Logarithm 12

Calculus Level 4

Given the functions

f ( x , y ) = x 2 e − ( x 2 + y 2 2 ) f(x,y) = \frac{x}{2} e^{ - \big(\frac{x^2 + y^2}{2} \big)}

g ( z ) = log ⁡ 2 ( z 5 ) g(z) = \log_2 (\frac{z}{5})

and the matrices

A = [ ∂ f ( x , y ) ∂ x ∣ ( 0 , 0 ) ∂ 2 f ( x , y ) ∂ x ∂ y ∣ ( 1 , 1 ) ∂ 2 f ( x , y ) ∂ y ∂ x ∣ ( 0 , 0 ) ∂ f ( x , y ) ∂ y ∣ ( 1 , 1 ) ] \begin{bmatrix} \\ {{\partial f(x,y) \over \partial x} \bigg|_{(0,0)}} & {{\partial^2 f(x,y) \over \partial x \partial y} \bigg|_{(1,1)}} \\ \\ {{\partial^2 f(x,y) \over \partial y \partial x} \bigg|_{(0,0)}} & {{\partial f(x,y) \over \partial y} \bigg|_{(1,1)}} \\ \\ \end{bmatrix}

B = [ g ′ ( 1 ) g ′ ( 2 ) g ′ ′ ( 1 ) g ′ ′ ( 2 ) ] \begin{bmatrix} g'(1) & g'(2) \\ g''(1) & g''(2) \\ \end{bmatrix}

find k = ln ⁡ ( 2 ) . d e t B d e t A k = \ln (2) . \frac{det B}{det A}

Notations:

  • ∂ f ( x , y ) ∂ x ∣ ( a , b ) {{\partial f(x,y) \over \partial x} \bigg|_{(a,b)}} denotes the first-order partial derivative of f ( x , y ) f(x,y)

with respect to x x at the point ( a , b , f ( a , b ) ) (a,b,f(a,b))

  • ∂ f ( x , y ) ∂ y ∣ ( a , b ) {{\partial f(x,y) \over \partial y} \bigg|_{(a,b)}} denotes the first-order partial derivative of f ( x , y ) f(x,y)

with respect to y y at the point ( a , b , f ( a , b ) ) (a,b,f(a,b))

  • ∂ 2 f ( x , y ) ∂ x ∂ y ∣ ( a , b ) {{\partial^2 f(x,y) \over \partial x \partial y} \bigg|_{(a,b)}} denotes the second-order partial derivative of f ( x , y ) f(x,y)

with respect to y y and x x at the point ( a , b , f ( a , b ) ) (a,b,f(a,b))

  • ∂ 2 f ( x , y ) ∂ y ∂ x ∣ ( 0 , 0 ) {{\partial^2 f(x,y) \over \partial y \partial x} \bigg|_{(0,0)}} denotes the second-order partial derivative of f ( x , y ) f(x,y)

with respect to x x and y y at the point ( a , b , f ( a , b ) ) (a,b,f(a,b))

  • g ′ ( a ) g'(a) denotes the first derivative of g ( z ) g(z) at the point ( a , g ( a ) ) (a, g(a))

  • g ′ ′ ( a ) g''(a) denotes the second derivative of g ( z ) g(z) at the point ( a , g ( a ) ) (a, g(a))

  • log ⁡ 2 ( ⋅ ) \log_2 (\cdot) denotes the binary logarithm of ( ⋅ ) (\cdot)

  • ln ⁡ ( ⋅ ) \ln (\cdot) denotes the natural logarithm of ( ⋅ ) (\cdot)

  • d e t A det A denotes the determinant of matrix A

  • d e t B det B denotes the determinant of matrix B

k = − e 2 ln ⁡ ( 2 ) k = - \frac{e}{2 \ln (2)} k = − e ( ln ⁡ ( 2 ) ) 2 k = - \frac{e}{ (\ln (2))^2} k = − e ln ⁡ ( 2 ) k = - \frac{e}{\ln (2)} k = + e ln ⁡ ( 2 ) k = + \frac{e}{\ln (2)} k = + 2 e ln ⁡ ( 2 ) k = + \frac{2 e}{\ln (2)}

Victor Porto by Victor Porto , 22, Brazil

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1 solution

Tom Engelsman
Dec 21, 2020

For the matrix A : A:

f x = 1 − x 2 2 ⋅ e − ( x 2 + y 2 ) / 2 ⇒ f x ∣ ( x , y ) = ( 0 , 0 ) = 1 / 2 ; f_{x} = \frac{1-x^2}{2} \cdot e^{-(x^2+y^2)/2} \Rightarrow f_{x}|_{(x,y)=(0,0)} = 1/2;

f x y = y ( x 2 − 1 ) 2 ⋅ e − ( x 2 + y 2 ) / 2 ⇒ f x y ∣ ( x , y ) = ( 1 , 1 ) = 0 ; f_{xy} = \frac{y(x^2-1)}{2} \cdot e^{-(x^2+y^2)/2} \Rightarrow f_{xy}|_{(x,y)=(1,1)} = 0;

f y x = y ( x 2 − 1 ) ⋅ e − ( x 2 + y 2 ) / 2 ⇒ f y x ∣ ( x , y ) = ( 0 , 0 ) = 0 ; f_{yx} = y(x^2-1) \cdot e^{-(x^2+y^2)/2} \Rightarrow f_{yx}|_{(x,y)=(0,0)} = 0;

f y = − x y 2 ⋅ e − ( x 2 + y 2 ) / 2 ⇒ f y ∣ ( x , y ) = ( 1 , 1 ) = − 1 2 e f_{y} = -\frac{xy}{2} \cdot e^{-(x^2+y^2)/2} \Rightarrow f_{y}|_{(x,y)=(1,1)} = -\frac{1}{2e}

or A = [ 1 2 0 0 − 1 2 e ] . \boxed{A = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2e} \end{bmatrix}}.

For the matrix B : B:

g ′ ( x ) = 1 x ln ⁡ ( 2 ) g'(x) = \frac{1}{x\ln(2)} and g ′ ′ ( x ) = − 1 x 2 ln ⁡ ( 2 ) g''(x) = -\frac{1}{x^2\ln(2)}

or B = [ 1 ln ⁡ ( 2 ) 1 2 ln ⁡ ( 2 ) − 1 ln ⁡ ( 2 ) − 1 4 ln ⁡ ( 2 ) ] . \boxed{B = \begin{bmatrix} \frac{1}{\ln(2)} & \frac{1}{2\ln(2)} \\ -\frac{1}{\ln(2)} & -\frac{1}{4\ln(2)} \end{bmatrix}}.

Finally:

k = ln ⁡ ( 2 ) ⋅ d e t ( B ) d e t ( A ) = ln ⁡ ( 2 ) ⋅ [ 1 4 ln ⁡ 2 ( 2 ) ÷ − 1 4 e ] = ln ⁡ ( 2 ) ⋅ − e ln ⁡ 2 ( 2 ) = − e ln ⁡ ( 2 ) . k = \ln(2) \cdot \frac{det(B)}{det(A)} = \ln(2) \cdot [\frac{1}{4\ln^{2}(2)} \div -\frac{1}{4e}] = \ln(2) \cdot -\frac{e}{\ln^{2}(2)} = \boxed{-\frac{e}{\ln(2)}}.

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