Dr. Z’s Math251 Handout #14.4 [Tangent Planes and Linear Approximations]By Doron ZeilbergerProblem Type 14.4a: Find an equation of the tangent plane to the given surface at the specifiedpoint.z = f(x, y) , (x0, y0, z0) .Example Problem 14.4a: Find an equation of the tangent plane to the given surface at thespecified point.z = 9x2+ y2+ 6x − 3y + 5 , (1, 2, 18) .Steps Example1. First make sure that z0= f(x0, y0)or refuse to do the problem. Then takefx=∂f∂xand fy=∂f∂y.1. 9 ·12+ 22+ 6 ·1 −3 ·2 + 5 = 18, so thepoint (1, 2, 18) indeed lies on the surface.Nowfx=∂∂x(9x2+y2+6x−3y+5) = 18x+6 ,fy=∂∂y(9x2+y2+6x−3y+5) = 2y−3 .2. Plug in x = x0, y = y0into fxand fythat you have just found.2.fx(1, 2) = 18 · 1 + 6 = 24 .fy(1, 2) = 2 · 2 − 3 = 1 .3. An equation for the tangent plane forthe given surface at the given point isz−z0= fx(x0, y0)(x−x0)+fy(x0, y0)(y−y0) .Plug-in the x0, y0, z0from the data of theproblem and fx(x0, y0), fy(x0, y0) from step2.3. z − 18 = 24(x − 1) + (y − 2). Or, inexpanded form: z = 24x + y − 8.1Problem Type 14.4b: Explain why the function is differentiable at the given point. Then findthe linearization of that function at the given point.z = f(x, y) , (a, b) .Example Problem 14.4b: Explain why the function is differentiable at the given point. Thenfind the linearization of that function at the given point.z = exsin(xy) , (0, π/2) .Steps Example1. Find the first partial derivatives fxandfy.1.fx=∂∂xexsin(xy) = (∂∂xex) sin(xy)+ex(∂∂xsin(xy))exsin(xy) + exy cos(xy) ,fy=∂∂yexsin(xy) = ex(∂∂ysin(xy)) = exx cos(xy) .2. If both fxand fyare continuous at thedesignated point (a, b) (i.e. they are de-fined and do not blow up), then the func-tion is differentiable at that point, and itis OK to have a linearlization.L(x, y) =f(a, b) +fx(a, b)(x−a)+fy(a, b)(y −b) .2. fx= exsin(xy) + exy cos(xy) and fy=exx cos(xy) are both continuous at the point(0, π/2) (they are continuous everywherefor that matter), so the function is differ-entiable there. Nowf(0, π/2) = e0sin(0) = 0 ,fx(0, π/2) = e0sin(0)+e0·(π/2) cos(0) = π/2 ,fy(0, π/2) = e0· 0 · cos(0) = 0 .The linearization isL(x, y) = 0+(π/2)·(x−0)+0·(y−π/2) = (π/2)x .Ans.: The linearization of the functionexsin(xy) at the point (0, π/2) is (π/2)x.2Problem Type 14.4c: Use the linear apprimation of the function f(x, y) at (a, b) to approximatef(a1, b1), where (a1, b1) is “near” (a, b).Example Problem 14.4b: Use the linear apprimation of the function f(x, y) =p20 − x2− 2y2at (3, 1) to approximate f(3.05, .97).Steps Example1. The beginning is exactly as before.Just find the linearization of the functionat the designated point (a, b).So first find the first partial derivatives fxand fy.1.fx=∂∂x(20−x2−2y2)1/2= (1/2)(20−x2−2y2)−1/2·(−2x)=−xp20 − x2− 2y2,fy=∂∂y(20−x2−2y2)1/2= (1/2)(20−x2−2y2)−1/2·(−4y)=−2yp20 − x2− 2y2.32. If both fxand fyare continuous at thedesignated point (a, b) (i.e. they are de-fined and do not blow up), then the func-tion is differentiable at that point, and itis OK to have a linearlization.L(x, y) = f(a, b)+fx(a, b)(x−a)+fy(a, b)(y−b) .2. fx=−x√20−x2−2y2and fy=−2y√20−x2−2y2are both continuous at the point (3, 1)(the argument of the square-root is 9 whichis positive and nothing blows up). Nowf(3, 1) =p20 − 32− 2 · 12= 3 .fx(3, 1) =−3√20 − 32− 2 · 12= −1 ,fy(3, 1) =−2 · 1√20 − 32− 2 · 12= −2/3 .and the linearization isL(x, y) = 3 − (x − 3) − (2/3)(y − 1) .So the linear approximation, valid near(3, 1) isf(x, y) ≈ 3 − (x − 3) − (2/3) · (y − 1) .3. Plug-in (a1, b1) into this approxima-tion.3.f(3.05, .97) ≈ 3−(3.05−3)−(2/3)·(.97−1) =3 − .05 + .02 = 2.97 .Ans.: f (3.05, .97) ≈ 2.97