Help in evaluating the chain rule?

  • Thread starter hivesaeed4
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  • #1
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$${x = r \cos \theta}$$,
$${y = r \sin \theta}$$, $${r^2 = x^2 + y^2}$$ and $${\theta = \tmop{ \arctan} (y / x)}$$ (with some caveats for the last formula).

Suppose $${u = u (x, y)}$$.

Show that

$${\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta}$$

$${\frac{\partial u}{\partial \theta} = - \frac{\partial u}{\partial x} r \sin \theta + \frac{\partial u}{\partial y} r \cos \theta}$$


Now suppose that $${u = u (r, \theta)}$$. Show that

$${\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \cos \theta - \frac{\partial u}{\partial \theta} \frac{\sin \theta}{r}}$$

$${\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \sin \theta + \frac{\partial u}{\partial \theta} \frac{\cos \theta}{r}}$$


Now I've proved the italic part but am stuck on the bold parts basically how are they calculated. I've tried the same method as used in the italic part ut am getting different answers.
For e.g in the last bold part instead of cos(theta)/r I got (cos(theta))^2.

Help?
 

Answers and Replies

  • #2
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$${x = r \cos \theta}$$,
$${y = r \sin \theta}$$, $${r^2 = x^2 + y^2}$$ and $${\theta = \tmop{ \arctan} (y / x)}$$

Okay, I'll show the first one, do you know the chain rule?
(∂u/∂r) = (∂u/∂x)(∂x/∂r)+(∂u/∂y)(∂y/∂r)
(∂u/∂θ) = (∂u/∂x)(∂x/∂θ)+(∂u/∂y)(∂y/∂θ)
Okay, if you agree with that, then do you agree that:

(∂x/∂r) = cosθ and
(∂y/∂r) = sinθ and
(∂x/∂θ) = (apply product rule) = (∂r/∂θ)*(cosθ) + (r)(∂cosθ/∂θ) = -r*sin(θ)
because (∂r/∂θ)=0 and (∂cosθ/∂θ) = -sin(θ) (where here I applied the chain rule, "outside is cos() and "inside" is theta. Derivative of outside is sine and derivative of theta with respect to theta is 1.
(∂y/∂θ) = r*cos(θ), for similar reasons.

If you have any questions about this, let me know. Also, are you concerned about the proof of the chain rule because there are different levels of the proof depending on the level of mathematics you're at.... let me guess, calculus 2 final next week?
 
  • #3
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I don't know how you did it, but your guess was spot on (regarding the cal-II final). Anyway I do need the proof let's say at a level expected for any cal-II student.

Now as far as the
(∂x/∂r) = cosθ and
(∂y/∂r) = sinθ and
(∂x/∂θ) = (apply product rule) = (∂r/∂θ)*(cosθ) + (r)(∂cosθ/∂θ) = -r*sin(θ)
because (∂r/∂θ)=0 and (∂cosθ/∂θ) = -sin(θ) (where here I applied the chain rule, "outside is cos() and "inside" is theta. Derivative of outside is sine and derivative of theta with respect to theta is 1.
(∂y/∂θ) = r*cos(θ), for similar reasons.


I apologize cause I stated the wrong part of the question. My bad.

The question is :

Now suppose that $${u = u (r, \theta)}$$. Show that

$${\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \cos \theta - \frac{\partial u}{\partial \theta} \frac{\sin \theta}{r}}$$

$${\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \sin \theta + \frac{\partial u}{\partial \theta} \frac{\cos \theta}{r}}$$


How do I prove these?
 

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