Math S21a: Multivariable calculus Oliver Knill, Summer 2011Lecture 10: LinearizationIn single variable calculus, you have seen the following definition:The linear approximation of f(x) at a point a is the linear functionL(x) = f(a) + f′(a)(x − a) .y=LHxLy=fHxLThe gr aph of the function L is close to the graph of f at a. We generalize this now to higherdimensions:The linear approximation of f(x, y) at (a, b) is the linear functionL(x, y) = f(a, b) + fx(a, b)(x − a) + fy(a, b)(y − b) .The linear approximation of a function f(x, y, z) at (a, b, c) isL(x, y, z ) = f(a, b, c) + fx(a, b, c)(x − a) + fy(a, b, c)(y − b) + fz(a, b, c)(z − c) .Using the gradient∇f(x, y) = hfx, fyi, ∇f(x, y, z) = hfx, fy, fzi ,the linearization can be written more compactly asL(~x) = f (~x0) + ∇f (~a) · (~x −~a) .How do we justify the linearization? If the second variable y = b is fixed, we have a one-dimensionalsituation, where the only variable is x. Now f(x, b) = f (a, b) + fx(a, b)(x − a) is the linear ap-proximation. Similarly, if x = x0is fixed y is the single variable, then f(x0, y) = f(x0, y0) +fy(x0, y0)(y − y0). Knowing the linear a pproximations in both the x and y variables, we can getthe general linear approximation by f(x, y) = f(x0, y0) + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0).1 What is t he linear approximation of the function f (x, y) = sin(πxy2) at the point (1, 1)? Wehave (fx(x, y), yf(x, y) = (πy2cos(πxy2), 2yπ cos(πxy2)) which is at the point (1, 1) equal to∇f(1, 1) = hπ cos(π), 2π cos(π)i = h−π, 2πi.2 Linearization can be used to estimate functions near a point. In the previous example,−0.00943 = f (1+0.01, 1+0.01) ∼ L(1+0.01, 1+0.01) = −π0.01−2π0.01+3π = −0.00942 .3 Here is an example in three dimensions: find t he linear approximation to f (x, y, z) = xy +yz + zx at the point (1, 1, 1). Since f (1, 1, 1) = 3, and ∇f (x, y, z) = (y + z, x + z, y +x), ∇f(1, 1, 1) = (2, 2, 2). we have L(x, y, z) = f (1 , 1, 1) + (2, 2, 2) · (x − 1, y − 1, z − 1) =3 + 2(x − 1) + 2(y − 1) + 2(z − 1) = 2x + 2y + 2z − 3.4 Estimate f (0.01, 24.8, 1.02) for f (x, y, z) = ex√yz.Solution: take (x0, y0, z0) = (0, 25, 1), where f(x0, y0, z0) = 5. The gradient is ∇f(x, y, z) =(ex√yz, exz/(2√y), ex√y). At the point (x0, y0, z0) = (0, 25, 1) the gradient is the vector(5, 1/10, 5). The linear approximation is L(x, y, z) = f (x0, y0, z0) + ∇f(x0, y0, z0)(x −x0, y −y0, z −z0) = 5 + (5, 1/10, 5)(x −0, y −25, z −1) = 5x + y/10 + 5z −2.5. We can approximatef(0.01, 24.8, 1.02) by 5 + (5, 1/10, 5) · (0.01, −0.2, 0.02) = 5 + 0.05 − 0.02 + 0.10 = 5.13. Theactual value is f(0.01 , 24.8, 1.02) = 5.1306, very close to the estimate.5 Find the tangent line to the graph of the function g(x) = x2at the point (2, 4).Solution: the level curve f (x, y) = y − x2= 0 is the graph of a function g(x) = x2andthe tangent at a point (2, g(2)) = (2, 4) is obtained by computing the gradient ha, bi =∇f(2, 4) = h−g′(2), 1i = h−4, 1i and forming −4x + y = d, where d = −4 · 2 + 1 · 4 = −4.The answer is −4x + y = −4 which is the line y = 4x − 4 of slope 4.6 The Barth surface is defined as the level surface f = 0 off(x, y, z) = (3 + 5t)(−1 + x2+ y2+ z2)2(−2 + t + x2+ y2+ z2)2+ 8(x2− t4y2)(−(t4x2) + z2)(y2− t4z2)(x4− 2x2y2+ y4− 2x2z2− 2y2z2+ z4) ,where t = (√5 + 1)/2 is a constant called the golden ratio. If we replace t with 1/t =(√5 − 1)/2 we see the surface to the middle. For t = 1, we see to the right the surfacef(x, y, z) = 8. Find the tangent plane of the later surface at the point (1, 1 , 0). Answer:We have ∇f (1, 1, 0) = h64, 64, 0 i. The surface is x + y = d for some constant d. By pluggingin (1, 1, 0) we see that x + y = 2 .7 The quartic surfacef(x, y, z) = x4− x3+ y2+ z2= 0is called the piriform. What is the equation for t he tangent plane at the point P = (2, 2, 2)of this pair shaped surface? We get h a, b, ci = h20, 4, 4i a nd so the equation of the plane20x + 4y + 4z = 56, where we have obtained the constant to the right by plugging in thepoint (x, y, z) = (2, 2, 2).Remark: some books use differentials etc to describe linearizations. This is 19 century notationand terminology and should be avoided by a ll means. For us, the linearlization of a function ata point is a linear function in the same number of variables. 20th century mathematics hasinvented the notion of differential forms which is a valuable mathematical notion, but it is aconcept which becomes only useful in follow-up courses which build on multivariable calculus likeRiemannian geometry. The notion of ”differentials” comes from a time when calculus was stillfoggy in some areas. Unfortunately it has survived and appears even in some calculus books.Homework1 If 2x + 3y + 2z = 9 is the tangent plane to the graph of z = f(x, y) at the point (1, 1, 2).Extimate f(1.01, 0.98).2 Estimate 10001/5using linear approximation3 Find f ( 0.01, 0.999) for f(x, y) = cos(πxy)y + sin(x + πy).4 Find the linear approximation L(x, y) of the functionf(x, y) =q10 − x2− 5y2at (2, 1) and use it to estimate f(1.95, 1.04).5 Sketch a contour map of the functionf(x, y) = x2+ 9y2find the gradient vector ∇f = hfx, fyi of f at the point (1, 1). Draw it together with thetangent line ax + by = d to the curve at (1,
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