Review of Partial Derivatives - Bruce E. Shapiro

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Polar Coordinates In polar coordinates, instead of using the distances x and y from the origin to locate a point, we use a single distance r and an orientation angle with respect to the x-axis. x=rcosθ r y=rsinθ θ The two coordinate systems are related to each other as follows: x = rcosθ r = x 2 + y 2 y = rsinθ y θ = arctan x We can also define unit vectors and r θ at any given point in space; they can be related to the cartesian unit vectors i r and r j from the following geometry. r j r θ θ θ r r i P θ Since all the vectors shown are unit vectors, we observe that: x component of ˆr is cosθ ; y component of ˆr is sinθ ; x component of ˆθ is −sinθ y component of ˆθ is cosθ Page 8
Therefore rˆ = cosθi ˆ + sinθj ˆ θˆ =− sinθi ˆ + cosθj ˆ Exercise: Find expressions for î and ĵ in terms of ˆr and ˆθ . Solution. Multiply the first expression by sinθ and the second expression by cosθ ˆsin sin cos ˆ 2 r θ = θ θi + sin θjˆ θˆ cos θ cos θ sin θiˆ 2 =− + cos θjˆ Adding the two expressions, ˆsin ˆ cos sin 2 ˆ 2 r θ + θ θ = θj + cos θjˆ = jˆ which gives an expression for ĵ in terms of ˆr and ˆθ . To get a similar expressionf or î , multiply the expression for ˆr by cosθ and the expression for ˆθ by sinθ , 2 rˆ cosθ = cos θiˆ + cosθsinθj ˆ ˆsin sin 2 θ θ =− θiˆ + cosθsinθj ˆ Subtract the bottom expression from the top, ˆ cos ˆsin cos 2 ˆ 2 r θ − θ θ = θi + sin θiˆ = iˆ Summarizing the conversion expressions, we have r r i = rˆ cosθ −θˆsin θ r = cosθi + sinθj and r r r jˆ = rˆsinθ + θˆ cos θ θ =− sinθi + cosθj Example. Find an expression for the vector field F r = 3xiˆ+ 4x 2 yjˆ in polar coordinates. Solution. r F = 3xiˆ 2 + 4x yjˆ = 3( rcos θ)(ˆ rcosθ − θˆsin θ ) + 4 ( r cos θ ) ( r sin θ )(ˆ r cos θ + θˆsin θ ) 2 3 3 3 2 2 = 3rcos θrˆ − 3rcosθsinθθ ˆ + 4r cos θsinθrˆ + 4r cos θsin θθˆ 2 3 3 3 2 2 = ( 3rcos θ + 4r cos θsin θ)ˆ r + ( 4r cos θsin θ − 3rcosθsin θ) θ ˆ 2 2 2 = rcos θ( 3+ 4r cosθsin θ)ˆ r + rcosθsin θ( 4r cosθsin θ −3) θ ˆ Exercise: Recall that the gradient in two dimensions in cartesian coordinates is ∇ = ∂ f f x y i ∂ + ∂ f ( , ) ˆ jˆ . Find an expression for the gradient in polar coordinates, i .e., find x ∂y the functions g and h so that ∇ f(, r θ ) = rg ˆ (, r θ ) + θ ˆ h (, r θ ). Hint: you must use the chain rule and the expressions for transformation of coordinates. 2 Page 9
Page 1 and 2: Review of Partial Derivatives Defin
Page 3 and 4: Chain Rule If zt () = f( xt (), yt
Page 5 and 6: We can also calculate the cross pro
Page 7: The Curl Operator is defined as “

Review of Partial Derivatives - Bruce E. Shapiro

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