for the y coordinate.

I used the same exact procedure, but when I add dy to y(2,pi/6) the value increases to 1.00931, instead of decreasing to approx .99878.

My gut tells me I made a sign error somewhere or something, but I can't seem to find it. What am I doing wrong?

Thanks.

No sign error.

Near (r, θ) = (2, π/6):

increasing r increases x as does decreasing θ.

Increasing r also increases y. However, decreasing θ will decrease y.​

If you look at the per cent change "predicted" by the differential compared to the actual percent change in the x coordinate and make the same comparison for the y coordinate, I think you will find similar results. It's just that per cent change in y is near zero.

 

[ElijahRockers]

y(2,pi/6) = 1 and y(2.2,(pi/6-pi/60)) = .998779

when i calculate 2.2sin(pi/6-pi/60) - 2sin(pi/6) i get -.0012209006. this is Delta y.

so dy is $dy=sin\theta dr+rcos\theta d\theta$
which is .0093100318 = dy.

why is dy +ve and Delta y -ve?

I thought I understood the process, but shouldn't they both be the same sign, at least? Even if they are different in magniutde? My estimation is saying the change is in the wrong direction, that doesn't make sense to me.

Thanks in advance.

Edit: PS differentials are probably my weakest area of all. they give me more trouble than trigonometric substituion.

 

[Office_Shredder]

Don't try to take r changing from 2 to 2.2, that's too big of a jump for the derivatives to give a good look at how x and y change. Just let theta change from pi/6 to pi/6-pi/60 and let r=2.2 always

 

[ElijahRockers]

Alright... well I get a hunch that he wants us to vary 'r' too. i will have time to ask him tomorrow. Seems like a poorly designed question if you ask me.

 

[ElijahRockers]

I asked him in class, you were right. When I brought the issue up in class, he worked through it. Coming to the same conclusion I did, he accosted me with a sly grin and said "Well, that's why it's just an approximation."

Thanks for your help, a question like this shouldn't have given me this much trouble, but I had to make sure, because differentials have been a weak spot of mine for awhile.