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Let u(1) = (x(1), y(y), z(1)) be a curve in 3-space, i.e. a function u: R→ R³, and consider its derivative (dx dy du di (a) Suppose that the dot product of du/dt and the gradient Vf of some 3-variable function f = f(x, y, z) is always positive: -(1)-Vf(u(t)) > 0 du dt Show that the single variable function g(t) = f(x(1), y(1), 2(1)) is an increasing function of 1. (b) Suppose instead that for some value t = to we have that du (o)-Vf(u(fo)) = 0 Show that g'(to) = 0 and interpret the situation geometrically in terms of the curve u(t) and the

Let u(1) = (x(1), y(y), z(1)) be a curve in 3-space, i.e. a function u: R→ R³, and consider its derivative (dx dy du di (a) Suppose that the dot product of du/dt and the gradient Vf of some 3-variable function f = f(x, y, z) is always positive: -(1)-Vf(u(t)) > 0 du dt Show that the single variable function g(t) = f(x(1), y(1), 2(1)) is an increasing function of 1. (b) Suppose instead that for some value t = to we have that du (o)-Vf(u(fo)) = 0 Show that g'(to) = 0 and interpret the situation geometrically in terms of the curve u(t) and the

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let u(t) = (x(t), y(y), z(t)) be a curve in 3-space, i.e. a function u : R → R³, and consider its derivative du (dx dy (t) = -(t), -(t), dt dt dt dz 4/5). (a) Suppose that the dot product of du/dt and the gradient Vf of some 3-variable function f = f(x, y, z) is always positive: du dt -(t)-Vf(u(t))>0 1 Show that the single variable function g(t) = f(x(t), y(t), z(t)) is an increasing function of t. (b) Suppose instead that for some value t = to we have that du dt Show that g'(to) = 0 and interpret the situation geometrically in terms of the curve u(t) and the level surface f(x, y, z) = c for the constant c = f(u(t)). You can draw a picture or describe the geometry in words. (to) · Vƒ(u(to)) = 0
Transcribed Image Text:Let u(t) = (x(t), y(y), z(t)) be a curve in 3-space, i.e. a function u : R → R³, and consider its derivative du (dx dy (t) = -(t), -(t), dt dt dt dz 4/5). (a) Suppose that the dot product of du/dt and the gradient Vf of some 3-variable function f = f(x, y, z) is always positive: du dt -(t)-Vf(u(t))>0 1 Show that the single variable function g(t) = f(x(t), y(t), z(t)) is an increasing function of t. (b) Suppose instead that for some value t = to we have that du dt Show that g'(to) = 0 and interpret the situation geometrically in terms of the curve u(t) and the level surface f(x, y, z) = c for the constant c = f(u(t)). You can draw a picture or describe the geometry in words. (to) · Vƒ(u(to)) = 0
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