Chain Rule (§14.5)Directional Derivatives (§14.6)Gradient (§14.6)Math 241, Spring 2014Math 241, Spring 2014Jayadev S. AthreyaSpring 2014Math 241, Spring 2014Chain Rule (§14.5)Derivatives along curvesSuppose we have a parametric curve (x(t), y(t)) ∈ R2, and afunction f : R2→ R. We can make the functionh(t) = f (x(t), y(t)).How do we compute h0(t)?Math 241, Spring 2014Chain Rule (§14.5)One-variable chain ruleh(t) = f (x(t)), thenh0(t) = f0(x(t))x0(t)Math 241, Spring 2014Chain Rule (§14.5)Chain ruleh0(t) = fx(x(t), y(t))x0(t) + fy(x(t), y(t))y0(t)Math 241, Spring 2014Chain Rule (§14.5)How do we get to this?A linear approximation to f near (a , b) can be written as:∆f ≈ fx(a, b)∆x + fy(a, b)∆y∆f = f (x, y) − f (a, b), ∆x = x − a, ∆y = y − bMath 241, Spring 2014Chain Rule (§14.5)Approximating hh(t + ∆t) ≈ h(t) + h0(t)∆th(t + ∆t) = f (x(t + ∆t), y(t + ∆t))x(t + ∆t) ≈ x(t) + x0(t)∆t, y(t + ∆t) ≈ y(t) + y0(t)∆tMath 241, Spring 2014Chain Rule (§14.5)Combining Approximationsh(t+∆t) = f (x(t+∆t), y(t+∆t)) ≈ f (x(t)+x0(t)∆t, y(t)+y0(t)∆t)h(t + ∆t) − h(t) ≈ f (x(t) + x0(t)∆t, y(t) + y0(t)∆t) − f (x(t), y(t))≈ fx(x(t), y(t))(x0(t)∆t) + fy(x(t), y(t))(y0(t)∆t)= ∆t(fx(x(t), y(t))x0(t) + fy(x(t), y(t))y0(t))Math 241, Spring 2014Chain Rule (§14.5)Combining Approximationsh(t+∆t) = f (x(t+∆t), y(t+∆t)) ≈ f (x(t)+x0(t)∆t, y(t)+y0(t)∆t)h(t + ∆t) − h(t) ≈ f (x(t) + x0(t)∆t, y(t) + y0(t)∆t) − f (x(t), y(t))≈ fx(x(t), y(t))(x0(t)∆t) + fy(x(t), y(t))(y0(t)∆t)= ∆t(fx(x(t), y(t))x0(t) + fy(x(t), y(t))y0(t))Math 241, Spring 2014Chain Rule (§14.5)Combining Approximationsh(t+∆t) = f (x(t+∆t), y(t+∆t)) ≈ f (x(t)+x0(t)∆t, y(t)+y0(t)∆t)h(t + ∆t) − h(t) ≈ f (x(t) + x0(t)∆t, y(t) + y0(t)∆t) − f (x(t), y(t))≈ fx(x(t), y(t))(x0(t)∆t) + fy(x(t), y(t))(y0(t)∆t)= ∆t(fx(x(t), y(t))x0(t) + fy(x(t), y(t))y0(t))Math 241, Spring 2014Chain Rule (§14.5)Combining Approximationsh(t+∆t) = f (x(t+∆t), y(t+∆t)) ≈ f (x(t)+x0(t)∆t, y(t)+y0(t)∆t)h(t + ∆t) − h(t) ≈ f (x(t) + x0(t)∆t, y(t) + y0(t)∆t) − f (x(t), y(t))≈ fx(x(t), y(t))(x0(t)∆t) + fy(x(t), y(t))(y0(t)∆t)= ∆t(fx(x(t), y(t))x0(t) + fy(x(t), y(t))y0(t))Math 241, Spring 2014Chain Rule (§14.5)Notationz = f (x, y ), x = x(t), y = y(t)dzdt=∂z∂xdxdt+∂z∂ydydtMath 241, Spring 2014Chain Rule (§14.5)ExampleMath 241, Spring 2014Chain Rule (§14.5)Higher dimensionsChain rule works in any dimension: f : Rn→ R,x1, . . . , xn: R → R,h(t) = f (x1(t), . . . xn(t)) = f (x(t)),h0(t) = fx1(x(t))x01(t) + . . . fxn(x(t))x0n(t).Math 241, Spring 2014Chain Rule (§14.5)Higher dimensionsChain rule works in any dimension: f : Rn→ R,x1, . . . , xn: R → R,h(t) = f (x1(t), . . . xn(t)) = f (x(t)),h0(t) = fx1(x(t))x01(t) + . . . fxn(x(t))x0n(t).Math 241, Spring 2014Chain Rule (§14.5)Higher dimensional inputsx = x(u, v), y = y(u, v ), z = z(x, y) = z(x(u, v ), y(u, v)) Then∂z∂u=∂z∂x∂x∂u+∂z∂y∂y∂u∂z∂v=∂z∂x∂x∂v+∂z∂y∂y∂vPartial derivatives reduce this to case of 1-variable inputs.Math 241, Spring 2014Chain Rule (§14.5)Geometric examplez = z(x, y ) = x2+ y2, x = x(r , θ) = r cos θ, y = y(r, θ) = r sin θ∂z∂θ=∂z∂x∂x∂θ+∂z∂y∂y∂θ= 2x(−r sin θ) + 2y(r cos θ)= 2r cos θ(−r sin θ) + 2r sin θ(r cos θ) = 0Math 241, Spring 2014Chain Rule (§14.5)Geometric examplez = z(x, y ) = x2+ y2, x = x(r , θ) = r cos θ, y = y(r , θ) = r sin θ∂z∂θ=∂z∂x∂x∂θ+∂z∂y∂y∂θ= 2x(−r sin θ) + 2y(r cos θ)= 2r cos θ(−r sin θ) + 2r sin θ(r cos θ) = 0Math 241, Spring 2014Chain Rule (§14.5)Geometric examplez = z(x, y ) = x2+ y2, x = x(r , θ) = r cos θ, y = y(r , θ) = r sin θ∂z∂θ=∂z∂x∂x∂θ+∂z∂y∂y∂θ= 2x(−r sin θ) + 2y(r cos θ)= 2r cos θ(−r sin θ) + 2r sin θ(r cos θ) = 0Math 241, Spring 2014Chain Rule (§14.5)Geometric examplez = z(x, y ) = x2+ y2, x = x(r , θ) = r cos θ, y = y(r , θ) = r sin θ∂z∂r=∂z∂x∂x∂r+∂z∂y∂y∂r= 2x(cos θ) + 2y(sin θ)= 2r cos θ(cos θ) + 2r sin θ(sin θ) = 2r(sin2θ + cos2θ) = 2rMath 241, Spring 2014Chain Rule (§14.5)Geometric examplez = z(x, y ) = x2+ y2, x = x(r , θ) = r cos θ, y = y(r , θ) = r sin θ∂z∂r=∂z∂x∂x∂r+∂z∂y∂y∂r= 2x(cos θ) + 2y(sin θ)= 2r cos θ(cos θ) + 2r sin θ(sin θ) = 2r(sin2θ + cos2θ) = 2rMath 241, Spring 2014Chain Rule (§14.5)Geometric examplez = z(x, y ) = x2+ y2, x = x(r , θ) = r cos θ, y = y(r , θ) = r sin θ∂z∂r=∂z∂x∂x∂r+∂z∂y∂y∂r= 2x(cos θ) + 2y(sin θ)= 2r cos θ(cos θ) + 2r sin θ(sin θ) = 2r(sin2θ + cos2θ) = 2rMath 241, Spring 2014Directional Derivatives (§14.6)Other directionsPartial derivatives are rates of change along lines parallel to thex and y-axes. What about other lines?Math 241, Spring 2014Directional Derivatives (§14.6)Lines through (a, b)A line through (a, b) in direction v = (v1, v2) is given byx(t) = a + v1t, y(t) = b + v2th(t) = f (x(t), y (t))(Dvf )(a, b) = h0(0)Math 241, Spring 2014Directional Derivatives (§14.6)Applying Chain Rulex(t) = a + v1t, y(t) = b + v2th(t) = f (x(t), y (t))h0(0) = fx(a, b)x0(0) + fy(a, b)y0(0) = v1fx(a, b) + v2fy(a, b)(Dvf )(a, b) = v · (fx(a, b), fy(a, b))Math 241, Spring 2014Gradient (§14.6)The GradientGiven f : R2→ R, we define the gradient ∇f of f by∇f (a, b) = (fx(a, b), fy(a, b)) = fx(a, b)i + fy(a, b)j(Dvf )(a, b) = v · ∇f (a, b)Math 241, Spring 2014Gradient (§14.6)The GradientGiven f : R2→ R, we define the gradient ∇f of f by∇f (a, b) = (fx(a, b), fy(a, b)) = fx(a, b)i + fy(a, b)j(Dvf )(a, b) = v · ∇f (a, b)Math 241, Spring 2014Gradient (§14.6)Three dimensionsFor f : R3→ R,∇f (a, b, c) = (fx(a, b, c), fy(a, b, c), fz(a, b, c))= fx(a, b, c)i + fy(a, b, c)j + fz(a, b, c)kFor v ∈ R3(Dvf )(a, b, c) = v · ∇f (a, b, c)Math 241, Spring 2014Gradient (§14.6)Three dimensionsFor f : R3→ R,∇f (a, b, c) = (fx(a, b, c), fy(a, b, c), fz(a, b, c))= fx(a, b, c)i + fy(a, b, c)j + fz(a, b, c)kFor v ∈ R3(Dvf )(a, b, c) = v · ∇f (a, b, c)Math 241, Spring 2014Gradient (§14.6)n-dimensionsFor f : Rn→ R,∇f (x) = (fx1(x), fx2(x), . . . , fxn(x))For v ∈ Rn(Dvf )(x) = v · ∇f (x)Math 241, Spring 2014Gradient (§14.6)n-dimensionsFor f : Rn→ R,∇f (x) = (fx1(x), fx2(x), . . . , fxn(x))For v ∈ Rn(Dvf )(x) = v · ∇f (x)Math 241, Spring 2014Gradient (§14.6)Rates of changeQuestionIn what
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