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# Slide_07 - MA1104 Multivariable Calculus Lecture 7 Dr KU...

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MA1104 Multivariable Calculus Lecture 7 Dr KU Cheng Yeaw Thursday Feb 5, 2009

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Recap Last lecture ... we learned about chain rule for functions of several variables and its application in implicit differentiation we introduced Directional Derivatives which generalizes partial derivatives. To this end, we define the gradient vector O f = h f x , f y , f z i (if f is a function of three variables) or O f = h f x , f y i (if f is a function of two variables).
Overview In this lecture, we will see the significance of O f one of the main uses of ordinary derivatives is in finding maximum and minimum values of functions of a single variable. Here, we will see how to use partial derivatives to locate maxima and minima of functions of two variables

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Significance of O f Notice in this course, we seek to know the geometric meaning of mathematical notion we develop. So, what geometric significance does O f have?
Level Curve and O f To see one of the significance of O f ( x 0 , y 0 ) , consider the level curve f ( x, y ) = k for some k R on which the point ( x 0 , y 0 ) lies. Represent this curve (in 2D) by a parametric equation r ( t ) = h x ( t ) , y ( t ) i such that r ( t 0 ) = h x (0) , y (0) i = h x 0 , y 0 i . By the Chain Rule, df ( x ( t ) , y ( t )) dt = ∂f ∂x dx dt + ∂f ∂y dy dt

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Since f ( x ( t ) , y ( t )) = k for all t , df ( x ( t ) , y ( t )) dt = 0 . So 0 = ∂f ∂x dx ( t ) dt + ∂f ∂y dy ( t ) dt = O f ( x, y ) ·h dx ( t ) dt , dy ( t ) dt i = O f ( x, y ) · r 0 ( t ) .
At point ( x 0 , y 0 ) , i.e. t = t 0 , 0 = O f ( x 0 , y 0 ) · r 0 ( t 0 ) . Recall that r 0 ( t 0 ) is tangent vector to the level curve at ( x 0 , y 0 ) . Therefore Level Curve vs O f Suppose f ( x, y ) is differentiable function of x and y at ( x 0 , y 0 ) . Suppose O f ( x 0 , y 0 ) 6 = 0 . Then O f ( x 0 , y 0 ) is perpendicular/normal to the level curve f ( x, y ) = k at the point ( x 0 , y 0 ) where f ( x 0 , y 0 ) = k .

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Again, I repeat O f ( x 0 , y 0 ) · r 0 ( t 0 ) = 0 where r ( t ) is a parametric equation of the level curve f ( x, y ) = k such that f ( x 0 , y 0 ) = k and r ( t 0 ) = h x 0 , y 0 i . Notice k is determined by x 0 and y 0 and f . Using a similar argument, we can prove that this phenomenon also holds for level surfaces F ( x, y, z ) = k .

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Level Surface vs O f Suppose F ( x, y, z ) is differentiable function of x , y and z at ( x 0 , y 0 , z 0 ) . Suppose S is the level surface F ( x, y, z ) = k con- taining ( x 0 , y 0 , z 0 ) . Let C be any curve that lies on S and passes through ( x 0 , y 0 , z 0 ) . Let r ( t ) be a parametric equation of C such that r ( t 0 ) = ( x 0 , y 0 , z 0 ) . Suppose O F ( x 0 , y 0 , z 0 ) 6 = 0 . Then O F ( x 0 , y 0 , z 0 ) · r 0 ( t 0 ) = 0 , That is, the O F ( x 0 , y 0 , z 0 ) is perpendicular/normal to tangent vec- tor r 0 ( t 0 ) to any curve C on the surface S that passes through ( x 0 , y 0 , z 0 ) .
O F ( x 0 , y 0 , z 0 ) · r 0 ( t 0 ) = 0 . F ( x 0 , y 0 , z 0 ) = k.

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Consequently, the tangent plane to the level surface F ( x, y, z ) = k at ( x 0 , y 0 , z 0 ) is given by the equation Tangent Plane to Level Surface O F ( x 0 , y 0 , z 0 ) · h x - x 0 , y - y 0 , z - z 0 i = 0 F x ( x 0 , y 0 , z 0 )( x - x 0 )+ F y ( x 0 , y 0 , z 0 )( y - y 0 )+ F z ( x 0 , y 0 , z 0 )( z - z 0 ) = 0 .
Maximizing Directional Derivatives Lets return to the directional derivative D u f ( x, y, z ) and ask some questions. We know that D u f ( x, y, z ) = O f ( x, y, z ) · u is a scalar function of x , y and z (because it is a dot product of two vectors).

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