If V=x y /(x^2+y^2)^2 and x=r cosθ, y=r sinθ, show that (∂^2 V)/(∂r^2)+(1)/(r) (∂V)/(∂r)+(1)/(r^

Name: Problem 5, Chapter 23: Engineering Mathematics
Uploaded: 2022-01-14T00:27:17-08:00
Duration: 1 min 6 s
Channel: Carson Merrill
Description: If V=x y /(x^2+y^2)^2 and x=r cosθ, y=r sinθ, show that (∂^2 V)/(∂r^2)+(1)/(r) (∂V)/(∂r)+(1)/(r^2) (∂^2 V)/(∂0^2)=0

step 1 For the following problem, we want to find the partial derivative of v with respect to R and the

Question

If $V=x y /\left(x^{2}+y^{2}\right)^{2}$ and $x=r \cos \theta, y=r \sin \theta$, show that $\frac{\partial^{2} V}{\partial r^{2}}+\frac{1}{r} \frac{\partial V}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} V}{\partial 0^{2}}=0$

    If $V=x y /\left(x^{2}+y^{2}\right)^{2}$ and $x=r \cos \theta, y=r \sin \theta$, show that $\frac{\partial^{2} V}{\partial r^{2}}+\frac{1}{r} \frac{\partial V}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} V}{\partial 0^{2}}=0$

Engineering Mathematics

K. A. Stroud, Dexter… 5th Edition

Chapter 23, Problem 5

Instant Answer

Solved by Expert Carson Merrill

01/14/2022

Step 1

We know that $x = r\cos\theta$ and $y = r\sin\theta$. Substituting these into $V$, we get: \[V = r\cos\theta \cdot r\sin\theta / (r\cos\theta)^{2}+(r\sin\theta)^{2})^{2} = r^{2}\cos\theta\sin\theta / r^{4} = \frac{\sin\theta\cos\theta}{r^{2}}.\] Show more…

Show all steps

Thanks for your feedback!

If $V=x y /\left(x^{2}+y^{2}\right)^{2}$ and $x=r \cos \theta, y=r \sin \theta$, show that $\frac{\partial^{2} V}{\partial r^{2}}+\frac{1}{r} \frac{\partial V}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} V}{\partial 0^{2}}=0$

Play audio

Feedback

Carson Merrill and 63 other educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Polar Coordinate Transformation

This concept involves rewriting expressions and operations defined in Cartesian coordinates into polar coordinates, where x = r cos? and y = r sin?. It is useful for problems exhibiting radial symmetry and for simplifying the expressions of functions and differential operators.

Laplacian Operator in Polar Coordinates

The Laplacian operator, central to potential theory and many areas of physics, takes a specific form in polar coordinates: ?²/?r² + (1/r)?/?r + (1/r²)?²/??². Understanding this formulation is essential for converting and solving partial differential equations expressed in Cartesian coordinates.

Harmonic Functions

Harmonic functions satisfy Laplace's equation, meaning their Laplacian equals zero. They frequently appear in contexts such as electrostatics, fluid mechanics, and heat conduction, representing equilibrium states and potential fields in these systems.

Please give Ace some feedback