Let f(x,y)=ln(x^2+y^2+1) +e^(2xy): find the gradient of f at the point (0,-2) b)

ID: 3002959 • Letter: L

Question

Let f(x,y)=ln(x^2+y^2+1) +e^(2xy):

find the gradient of f at the point (0,-2)

b)Find the directional derivative of f at the point (0,-2)

c)Find the maximum value of the directional derivative at the point (0,-2)

Explanation / Answer

b)Find the directional derivative of f at the point (0,-2) ?f = ==> ?f(0, 1, -2) = . The direction vector v from (0, 1, -2) toward the origin is . Normalizing yields u = /v5. So, Dv f(0, 1, -2) = · /v5 = -12/v5. c)Find the maximum value of the directional derivative at the point (0,-2) Is that supposed to be z = y/(x² + y²)? I'll assume that it is. ?z/?x = -y(x² + y²)^(-2)(2x) = -2xy/(x² + y²)² ?z/?y = [(x² + y²) - y(2y)]/(x² + y²)² = (x² - y²)/(x² + y²)². Using these partials, grad(z)(1,1) = (-2/2², 0/2²) = (-1/2, 0). The maximum value of the directional derivative occurs in the direction of the gradient. The maximal value is the magnitude of grad(z) at the given point. ||grad(z)(1,1)|| = 1/2

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Let f(x,y)=ln(x^2+y^2+1) +e^(2xy): find the gradient of f at the point (0,-2) b)

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