Partial differentiation problem, multiple variables (chain rule?)

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Homework Statement



if z = x2 + 2y2 , x = r cos θ , y = r sin θ , find the partial derivative

[itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex]

Homework Equations



z = x2 + 2y2
x = r cos θ
y = r sin θ

The Attempt at a Solution



The textbook says that the equation should be re-written to include only the variables θ and x, and then differentiated with respect to θ.

Substituting y = r sin θ :

z = x2 + 2r2 sin2 θ

then [itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = 4r2sin θ cos θ

However the solutions in the book give

[itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = 4r2 tan θ

What am I missing here?

Thanks in advance.
 

Answers and Replies

  • #2
pasmith
Homework Helper
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Homework Statement



if z = x2 + 2y2 , x = r cos θ , y = r sin θ , find the partial derivative

[itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex]

Homework Equations



z = x2 + 2y2
x = r cos θ
y = r sin θ

The Attempt at a Solution



The textbook says that the equation should be re-written to include only the variables θ and x, and then differentiated with respect to θ.

Substituting y = r sin θ :

z = x2 + 2r2 sin2 θ

then [itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = 4r2sin θ cos θ

However the solutions in the book give

[itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = 4r2 tan θ

What am I missing here?

Thanks in advance.

You need to differentiate [itex]y^2 = r^2 \sin^2 \theta[/itex] using the product rule: [itex]r[/itex] is not independent of [itex]\theta[/itex], since [tex]
r = \frac{x}{\cos \theta}[/tex]
 
Last edited:
  • #3
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0
Thanks! I used the product rule to differentiate, but I still think I'm missing something. My working is as follows:

z = x2 + 2r2 sin2 θ

[itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = [itex]\frac{\partial}{\partial \theta}2r^{2}sin^{2}\theta[/itex]

[itex]= 4r\frac{\partial r}{\partial \theta}sin^{2}\theta + 2r^{2}2sin\theta cos\theta[/itex] (chain rule for dr/dθ)

since [itex] r = \frac{x}{cos \theta}[/itex], [itex]\frac{\partial r}{\partial \theta} = \frac{-x sin \theta}{cos^{2} \theta}[/itex]

giving [itex] -4r \frac{x sin \theta}{cos^{2} \theta} sin^{2} \theta + 4r^{2} sin \theta cos \theta [/itex]

I can replace x with r / cos θ, but I don't see how it reduces to the given solution of 4r2 tan θ.

Thanks again
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
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Homework Statement



if z = x2 + 2y2 , x = r cos θ , y = r sin θ , find the partial derivative

[itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex]

Homework Equations



z = x2 + 2y2
x = r cos θ
y = r sin θ

The Attempt at a Solution



The textbook says that the equation should be re-written to include only the variables θ and x, and then differentiated with respect to θ.

Substituting y = r sin θ :

z = x2 + 2r2 sin2 θ

then [itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = 4r2sin θ cos θ

However the solutions in the book give

[itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = 4r2 tan θ

What am I missing here?

Thanks in advance.

You are missing the fact that ##x## is held constant. One way to do it is:
[tex] dz = 2 x\, dx + 4 y \,dy\\
dx = \cos(\theta) \, dr - r \sin(\theta)\, d \theta\\
dy = \sin(\theta)\, dr + r \cos(\theta)\, d \theta[/tex]
But ##dx = 0 \Longrightarrow dr = r \tan(\theta) \, d \theta##, so
[tex] dy = r \sin(\theta)\tan(\theta)\, d \theta + r \cos(\theta) \,d \theta
= r \left( \frac{\sin^2(\theta)}{\cos(\theta)} + \cos(\theta)\right)\, d \theta
= \frac{r}{\cos(\theta)}\, d \theta [/tex]
Thus
[tex] dz = 4 y dy = 4 r \sin(\theta) (r/\cos(\theta)) \,d \theta = 4 r^2 \tan(\theta) \, d \theta [/tex]
The partial ##(\partial z/\partial \theta)_{x} ## is the coefficient of ##d \theta## in the above.
 
  • #5
pasmith
Homework Helper
2,075
696
Thanks! I used the product rule to differentiate, but I still think I'm missing something. My working is as follows:

z = x2 + 2r2 sin2 θ

[itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = [itex]\frac{\partial}{\partial \theta}2r^{2}sin^{2}\theta[/itex]

[itex]= 4r\frac{\partial r}{\partial \theta}sin^{2}\theta + 2r^{2}2sin\theta cos\theta[/itex] (chain rule for dr/dθ)

since [itex] r = \frac{x}{cos \theta}[/itex], [itex]\frac{\partial r}{\partial \theta} = \frac{-x sin \theta}{cos^{2} \theta}[/itex]

The derivative of [itex]u^{-1}[/itex] with respect to [itex]u[/itex] is [itex]-u^{-2}[/itex]. The derivative of [itex]\cos \theta[/itex] with respect to [itex]\theta[/itex] is [itex]-\sin \theta[/itex]. The two minus signs cancel.

I can replace x with r / cos θ

You can't, but you can replace [itex]x[/itex] with [itex]r \cos \theta[/itex] and do some trigonometric simplifications; the first step is to express everything in terms of sines and cosines.
 
  • #6
64
6
Thanks! I used the product rule to differentiate, but I still think I'm missing something. My working is as follows:

z = x2 + 2r2 sin2 θ

[itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = [itex]\frac{\partial}{\partial \theta}2r^{2}sin^{2}\theta[/itex]

[itex]= 4r\frac{\partial r}{\partial \theta}sin^{2}\theta + 2r^{2}2sin\theta cos\theta[/itex] (chain rule for dr/dθ)

since [itex] r = \frac{x}{cos \theta}[/itex], [itex]\frac{\partial r}{\partial \theta} = \frac{-x sin \theta}{cos^{2} \theta}[/itex]

giving [itex] -4r \frac{x sin \theta}{cos^{2} \theta} sin^{2} \theta + 4r^{2} sin \theta cos \theta [/itex]

I can replace x with r / cos θ, but I don't see how it reduces to the given solution of 4r2 tan θ.

Thanks again

Try saying z = x2 + 2[itex]\frac{x^2}{cos^2θ}[/itex]sin2θ = x2(1 + 2tan2θ)

Now with this for z you can perform [itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] quite easily.

Hint* Remember, that after you perform the derivation to look for anywhere you can make a substitution to remove 'x'.
 
  • #7
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The derivative of [itex]u^{-1}[/itex] with respect to [itex]u[/itex] is [itex]-u^{-2}[/itex]. The derivative of [itex]\cos \theta[/itex] with respect to [itex]\theta[/itex] is [itex]-\sin \theta[/itex]. The two minus signs cancel.



You can't, but you can replace [itex]x[/itex] with [itex]r \cos \theta[/itex] and do some trigonometric simplifications; the first step is to express everything in terms of sines and cosines.

That was a typo, my bad. So after sorting the minus sign, I'm left with essentially what I had before, but I can't see anyway of reducing [itex] 4r^{2} tan \theta sin^{2} \theta + 4r^{2} sin \theta cos \theta[/itex] to [itex] 4r^{2} tan \theta[/itex] without ending up with a huge mess.

Sorry for being dense :P
 
  • #8
13
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Figured it out. Thanks guys. Turns out I laid it out the way jaytech said, but didn't use the product rule properly so I abandoned that method and tried it another way, which lead to that whole mess. Whoops!

Thanks to everyone who helped!
 
  • #9
64
6
You should try the steps I previously mentioned. Then reflect on why it works..
 
  • #10
pasmith
Homework Helper
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That was a typo, my bad. So after sorting the minus sign, I'm left with essentially what I had before, but I can't see anyway of reducing [itex] 4r^{2} tan \theta sin^{2} \theta + 4r^{2} sin \theta cos \theta[/itex] to [itex] 4r^{2} tan \theta[/itex] without ending up with a huge mess.

Sorry for being dense :P

[tex]\sin \theta \cos \theta = \tan\theta \cos^2 \theta[/tex]
 
  • #11
13
0
I did use the steps you suggested, jaytech. As for why it works, I imagine that I should be able to reach the solution from any starting point, with proper application of the chain rule/product rule, but arranging it as you suggested means I can skip over a lengthy simplification process after the operation. Any other insight you care to offer? :)
 

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