Question

Find the directional derivative of f(x; y; z) = 1/(√ x^2+y^2+z^2) at P : (3; 0; 4) in the direction of a = [1; 1; 1].

Answer 1

Find the directional derivative of f(x; y; z) = 1/(√ x^2+y^2+z^2) at P : (3; 0; 4) in the direction of a = [1; 1; 1].

The directional derivative is:

Du.f(x,y,z) = fx(x,y,z).a + fy(x,y,z).b + fz(x,y,z).c,  where,

f(x, y, z) = 1/(√ x^2+y^2+z^2) and u = (a, b, c) = (1, 1, 1)

fx = -x/(x^2+y^2+z^2)^(3/2), fy = -y/(x^2+y^2+z^2)^(3/2), fz = -z/(x^2+y^2+z^2)^(3/2)

Du.f(x,y,z) = fx(x,y,z).a + fy(x,y,z).b + fz(x,y,z).c

Du.f(x,y,z) = fx(x,y,z).1 + fy(x,y,z).1 + fz(x,y,z).1

Du.f(x,y,z) = fx(x,y,z) + fy(x,y,z) + fz(x,y,z)

Du.f(x,y,z) = -x/(x^2+y^2+z^2)^(3/2)  –  y/(x^2+y^2+z^2)^(3/2)  –  z/(x^2+y^2+z^2)^(3/2)

Du.f(x,y,z) =( -x –  y  –  z) / (x^2+y^2+z^2)^(3/2)

Du.f(x,y,z) = -(x +  y  +  z) / (x^2+y^2+z^2)^(3/2)

At P(3, 0, 4)

D = -(3 +  0  +  4) / (9 + 0 +16)^(3/2)

D = -(7) / (125)

 

Comments

Thanks for answering these. I started to work them out, then realised I needed to brush up on vectors!

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Yeh, I just noticed these questions earlier on. Had to do a bit of brushing up myself

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If you look amongst the unanswered questions, there are more vector problems: grad, div and curl raising their heads again! Curly heads, no doubt!