Respuesta :
Answer-
The directional derivative of f(x,y) at (0,6) in the direction θ = 2π/3 is, 7.73
Solution-
The unit vector in the direction of the given angle is
[tex]\hat{u}= < \cos \frac{2\pi }{3} , \sin \frac{2\pi }{3}> \ = \ <\frac{-1}{2} , \frac{\sqrt{3} }{2} >[/tex]
Given that,
[tex]f(x,y)=2ye^{-x}[/tex]
Gradient of f(x,y)
[tex]\triangledown f(x,y) = < -2ye^{-x} ,2e^{-x} >[/tex]
Gradient of f(x,y) at (0,6)
[tex]\triangledown f(x,y)_{(0,6)}[/tex] = <(-2)(6)(1) , (2)(1)> = <-12 , 2>
The directional derivative of f(x,y) at (0,6) is
[tex]\triangledown f(0,6)\cdot \hat{u} \ = \ <-12 , 2>\cdot<\frac{-1}{2},\frac{\sqrt{3}}{2} > = (-12)(\frac{-1}{2} )+(2)(\frac{\sqrt{3}}{2})= 6+\sqrt{3} = 7.73[/tex]
Directional derivatives are derivatives of a function in particular directions. The directional derivative of given function at (0,6) in θ = 2π/3 is 7.732 approximately.
What is directional derivative of a function?
Directional derivative of a function at a given point is the rate of growth of that function on that point in the given particular direction.
For a function [tex]f(x,y,z)[/tex] , its directional derivative on point [tex](x_1, y_1, z_1)[/tex] in direction of unit vector [tex]\hat{u}[/tex] is evaluated as:
[tex]\nabla_{\hat{u}}f(x_1, y_1, z_1) = \dfrac{\partial f}{\partial x}|\hat{u}_x| + \dfrac{\partial f}{\partial y}|\hat{u}_y| + \dfrac{\partial f}{\partial z}|\hat{u}_z| _{x=x_1,y=y_1,z=z_1}[/tex]
where [tex]\hat{u}_x, \hat{u}_y, \: \rm and \: \: \hat{u}_z[/tex] components of the considered unit vectors in direction of coordinate axes, and that |u| like symbol means magnitude of that component.
For the given case, the function is [tex]f(x,y) = 2ye^{-x}[/tex]
The point at which directional derivative of given function is to be found is (0,6), and direction is [tex]\theta = \dfrac{2\pi}{3}[/tex]
Getting the unit vector in the given direction, we get:
[tex]\hat{u_x} = 1.\cos(\dfrac{2\pi}{3}}) = -1/2\\\\\hat{u_y} = 1.\sin(\dfrac{2\pi}{3}}) = \sqrt{3}/2[/tex]
Also, we get partial derivatives of given function as:
[tex]\dfrac{\partial f}{\partial x} = -2ye^{-x}\\\\\dfrac{\partial f}{\partial y} = 2e^{-x}[/tex]
Thus, the directional derivative is found as:
[tex]\nabla_{\hat{u}}f(x_1, y_1) = \dfrac{\partial f}{\partial x}|\hat{u}_x| + \dfrac{\partial f}{\partial y}|\hat{u}_y| \\\\\nabla_{\hat{u}}f(x_1, y_1) = - 2ye^{-x} \times-\dfrac{1}{2} + 2e^{-x} \dfrac{\sqrt{3}}{2} |_{(x=0,y=6)}\\\\\nabla_{\hat{u}}f(0,6) = 6 + \sqrt{3} \approx 7.732[/tex]
Thus, The directional derivative of given function at (0,6) in θ = 2π/3 is 7.732 approx
Learn more about directional derivatives here:
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