The number we get in this way coincides with so called thedirectionalderivativeFv(x, y, z) =limt→0F(x+tv1, y+tv2, z+tv3) -F(x, y, z)twhich is known to exist wheneverFis differentiable. In particular, whenvis a unit vector, we call the numberFv(x, y, z)as therate of changeofFinthe direction of the unit vectorv. In general, therate of changeofFin thedirection of a nonzero vectorvis the rate of change ofFin the direction ofthe vectorv/||v||.At a point(a, b, c), we can ask in which direction, the rate of changeofFis maximal/minimal. The answer is simple. The rate of change ofFat(a, b, c)maximizes (i.e.Fincreases most) in the direction of the vec-tor∇F(a, b, c)and minimizes (i.e.Fdecreases most) in the direction of-∇F(a, b, c)if∇F(a, b, c)is not zero. If∇F(a, b, c)is zero, the point(a, b, c)is called a critical point ofFand in this case, we need to know more infor-mations, namely the second derivativesFxx, Fxy, Fyyto say more aboutF.The gradient vector is extremely useful when we compute the tangentplane to the surface given by the equationS:F(x, y, z) =0.14
We can find the tangent plane to the surfaceSat the point(a, b, c)as fol-lows. First, we compute the gradient vector∇F(a, b, c)at(a, b, c). Suppose∇F(a, b, c) =⟨A, B, C⟩is not zero vector. Then, thetangent plane to the surfaceSat(a, b, c)is given byA(x-a) +B(y-b) +C(z-c) =0.Namely, it is the plane which contains(a, b, c)and has normal direction∇F(a, b, c) =⟨A, B, C⟩.Examples (work out by yourself)Compute the gradient vector for each the following functionsF(x, y, z).•The functionF(x, y, z) =x2+y2+z2.•The functionF(x, y, z) =x2-y2-z2.•The functionF(x, y, z) =3x2-y2.•The functionF(x, y, z) =xyz.•The functionF(x, y, z) =x+y+z.•The functionF(x, y, z) =2x+y+3z.•The functionF(x, y, z) =z(sinx+cosy).•The functionF(x, y, z) =e-xsiny+excosy.•The functionF(x, y, z) =e-x2-y2-z2(x2+y2+z2).Problems and Explanations (think by yourself first beforelooking at explanations)14.6.5 Problem.Letf(x, y, z) =yz+exyz. Find the rate of change offat thepointP(0, 1, 2)towards the pointQ(-2,-1, 0).15
Explanation: We compute the gradient vector∇foffas∇f=⟨fx, fy, fz⟩=⟨yzexyz, z+xzexyz, y+xyexyz⟩which is at the pointP(0, 1, 2)equal to⟨2, 2, 1⟩.We compute the direction-→PQfromP(0, 1, 2)towards the pointQ(-2,-1, 0)as-→PQ=⟨-2,-1, 0⟩-⟨0, 1, 2⟩=⟨-2,-2,-2⟩.We compute the unit vector-→PQ/||-→PQ||as-→PQ||-→PQ||=⟨-2,-2,-2⟩√4+4+4=⟨-1,-1,-1⟩√1+1+1=1√3⟨-1,-1,-1⟩.Hence, the rate of change offat the pointPin the direction of-→PQis∇f·-→PQ||-→PQ||=⟨2, 2, 1⟩ ·1√3⟨-1,-1,-1⟩=1√3(-2-2-1) =-5√18.104.22.168 Problem.The temperature at a point(x, y, z)in space is given byT(x, y, z) =xyexz. What is the rate of change of temperature at the point(2, 1, 1)in the direction toward the point(4, 0, 3)?Explanation: We compute the gradient vector∇fas∇f=⟨fx, fy, fz⟩=⟨yexz+xyzexz, xexz, x2yexz⟩which is at the point(2, 1, 1)equal to⟨3e2, 2e2, 4e2⟩.