Podcast
Questions and Answers
Given $r^2 = x^2 + y^2 + z^2$, which of the following expressions correctly represents $\frac{\partial r}{\partial x}$?
Given $r^2 = x^2 + y^2 + z^2$, which of the following expressions correctly represents $\frac{\partial r}{\partial x}$?
- $\frac{1}{2r}$
- $\frac{x}{r}$ (correct)
- $\frac{r}{x}$
- $x^2 + y^2 + z^2$
If $\vec{r} = xi + yj + zk$, what is the divergence of $\vec{r}$?
If $\vec{r} = xi + yj + zk$, what is the divergence of $\vec{r}$?
- 2
- 1
- 0
- 3 (correct)
Given $\nabla^2 r^n = n(n + 1)r^{n-2}$, for what value of 'n' does $\nabla^2 r^n = 0$?
Given $\nabla^2 r^n = n(n + 1)r^{n-2}$, for what value of 'n' does $\nabla^2 r^n = 0$?
- 0
- 2
- -1 (correct)
- 1
Given $\nabla^2 r^n = n(n + 1)r^{n-2}$, what is the value of $\nabla^2 r^2$?
Given $\nabla^2 r^n = n(n + 1)r^{n-2}$, what is the value of $\nabla^2 r^2$?
If $x = r \cos \theta$ and $y = r \sin \theta$, which integral expression represents the area element in polar coordinates?
If $x = r \cos \theta$ and $y = r \sin \theta$, which integral expression represents the area element in polar coordinates?
Which vector operation is equivalent to $\nabla^2 r^n$?
Which vector operation is equivalent to $\nabla^2 r^n$?
What is the correct expansion of $div(n r^{n-2} \vec{r})$?
What is the correct expansion of $div(n r^{n-2} \vec{r})$?
If an integral in polar coordinates is given by $\int_0^{2\pi} \int_0^c 3hr dr d\theta$, where 'h' and 'c' are constants, what does 'c' represent?
If an integral in polar coordinates is given by $\int_0^{2\pi} \int_0^c 3hr dr d\theta$, where 'h' and 'c' are constants, what does 'c' represent?
Given $x = r \cos \theta$ and $y = r \sin \theta$, and the integral $\int \int 3h r d\theta dr$ is evaluated from $0$ to $2\pi$ for $\theta$ and $0$ to $c$ for $r$, what shape does this integral calculate the area of?
Given $x = r \cos \theta$ and $y = r \sin \theta$, and the integral $\int \int 3h r d\theta dr$ is evaluated from $0$ to $2\pi$ for $\theta$ and $0$ to $c$ for $r$, what shape does this integral calculate the area of?
Using the formula $\nabla^2 r^n = n(n + 1)r^{n-2}$, which expression correctly demonstrates that $\nabla^2 (\frac{1}{r}) = 0$?
Using the formula $\nabla^2 r^n = n(n + 1)r^{n-2}$, which expression correctly demonstrates that $\nabla^2 (\frac{1}{r}) = 0$?
Flashcards
Polar Coordinates
Polar Coordinates
In polar coordinates, x = r cos θ and y = r sin θ are the relationships between Cartesian and polar coordinates.
Laplacian of r
to the power n
Laplacian of r to the power n
∇²rⁿ = n(n + 1)rⁿ⁻²
Position vector
Position vector
r = xi + yj + zk, represents the position vector in Cartesian coordinates.
Magnitude squared
Magnitude squared
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Gradient of rⁿ
Gradient of rⁿ
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Divergence of nrⁿ⁻²r
Divergence of nrⁿ⁻²r
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Divergence of r
Divergence of r
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Laplacian of 1/r
Laplacian of 1/r
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Study Notes
- Given x = r cos θ and y = r sin θ, the double integral ∫∫3h r dr dθ over a region is converted to iterated integrals
- The integral is evaluated step-by-step: first with respect to r (from 0 to c) and then with respect to θ (from 0 to 2π)
- The final result of the double integral is 3πhc²
Proving the Laplacian of r^n
- Goal: Prove that ∇²rⁿ = n(n+1)rⁿ⁻² where r = xi + yj + zk, and show that ∇²(1/r) = 0
- r² = x² + y² + z² so that δr/δx = x/r, δr/δy = y/r, δr/δz = z/r
- grad rⁿ = nrⁿ⁻¹( r/r) = nrⁿ⁻² r
- ∇²rⁿ = div(grad rⁿ) = div(nrⁿ⁻² r)
- div(nrⁿ⁻² r) is expanded using the product rule for divergence: nrⁿ⁻² div r + grad(nrⁿ⁻²) ⋅ r
- Since div r = 3, the above expression simplifies to 3nrⁿ⁻² + n(n-2)rⁿ⁻³ (r /r ⋅ r)
- The dot product (r ⋅ r) is equal to r², further simplifying the expression to 3nrⁿ⁻² + n(n-2)rⁿ⁻⁴(r²) = 3nrⁿ⁻² + n(n-2)rⁿ⁻²
- Combining terms results in: ∇²(rⁿ) = n(n+1)rⁿ⁻²
- Setting n = -1 leads to: ∇²(1/r) = 0
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