Wrong answer on Differentiation

In summary, when calculating the partial derivative of r' with respect to y', it is incorrect to simply solve for r' and take the derivative of y'. This is because r' is a function of both r and \theta, which are both functions of time. Instead, the partial derivative must be calculated using the chain rule, taking into account the derivatives of both r and \theta with respect to y'. This can be seen when expanding the partial derivative of L with respect to y', where L is a function of r', \theta', r, and \theta.
  • #1
roeb
107
1

Homework Statement



In polar coords: y = rsin([tex]\theta[/tex])
y' = dy/dt = r'sin([tex]\theta[/tex]) + rcos([tex]\theta[/tex]) [tex]\theta[/tex]'

If I want [tex]\partial r' / \partial y' [/tex] can I just solve for r' and take the derivative of y'?

so

[tex]\partial r' / \partial y' = 1 / sin \theta[/tex]

I am hoping this is incorrect because I am getting the wrong answer on another part of a problem...

Homework Equations





The Attempt at a Solution

 
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  • #2


Starting with y = r sin([itex]\theta[/itex]),
[itex]\partial y/\partial r = \partial (r sin(\theta)) = sin(\theta)[/itex]

So [itex]\partial r/\partial y = 1/sin(\theta)[/itex]

Why do you have the primes on your partial derivative? I'm pretty sure that's an oversight on your part.

If you're trying to calculate dy/dt, and y is a function of both r and [itex]\theta[/itex], and r and [itex]\theta[/itex] are functions of t alone, then
dy/dt = [itex]\partial y/\partial r * dr/dt + \partial y/\partial \theta * d \theta/dt [/itex]
 
  • #3


Thanks for your reply,

The primes actually are indicating time derivatives, let me expand upon my question

I'm actually trying to do the following

[tex]\frac{\partial L}{\partial y'} = \frac{\partial L}{\partial r} \frac{\partial r}{\partial y'} + \frac{\partial L}{\partial \theta} \frac{\partial \theta}{ \partial y'}+ \frac{\partial L }{\partial r'} \frac{\partial r'}{\partial y'} + \frac{\partial L}{\partial \theta '} \frac{\partial \theta '}{\partial \y '}[/tex]

I am having a hard time find the various partial derivs with respect to y' (or maybe I'm not).

L is actually L = T-U where T = KE = 1/2 m (x'^2 + y'^2) and U doesn't matter in this case
so in other words, y' and x' are the velocities in y and x.

where of course
x = rcos(theta)
y = rsin(theta)
 
Last edited:

Related to Wrong answer on Differentiation

1. What is differentiation?

Differentiation is a mathematical process used to find the rate of change of a function with respect to its independent variable. It is often used in calculus to find the slope of a curve at a specific point.

2. What is a wrong answer on differentiation?

A wrong answer on differentiation could be any solution or calculation that does not accurately reflect the rate of change of the function. This could be due to a mistake in the application of differentiation rules or incorrect input values.

3. How do you know if your answer on differentiation is wrong?

If you are using a calculator or computer program to solve for the derivative, you can compare your answer to a known correct solution or use a graphing tool to visualize the function and its derivative. If you are solving by hand, you can double-check your work by reapplying the differentiation rules or using a graph to check for consistency.

4. What are some common mistakes made when solving for differentiation?

Some common mistakes when solving for differentiation include incorrectly applying the rules of differentiation, using incorrect input values, forgetting to take into account the chain rule, and making calculation errors such as forgetting to distribute or dropping negative signs.

5. How do you avoid getting a wrong answer on differentiation?

To avoid getting a wrong answer on differentiation, it is important to carefully follow the rules and steps of differentiation, double-check your work, and use a graph or calculator to verify your solution. It is also helpful to practice regularly to become familiar with the process and common mistakes to watch out for.

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