Find the derivative (Implicit)

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Homework Help Overview

The discussion revolves around finding the partial derivative \(\frac{\partial\theta}{\partial y}\) using implicit differentiation. The context involves relationships between the variables \(x\), \(y\), \(z\), \(r\), and \(\theta\) in a three-dimensional space defined by spherical coordinates.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of implicit differentiation and the challenges faced when differentiating the equation \(\cos\theta = \frac{z}{r}\) with respect to \(y\). There are inquiries about the correct application of differentiation rules and the handling of implicit functions.

Discussion Status

Some participants have provided insights into their previous attempts, such as finding \(\frac{\partial r}{\partial y}\), which seems to have been resolved correctly. However, there is ongoing uncertainty regarding the differentiation of \(\cos\theta\) and the right-hand side of the equation, with participants expressing confusion about specific steps in the process.

Contextual Notes

Participants are working under the constraints of implicit differentiation and are encouraged to clarify their understanding of the differentiation process. There is a mention of needing to keep certain variables constant while differentiating, which may be contributing to the confusion.

hoch449
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Homework Statement



Find [tex]\frac{\partial\theta}{\partial y}[/tex]

[tex]z=rcos\theta[/tex]
[tex]x=rsin\theta\cos\phi[/tex]
[tex]y=rsin\theta\sin\phi[/tex]
[tex]r^2=x^2 + y^2 + z^2[/tex]


The Attempt at a Solution



We know [tex]cos\theta=\frac{z}{r}=\frac{z}{\sqrt{x^2 + y^2 + z^2}}[/tex]

So implicit differentiation says to differentiate both sides with respect to y and this is where I begin to run into trouble.

Please be very specific when you try to explain how this is done lol.

Thanks!
 
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Do you know how to do implicit differentiation? Do you know how find the partial derivative of a function with respect to a variable?
 
An earlier part to this question was to find [tex]\frac{\partial r}{\partial y}[/tex] and I solved it correctly.

Here is how I did it.

[tex]r^2= x^2 + y^2 + z^2[/tex]

[tex]\frac{d}{dy}r^2= \frac{d}{dy}y^2[/tex]

[tex]\frac{d}{dr}r^2\frac{dr}{dy}=2y[/tex]

[tex]2r\frac{dr}{dy}=2y[/tex]

so therefore [tex]\frac{\partial r}{\partial y}= \frac{y}{r}[/tex]

I am just having some difficulty with the next part of the question.
 
hoch449 said:

The Attempt at a Solution



We know [tex]cos\theta=\frac{z}{r}=\frac{z}{\sqrt{x^2 + y^2 + z^2}}[/tex]

So implicit differentiation says to differentiate both sides with respect to y and this is where I begin to run into trouble.

In more specific terms, differentiate both sides of the equation with respect to y keeping all variables other than y and those that are explicit functions of y constant.
Where specifically did you run into trouble?
 
The Right Hand Side of the equation gives me the difficulty. I am sure I am making an elementary mistake.

[tex]\frac{\partial\theta}{\partial y}:[/tex]

[tex]\frac{d}{dy}cos\theta=\frac{d}{dy}(\frac{z}{r})[/tex]

[tex]\frac{d}{d\theta}(cos\theta)\frac{d\theta}{dy}=\frac{d}{dr}r^-1\frac{dr}{dy}[/tex]

[tex]-sin\theta\frac{d\theta}{dy}=-r^-2\frac{dr}{dy}[/tex]

I have a feeling that I have already made a mistake with the right side...
 

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