15.5.Ex4-9

# 15.5.Ex4-9 - 680 C H A P T E R 15 D I F F E R E N T I AT I...

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680 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) (a) Find d dt f ( c ( t )) without making any calculations. Explain. (b) Verify your answer to (a) using the Chain Rule. SOLUTION (a) The level curves of f ( x , y ) are the circles x 2 + y 2 = c 2 .S ince c ( t ) is a parametrization of the unit circle, f has constant value 1 on c .Thatis, f ( c ( t )) = 1, which implies that d dt f ( c ( t )) = 0. (b) We now ±nd d dt f ( c ( t )) using the Chain Rule: d dt f ( c ( t )) = f x dx dt + f y dy dt (1) We compute the derivatives involved in (1): f x = x ³ x 2 + y 2 ´ = 2 x , f y = y ³ x 2 + y 2 ´ = 2 y dx dt = d dt ( cos t ) =− sin t , dy dt = d dt ( sin t ) = cos t Substituting the derivatives in (1) gives d dt f ( c ( t )) = 2 x ( sin t ) + 2 y cos t Finally, we substitute x = cos t and y = sin t to obtain d dt f ( c ( t )) =− 2cos t sin t + 2sin t cos t = 0 . 4.

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15.5.Ex4-9 - 680 C H A P T E R 15 D I F F E R E N T I AT I...

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