(1 point) Introduction In this problem we will derive the Laplacian in polar coordinates using the identities (i) <= r c

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1 Point Introduction In This Problem We Will Derive The Laplacian In Polar Coordinates Using The Identities I R C 1 (65.88 KiB) Viewed 25 times

(1 point) Introduction In this problem we will derive the Laplacian in polar coordinates using the identities (i) <= r cos(6), (ii) y=r sin(0), (iii) x2 + y2 = r2, (iv) tan(0) = - 1 At the end of the problem you will find a complete list of the notations that you will need to enter and how to enter them. де де ә0 0 Part 1 - Find Әх ду деду де де де an dy To find and we will implicitly differentiate the above equations. Unlike calc 1, x and y are both independent variables, so that ax'ay' ax ду dx Implicitly differentiating both sides of (iii) with respect to x, without simplifying, gives us 0. 0 = 0 or Solving for we get ar д ac 0 Similarly, implicitly differentiating both sides of (iii) with respect to y, without simplifying, gives us 0 0 or Solving for we get ду д = 0 ae aa Now we need to find and a. ду . which take a bit more work. Implicitly differentiating both sides of (iv) with respect to x gives 0 0
Әв Әв Now we need to find and де ду which take a bit more work. Implicitly differentiating both sides of (iv) with respect to x gives 7 0 0 0 де Substitute (i) and (ii) into the right hand side of this equation for x and y respectively, and solve for да Әө 0 де Implicitly differentiating both sides of (iv) with respect to y gives о 0 ад Substitute (i) and (ii) into the right hand side of this equation for x and y respectively, and solve for ду де — 0 ду Part 2 - Find ди ди. деду By the chain rule, we know that ди ди Әи Әr = да Әr да т ди де де от and = ди де Әu де + де ду дө ду ду Substituting the results from part 1 into these equations we get ди Эх 0 ди 0 dy ә?и Part 3 - Find Әr2
22u Part 3 - Find дх2 а дги Әx2 ди дх (cos(в), а ar 0 sin(0) ди де От т де т sin(0) a) ( cos() и » ((+ ) Р) + (- = а cos(0) Әт Әи cos(0) де а + cos(0) от sin(0) au де sin(0) a т 20 ди cos(0) дт sin(0) ə + да ( sin(0) ди де т T Remembering to use the product rule when it's appropriate cos(0) ar (( ди cos(0) ar - о а cos(0) от sin(0) ди 00 = о sin(0) a де ди cos(0) Әr = о т sin(e) a де sin(e) ди Ә0 о т ә?u Part 4 - Find ду? ди ду? = я () - (ан (е) % сте ) (абото) сондө 2) sin( = sin cos(0) 0 + Ә0 ди cos() ди + Әr де ду ду. а sin(e). де = Әи sin(e). ar а + sin(e) От ( cos(0) ди де cos(0) ə + де ((+ ) - Әи sin(0) От + cos(0) a де cos(0) ди де т ә sin(0) ar ди sin(e) Әr ) 0 а sin(0) ar cos(O) au де о T
а sin(0) Әr cos(0) ди Ә0 — о r cos(0) Әв ди sin(e) ar о cos(0) ə Ә0 cos(0) ди т Ә0 - о r Part 5 - Putting it all together Using the results from parts 3 and 4 we see that д2 дх2 22 и = cos? (0) Әr2 2 sin(0) cos(0) 22u sin? (0) а?и + деде 2 Ә02 sin? (0) ди ar + 2 sin(0) cos(0) ди 2 де т Similarly а?и cos? (0) о?и + 2 Ә02 cos? (0) ди Әr 2 sin(0) cos(0) ди т2 де т arae т ә? 2 sin(0) cos(0) о?и = sin? (0) + ду? Әr2 Therefore, making use of some trig identities, Ә2 и Ә2u + 0 Әr2 ду? у?u= Notation Here is a list of the notations you will need to input your answers в = theta дө = thetax ду Ә0 ду thetay в = theta де = thetax да де ду thetay От: = IX Әr = ry ду х ди Әи Әr = uur utheta Ә0
Notation Here is a list of the notations you will need to input your answers ᎧᎾ ө = theta дө дх thetax thetay ду Ә0 ө: theta thetax ᏧᎾ = thetay дх ду Әr =rx дх ar ду II ry ди = ur ar ди дө - utheta Ә2 и urr Ә2u Ә02 uthetatheta Әr2 Ә2 и дүдө Ә2 и Әөдт urtheta - uthetar