Partial differentiation problem, multiple variables (chain rule?)

[bawbag]

Homework Statement

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

\left(\frac{\partial z}{\partial \theta}\right)_{x}

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 \left(\frac{\partial z}{\partial \theta}\right)_{x} = 4r²sin θ cos θ

However the solutions in the book give

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

What am I missing here?

Thanks in advance.

 

[pasmith]

Homework Statement

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

\left(\frac{\partial z}{\partial \theta}\right)_{x}

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 \left(\frac{\partial z}{\partial \theta}\right)_{x} = 4r²sin θ cos θ

However the solutions in the book give

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

What am I missing here?

Thanks in advance.

You need to differentiate y^2 = r^2 \sin^2 \theta using the product rule: r is not independent of \theta, since <br /> r = \frac{x}{\cos \theta}

 

[bawbag]

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

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

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

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

giving -4r \frac{x sin \theta}{cos^{2} \theta} sin^{2} \theta + 4r^{2} sin \theta cos \theta

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

 

[Ray Vickson]

Homework Statement

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

\left(\frac{\partial z}{\partial \theta}\right)_{x}

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 \left(\frac{\partial z}{\partial \theta}\right)_{x} = 4r²sin θ cos θ

However the solutions in the book give

\left(\frac{\partial z}{\partial \theta}\right)_{x} = 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:
dz = 2 x\, dx + 4 y \,dy\\<br /> dx = \cos(\theta) \, dr - r \sin(\theta)\, d \theta\\<br /> dy = \sin(\theta)\, dr + r \cos(\theta)\, d \theta
But $dx = 0 \Longrightarrow dr = r \tan(\theta) \, d \theta$, so
dy = r \sin(\theta)\tan(\theta)\, d \theta + r \cos(\theta) \,d \theta<br /> = r \left( \frac{\sin^2(\theta)}{\cos(\theta)} + \cos(\theta)\right)\, d \theta<br /> = \frac{r}{\cos(\theta)}\, d \theta
Thus
dz = 4 y dy = 4 r \sin(\theta) (r/\cos(\theta)) \,d \theta = 4 r^2 \tan(\theta) \, d \theta
The partial $(\partial z/\partial \theta)_{x}$ is the coefficient of $d \theta$ in the above.

 

[pasmith]

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

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

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

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

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

I can replace x with r / cos θ

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

 

[jaytech]

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

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

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

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

giving -4r \frac{x sin \theta}{cos^{2} \theta} sin^{2} \theta + 4r^{2} sin \theta cos \theta

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\frac{x^2}{cos^2θ}sin²θ = x²(1 + 2tan²θ)

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

Hint* Remember, that after you perform the derivation to look for anywhere you can make a substitution to remove 'x'.

 

[bawbag]

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



You can't, but you can replace x with r \cos \theta 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 4r^{2} tan \theta sin^{2} \theta + 4r^{2} sin \theta cos \theta to 4r^{2} tan \theta without ending up with a huge mess.

Sorry for being dense :P

 

[bawbag]

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!

 

[jaytech]

You should try the steps I previously mentioned. Then reflect on why it works..

 

[pasmith]

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 4r^{2} tan \theta sin^{2} \theta + 4r^{2} sin \theta cos \theta to 4r^{2} tan \theta without ending up with a huge mess.

Sorry for being dense :P

\sin \theta \cos \theta = \tan\theta \cos^2 \theta

 

[bawbag]

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? :)