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chain - CHAIN RULE HOMEWORK Section 13.5 2,6 12 44 46...

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CHAIN RULE Math21a, O. Knill HOMEWORK Section 13.5: 2,6, 12, 44, 46 1D CHAIN RULE. If f and g are functions of one variable t , then d/dtf ( g ( t )) = f ( g ( t )) g ( t ). For example, d/dt sin(log( t )) = cos(log( t )) /t . FINDING DERIVATIVES. Also the 1D chain rule was useful. For example, to find the derivative of log( x ) we can write 1 = d/dx exp( log ( x ) = d/dx exp(log( x )) = log ( x ) x so that log ( x ) = 1 /x . An other example: to find arccos ( x ), we write 1 = d/dx cos(arccos( x )) = sin(arccos( x )) arccos ( x ) = radicalBig 1 sin 2 (arccos( x )) arccos ( x ) = 1 x 2 arccos ( x ) so that arccos ( x ) = 1 / 1 x 2 . GRADIENT. Define f ( x, y ) = ( f x ( x, y ) , f y ( x, y )). It is called the gradient of f . It is the natural derivative of a function of several variables and a vector. THE CHAIN RULE. If vector r ( t ) is curve in space and f is a function of three variables, we get a function of one variables t mapsto→ f ( vector r ( t )). The chain rule is d/dtf ( vector r ( t )) = f ( vector r ( t )) · vector r ( t ) WRITING IT OUT. The chain rule is, when written out d dt f ( x ( t ) , y ( t )) = f x ( x ( t ) , y ( t )) x ( t ) + f y ( x ( t ) , y ( t )) y ( t ) . EXAMPLE. Let z = sin( x + 2 y ), where x and y are functions of t : x = e t , y = cos( t ). What is dz dt ? Here, z = f ( x, y ) = sin( x + 2 y ), z x = cos( x + 2 y ), and z y = 2 cos( x + 2 y ) and dx dt = e t , dy dt = sin( t ) and dz dt = cos( x + 2 y ) e t 2 cos( x + 2 y ) sin( t ). EXAMPLE. If f is the temperature distribution in a room and vector r ( t ) is the path of the spider Shelob , then f ( vector r ( t )) is the temperature, Shelob experiences at time t . The rate of change depends on the velocity vector r ( t
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