I Understanding the given transformation - PDE

chwala
Gold Member
Messages
2,825
Reaction score
413
TL;DR Summary
I am going through the notes on pde. I want to be certain that i am getting it right. From my understanding of determinants of 3 by 3 matrices ...my approach is indicated
My interest is on how they arrived at ##r^2 \sin θ##

My approach using the third line, is as follows

##\cos θ[r^2 \cos θ \sin θ \cos^2 ∅ + r^2 \sin θ \cos θ cos^2∅ ] + r\sin θ [r\sin^2 θ \cos^2∅ + r \sin^2 θ \sin^2 ∅]=##

##\cos θ[r^2 \cos θ \sin θ[\cos^2 ∅+ \sin^2 ∅]] + r\sin θ [r\sin^2 θ [cos^2 ∅+ \sin^2 ∅]]=r^2\cos^2θ\sin^2 θ + r^2\sin^2θ\sinθ=##

##r^2\sinθ(cos^2θ+\sin^2θ)=r^2\sin θ##


If correct then we could also use second row but with negative place value i.e ##-\sin θ\sin ∅ [ a-b] + r\cos θ sin ∅[c-d]...## to realize the same.

Cheers.


1716341578933.png
 
Physics news on Phys.org
Is your question about how to evaluate a 3 x 3 determinant? If so, you can choose to go across any row or down any column. To go across the 2nd row, the determinant is ##(-a_{2, 1})## times its cofactor + ##(+a_{2, 2})## times its cofactor + ##(-a_{2, 3}##) times its cofactor. For each of these terms the cofactor is the 2 x 2 determinant that consists of the entries not in that row and not in that column.
 
chwala said:
##\cos θ[r^2 \cos θ \sin θ \cos^2 ∅ + r^2 \sin θ \cos θ cos^2∅ ] + r\sin θ [r\sin^2 θ \cos^2∅ + r \sin^2 θ \sin^2 ∅]=##
The symbol ## \phi ## is called "phi" and is entered in ## \LaTeX ## as \phi (##\phi##). The Unicode character ∅ represents the empty set and should not be used as a substitute for ## \phi ##.
 
My intention was to distinguish phi and theta. Noted though.
 
chwala said:
My intention was to distinguish phi and theta. Noted though.
Maybe, but the way to do this is to use a symbol that represents phi to represent phi, not a symbol that represents the empty set. The fact that sometimes this symbol looks a bit like some representations of phi is irrelevant.

It's like writing the equation for the circumference of a circle as 2🥧r.
 
Also, due to a quirk of the ## \LaTeX ## implementation, the Unicode character ∅ is translated to the symbol \emptyset which renders as ## \emptyset ##: if you want a character that looks like ∅ you need to use \varnothing: ## \varnothing ##.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Back
Top