Find the derivative (Implicit)

In summary, you would 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.
  • #1
hoch449
13
0

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!
 
Physics news on Phys.org
  • #2
Do you know how to do implicit differentiation? Do you know how find the partial derivative of a function with respect to a variable?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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...
 

1. What is the purpose of finding the derivative implicitly?

The purpose of finding the derivative implicitly is to determine the rate of change of a function when it cannot be easily expressed in terms of a single variable. This is often the case for equations that involve multiple variables or have a complex structure.

2. How is the derivative found implicitly?

The derivative is found implicitly by using the chain rule and implicit differentiation. This involves treating the dependent variable as a function of the independent variable and taking the derivative of both sides of the equation with respect to the independent variable.

3. What are the steps for finding the derivative implicitly?

The steps for finding the derivative implicitly are:

  1. Identify the dependent variable and the independent variable in the equation.
  2. Use the chain rule to express the dependent variable as a function of the independent variable.
  3. Take the derivative of both sides of the equation with respect to the independent variable.
  4. Solve for the derivative of the dependent variable.

4. Can the derivative be found implicitly for any equation?

Yes, the derivative can be found implicitly for any equation, as long as the equation is differentiable. This means that the equation must have a well-defined tangent line at every point.

5. How is the derivative found implicitly for trigonometric functions?

The derivative can be found implicitly for trigonometric functions by using the chain rule and the identities for differentiating trigonometric functions. For example, the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x).

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
563
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
958
  • Calculus and Beyond Homework Help
Replies
8
Views
876
  • Calculus and Beyond Homework Help
Replies
3
Views
883
  • Calculus and Beyond Homework Help
Replies
3
Views
772
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
959
  • Calculus and Beyond Homework Help
Replies
6
Views
760
Back
Top