2.3 Partial Derivatives, Linear Approximation Prof. Tesler Math 20C - PowerPoint PPT Presentation

2.3 Partial Derivatives, Linear Approximation Prof. Tesler Math 20C Fall 2018 Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 1 / 28

Partial derivatives f ( x , y , z ) = sin ( x 2 + 4 xy + 3 z ) The partial derivative of f with respect to x means Treat x as a variable. Treat the other variables ( y and z ) as constants. Differentiate as a function of x . ∂ f ∂ x = cos ( x 2 + 4 xy + 3 z ) · ( 2 x + 4 y ) Result: Notation Partial derivatives One variable derivative ∂ : partial derivative symbol d ∂ f df ∂ x dx ∂ d ∂ x f dx f f ′ ( x ) or f x ( x , y , z ) f x Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 2 / 28

Partial derivatives f ( x , y , z ) = sin ( x 2 + 4 xy + 3 z ) The partial derivative of f with respect to y means Treat y as a variable. Treat the other variables ( x and z ) as constants. Differentiate as a function of y . Result: ∂ f ∂ y = 4 x cos ( x 2 + 4 xy + 3 z ) Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 3 / 28

Partial derivatives f ( x , y , z ) = sin ( x 2 + 4 xy + 3 z ) The partial derivative of f with respect to z means Treat z as a variable. Treat the other variables ( x and y ) as constants. Differentiate as a function of z . Result: ∂ f ∂ z = 3 cos ( x 2 + 4 xy + 3 z ) Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 4 / 28

Partial derivative at a point One variable f ′ ( 10 ) : Evaluate function f ′ ( x ) first, and then plug in value x = 10 . f ′ ( 10 ) = 3 ( 10 ) 2 = 300 f ( x ) = x 3 f ′ ( x ) = 3 x 2 Multiple variables f ( x , y ) = x 4 y f x ( x , y ) = 4 x 3 y f x ( 1 , 2 ) : Compute derivative as function: f x ( 1 , 2 ) = 4 ( 1 3 )( 2 ) = 8 and then plug in ( x , y ) = ( 1 , 2 ) : Several notations for this: � � ∂ f ∂ f ∂ f � � f x ( 1 , 2 ) = ∂ x ( 1 , 2 ) = = � � ∂ x ∂ x � � x = 1 , y = 2 ( 1 , 2 ) f y ( x , y ) = x 4 and f y ( 1 , 2 ) = 1 4 = 1 ∂ z ∂ z For z = x 4 y : ∂ x = 4 x 3 y ∂ y = x 4 Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 5 / 28

z = x y For z = x y , what are ∂ z ∂ x and ∂ z ∂ y ? ∂ z d ∂ x = y · x y − 1 dx ( x 3 ) = 3 x 2 ∂ z d dy ( 3 y ) = 3 y ln 3 ∂ y = x y ln ( x ) Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 6 / 28

Gradient The gradient of f ( x , y ) is � ∂ f � ∂ x , ∂ f = ∂ f ı + ∂ f ∇ f = ∇ f ( x , y ) = ∂ x ˆ ∂ y ˆ  ∂ y For f ( x , y ) = x 2 y 4 , we get ∇ f = 2 xy 4 , 4 x 2 y 3 � � . At point ( x , y ) = ( 1 , 10 ) : 2 · 1 · 10 4 , 4 · 1 2 · 10 3 � � = � 20000 , 4000 � ∇ f ( 1 , 10 ) = For a function of three variables: � ∂ f � ∂ x , ∂ f ∂ y , ∂ f ∇ f = ∇ f ( x , y , z ) = ∂ z This generalizes to any number of variables. Symbol “ ∇ ” is called Nabla . It’s an upside down Greek letter Delta, ∆ . Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 7 / 28

Formal definition of partial derivative z (a,b,f(a,b)) y (a,b,0) x Graph the surface z = f ( x , y ) . Consider point P = ( a , b , ? ) on surface. z = f ( x , y ) = f ( a , b ) , so the point on the surface is P = ( a , b , f ( a , b )) . Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 8 / 28

Formal definition of partial derivative z (a,b,f(a,b)) y (a,b,0) x ∂ f ∂ x : Compute derivative treating x as a variable and y as a constant. y = b = constant is a plane parallel to the xz plane ( y = 0 ). The graph of z = f ( x , b ) with x varying and y = b = constant gives the red curve on the surface. The tangent line in that plane has slope f x ( a , b ) : y = b and z = f ( a , b ) + f x ( a , b ) · ( x − a ) Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 9 / 28

Formal definition of partial derivative z (a,b,f(a,b)) y (a,b,0) x ∂ f ∂ y : Compute derivative treating y as a variable and x as a constant. x = a = constant is a plane parallel to the yz plane ( x = 0 ). The graph of z = f ( a , y ) with y varying and x = a = constant gives the green curve on the surface. The tangent line in that plane has slope f y ( a , b ) : and z = f ( a , b ) + f y ( a , b ) · ( y − b ) x = a Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 10 / 28

Formal definition of partial derivative z (a,b,f(a,b)) y (a,b,0) x f x ( a , b ) = rate of change of f w.r.t. x at ( a , b ) f ( a + ∆ x , b )− f ( a , b ) f ( x , b )− f ( a , b ) = lim = lim x − a ∆ x x → a ∆ x → 0 f y ( a , b ) = rate of change of f w.r.t. y at ( a , b ) f ( a , b + ∆ y )− f ( a , b ) f ( a , y )− f ( a , b ) = lim = lim ∆ y y − b ∆ y → 0 y → b Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 11 / 28

Tangent plane z (a,b,f(a,b)) y (a,b,0) x A tangent plane to a 3D surface z = f ( x , y ) generalizes a tangent line to a 2D curve. It’s a plane that just touches the surface at a given point. It approximates the function when ( x , y ) is near the starting point. Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 12 / 28

Tangent plane z (a,b,f(a,b)) y (a,b,0) x The tangent plane at point P contains both tangent lines. The formula of the tangent plane is: z = f ( a , b ) + f x ( a , b )( x − a ) + f y ( a , b )( y − b ) Holding x = a constant gives the green tangent line, and holding y = b constant gives the red tangent line. Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 13 / 28

Tangent plane — Vector formula z (a,b,f(a,b)) y (a,b,0) x z = f ( a , b ) + f x ( a , b )( x − a ) + f y ( a , b )( y − b ) � � = f ( a , b ) + f x ( a , b ) , f y ( a , b ) · � x − a , y − b � which gives an alternate formula z = f ( a , b ) + ∇ f ( a , b ) · � x − a , y − b � Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 14 / 28

Technicalities for the tangent plane to exist Left graph: no tangent plane at the top point. Right graph: no tangent plane at any point along the creases. Need f ( x , y ) and derivatives f x ( x , y ) and f y ( x , y ) to exist and be continuous at ( x , y ) = ( a , b ) , plus more technical conditions. Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 15 / 28

Example: z = f ( x , y ) = x 2 + 4 y 2 Find the equation of the tangent plane at ( a , b ) = ( 1 , 2 ) Need to fill in z . At ( x , y ) = ( 1 , 2 ) , z = 1 2 + 4 ( 2 2 ) = 17 . Find the tangent plane at ( 1 , 2 , 17 ) . Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 16 / 28

Example: z = f ( x , y ) = x 2 + 4 y 2 Find the equation of the tangent plane at ( 1 , 2 , 17 ) Slopes f ( x , y ) = x 2 + 4 y 2 f x ( x , y ) = 2 x f y ( x , y ) = 8 y f ( 1 , 2 ) = 1 2 + 4 ( 2 2 ) = 17 f x ( 1 , 2 ) = 2 ( 1 ) = 2 f y ( 1 , 2 ) = 8 ( 2 ) = 16 Tangent plane at ( a , b , f ( a , b )) = ( 1 , 2 , 17 ) z = f ( a , b ) + f x ( a , b )( x − a ) + f y ( a , b )( y − b ) z = f ( a , b ) + ∂ f ∂ x ( a , b )( x − a ) + ∂ f or ∂ y ( a , b )( y − b ) z = 17 + 2 ( x − 1 ) + 16 ( y − 2 ) Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 17 / 28

Other ways to write tangent plane formula z = 17 + 2 ( x − 1 ) + 16 ( y − 2 ) As a function L ( x , y ) = 17 + 2 ( x − 1 ) + 16 ( y − 2 ) f ( x , y ) is approximated by the tangent plane near the starting point: ≈ when ( x , y ) ≈ ( 1 , 2 ) f ( x , y ) L ( x , y ) � ��� �� ��� � � ��� �� ��� � z on surface z on tangent plane This is called local linearity . Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 18 / 28

Other ways to write tangent plane formula z = 17 + 2 ( x − 1 ) + 16 ( y − 2 ) 6 4 2 0 -2 -4 -6 60 40 z 20 0 x -20 -3 y -2 -1 0 1 2 3 z = f ( x , y ) = x 2 + 4 y 2 Surface: Tangent plane: z = L ( x , y ) = 17 + 2 ( x − 1 ) + 16 ( y − 2 ) Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 19 / 28

Other ways to write tangent plane formula z = 17 + 2 ( x − 1 ) + 16 ( y − 2 ) In terms of changes in x , y , z z − 17 = 2 ( x − 1 ) + 16 ( y − 2 ) ∆ z = 2 ∆ x + 16 ∆ y where ∆ x = x − a = x − 1 ∆ y = y − b = y − 2 ∆ z = z − f ( a , b ) = z − 17 General formula ∆ z = f x ( a , b ) ∆ x + f y ( a , b ) ∆ y = ∂ f ∂ x ( a , b ) ∆ x + ∂ f ∂ y ( a , b ) ∆ y Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 20 / 28

Other ways to write tangent plane formula z = 17 + 2 ( x − 1 ) + 16 ( y − 2 ) Vector version f ( x , y ) = x 2 + 4 y 2 has ∇ f ( x , y ) = � 2 x , 8 y � ∇ f ( 1 , 2 ) = � 2 ( 1 ) , 8 ( 2 ) � = � 2 , 16 � z = f ( a , b ) + ∇ f ( a , b ) · � x − a , y − b � z = 17 + ∇ f ( 1 , 2 ) · � x − 1 , y − 2 � z = 17 + � 2 , 16 � · � x − 1 , y − 2 � Vector version with changes in variables ∆ z = ∇ f ( a , b ) · � ∆ x , ∆ y � ∆ z = � 2 , 16 � · � ∆ x , ∆ y � where ∆ x = x − 1 , ∆ y = y − 2 , ∆ z = z − 17 . Prof. Tesler 2.3 Partial Derivatives, Linear Approximation Math 20C / Fall 2018 21 / 28