Directional Derivatives and the GradientJonathan LeeNovember 13, 2007These handouts are crudely extracted from my ongoing set of notes posted on my websiteat http://math.stanford.edu/~jlee/math51/.• given a function f : Rn→ R, a point p ∈ Rnand a unit vector v ∈ Rn, we can defineDvf(p), the directional derivative of f at a in the direction of v, to be g0(0), where wehave set g(t) = f(p + tv)• define the gradient of a function f : Rn→ R to be the vector of partial derivatives;that is,∇f =∂f∂x1,∂f∂x2, . . . ,∂f∂xn• directional derivatives can be expressed in terms of the gradient (which, as an overallderivative, encapsulates all the directional ones) by means ofDvf(p) = ∇f(p) · v– find the directional derivatives Dvf(p) forfunction point directionf(x, y) = eysin(x) p = (π3, 0) v =1√10(3, −1)f(x, y) =1x2+y2p = (3, −2) v = (1, −1)f(x, y) = ex− x2y p = (1, 2) v = (2, 1)f(x, y, z) = xyz p = (−1, 0, 2) v =1√5(0, 2, −1)f(x, y, z) = e−(x2+y2+z2)p = (1, 2, 3) v = (1, 1, 1)• the tangent plane of a function f : Rn→ R at a point p ∈ Rnis given by the parametricequationh(x) = f(p) + ∇f (p) · (x − p) ;in the case n = 2, we can use this to find non-zero vectors perpendicular to the plane• a surface in Rndefined by the equation f(x, y, z) = c has a perpendicular vector givenby ∇f; hence, if ∇f is non-z ero, an equation for the tangent plane at a point p is givenby∇f(p) · (v − p) = 01– find the plane tangent to the following surfaces at the given pointssurface pointx3+ y3+ z3= 7 p = (0, −1, 2)zeycos(x) = 1 p = (π, 0, −1)2xz + yz − x2y + 10 = 0 p = (1, −5, 5)2xy2= 2z2− xyz p = (2, −3,
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