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)