Answer to Question #140068 in Calculus for Haley

Question

f(x,y)=−2x^2+4y^2
find the value of the directional derivative at the point (3,4) in the direction given by the angle θ=2π/5. More specifically, find the directional derivative of f at the point (3,4) in the direction of the unit vector determined by the angle θ in polar coordinates.

Answer 1

Given,

$f(x,y)=-2x^2+4y^2$

unit vector in the direction of $\theta=\dfrac{2\pi}{5}$ ,

$\hat{u}=<cos(\dfrac{2\pi}{5}),sin(\dfrac{2\pi}{5})>$

$\hat{u}=< 0.30 ,0.95>$

Directional Derivative of F(x,y) is

$D_uf(x,y)=f_x(x,y)$$\times 0.30+f_y(x,y)\times 0.95$

$\to D_uf(x,y)=-4x\times0.30+$ $8y\times0.95$

$\to D_uf(3,4)=-4\times3\times0.30+8\times 4\times0.95$

$\to D_uf(3,4)=-3.6+30.4$

$\to D_uf(3,4)=26.8$

Hence the required directional derivative is 26.8