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MIT 18 01 - Study Notes

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18 02A Topic 30 Non independent variables chain rule Read TB 19 6 SN N 1 N 3 We ll get increasingly fancy We use the notation that fully specifies the role of all the variables w is the partial of w with respect ot x with y held constant x y This shows explicitly that x and y are independent variables Recall the chain rule If w f x y and x x u v y y u v w w x w y u x y u v y x u v v w x w y w v u x y v u y x v u Example 1 Given w x2 y 2 z 2 constrained by the relation z x2 y 2 w compute x y Method 1 Implicit differentiation Differentiate the formula for w x is the variable y is a constant and z is a function of x w z 2x 2z x y x y z Need to find differentiate the constraint relation implicitly x y w z 2x 2x 2z 2x x y x y Formalizing method 1 Let wx wy wz be the formal derivatives of w That is the derivatives when x y and z are thought of as independent I e w 2x wy 2y wz 2z x w x y z z wx wy wz wx 1 wy 0 wz x y x y x y x y x y z z x x y Slice y constant example 1 y Slice z constant example 2 continued 1 18 02A topic 30 2 Method 2 Total differentials dw wx dx wy dy wz dz 2x dx 2y dy 2z dz This is the usual approximation formula made infinitesimal If we used the constraint to eliminate z so that w w x y then we d have the formula w dw w dx dy x y y x This can be hard instead we use the constraint to remove dz Constraint dz 2x dx 2y dy dw 2x dx 2y dy 2z 2x dx 2y dy 2x 4xz dx 2y 4yz dy w w 2x 4xz 2y 4yz Compare this with above x y y x Note we get both differentials at once w Example 2 For the same functions find x z Now x and z are the independent variables and y is an intermediate variable w y y Method 1 2x 2y wx 1 wy wz 0 x z x z z x y x x y w 2x 2y 0 Constraint 0 2x 2y x z x z y x z y 2 2 2 2 2 Not surprising z constant x y is constant w x y z is constant Method 2 remove dy dw 2x dx 2y dy 2z dz wx dx wy dy wz dz 1 dz 2x dx 2y dy dy 2y dz xy dx 1 dz xy dx 2z dz Substitute dw 2x dx 2y 2y w x z 2x 2x dx 1 2z dz 0 dx 1 2z dz w 0 1 2z z x w Example 3 Let w x y z t xy zt Find x y z answer Variable x Constants y z Function of x t w t 2 2 3x y z x y z x y z t t t y Need differentiate xy zt implicitly y z x x y z x y z z y z w 3x2 y zy x y z 3 continued 2 18 02A topic 30 Example 4 3 3 2 Let w x y z t xy zt Find w x y z w y x z w z x y using differentials answer Independent variables x y z dependendent variables t w z 3 y z 2 t dw 3x2 y dx x3 dy 2zt dz z 2 dt xy zt y dx x dy t dz z dt Solve for dt dt yz dx xz dy zt dz Substitute in dw dw 3x2 y dx x3 dy 2zt dz z 2 yz dx xz dy zt dz 3x2 y zy dx x3 xz dy 2zt zt dz w w w 2 3 3x zy x xz 2zt zt x y z y x z z x y w w w dx dy dz Reason if w f x y z then dw x y z y x z z x y Thermodynamic variables p V T U S H pressure volume temperature internal energy entropy enthalpy Any two can be independent and then the others are dependent U V V When p T are independent have the law T p 0 p T T p p T Example 5 Express this law when V and T are the independent variables U V V answer We need to express in terms of derivatives p T p T T p with independent variables V T p U U etc To help with the algebra we use the shorthand pV V T T V T i e V T are always the independent variables Dependent variables are U and p so we look at dU and dp U U dU V dV dT UV dV UT dT T V T p p dp V T dV T V dT pV dV pT dT V V T Second eq dV p1V dp ppVT dT p1V and p p T p pV T Substitute for dV dU 1 pV UV dp UV ppVT UT dT Using the boxed formulas we can restate the law as continued 1 pV U V T U p pT pV T 1 pV p UV 1 pV 0 18 02A topic 30 4 Fanciest version Jacobian As before w f x y x x u v y y u v In matrix form the chain rule is x x w w w w u v v u y y u v v u x y y x u v v u w w xu xv yu yv x y y x The matrix is called the Jacobian matrix This is easy to derive using total differentials Example 6 Use the Jacobian to redo example 5 We do this in small steps Step 1 Chain rule x x y w w w w w w u v x y u v y u v v u x y v u y x x Step 2 Write in matrix form w w w w u v v u x y y x y v u x u v y u v x v u y v u Step 3 Decide which variables are x y and which are u v Old variables x y p T new variables u v V T p w w w w V T Step 4 Substitute into formula in step 2 V T T V T p T p T V T T 0 T 1 Step 5 Simplify the matrix V T V T p p w w w w V T T V T V T p V T p 0 1 T p 1 T 1 1 1 V Step 6 Call the matrix A find A …


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MIT 18 01 - Study Notes

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