18.02A Topic 26: Tangent plane approximation, directional derivatives.Read: TB: 19.1, 19.2 SN: TAFrom last time we have the tangent plane approximation.∆x = (x − x0), ∆y = (y − y0), ∆w = f(x, y) − f(x0, y0)∆w ≈∂w∂x0∆x +∂w∂y0∆yThe supplemental notes §TA give an analytic argument for thisand its genaralization to 3 dimensions:Suppose w = f(x, y, z) and∆x = (x − x0), ∆y = (y − y0), ∆z = (z − z0) then∆w = f(x, y) − f(x0, y0) ≈∂w∂x0∆x +∂w∂y0∆y +∂w∂z0∆zxyz....................................................................KKKKKKKKKKKKKKKKKKKKKKKKKKK•Gradient∂w∂x0= rate of change inbi direction,∂w∂y0= rate of change inbj direction.∂w∂x0,∂w∂y0= gradient of w = ∇wTo evaluate at P0we write ∇w(P0). (Will use this in a moment.)Directional derivativeFix a directionbu and a point P0in the plane.The directional derivative of w at P0in the directionbu isdefined asdwdsP0,bu= lim∆s→0∆w∆s.Here ∆w is the change in w caused by a step of length ∆s inthe direction ofbu (all in the xy-plane).It is a fact that:dwdsP0,bu= ∇w(P0) ·bu.xyw.......................SSS))RRRRRRRSSSSSSSSSSS••P0∆w∆s∆sbuProof: The tangent plane approximation and the bottompicture at right show∆w∆s≈∂w∂x0∆x∆s+∂w∂y0∆y∆s≈∂w∂x0cos φ +∂w∂y0sin φButbu = hcos φ, sin φi since it is a unit vector. Thus, the lastformula is just ∇w ·bu. In the limit the approximations becomeexact and we get the boxed equation. QEDxy•P0??bu=hcos φ,sin φi∆s∆x∆yφ(continued)118.02A topic 26 2Example: (Algebraic example) Let w = x3+ 3y2.Computedwdsat P0= (1, 2) in the direction of v = 3i + 4k.i) ∇w = h3x2+ 3y2i ⇒ ∇w|(1,2)= h15, 12i = 15 i + 12 j.ii)bu =v|v=35i +45j.iii)dwdsP0,bu= ∇w|(1,2)·bu = (15 i + 12 j) · (34i +45j) =935.Example: (Geometric example) Letbu be the direction of h1, −1i.Using the picture at right estimate∂w∂xP,∂w∂yp, anddwdsP,bu.By measuring from P to the next in level curve in thex direction we see that ∆x ≈ −.5.⇒∂w∂xP≈∆w∆x≈10−.5= −20.Similarly, we get∂w∂yP≈ 20..Measuring in the u direction we get ∆s ≈ −.3⇒dwdsP,bu≈∆w∆s≈10.3= −33.3.xyw = 25w = 15w = 5P•bu@@@@@@@@@@@∆xOO∆yoo11Direction of maximum change:bu =∇w|∇w|(Proof: angle = 0).The Gradient is perpendicular to level curves∇w ⊥ level curve (surface) w = c.Example: Consider the graph of y = ex. Find a vector perpindicularto the tangent to at the point (1, e)..................................•P0??bu?????????????−−→∇wOld method: Find the slope take the negative reciprocal and make the vector.New method: This graph is the level curve of w = y − exwith w = 0.∇w = h−ex, 1i ⇒ normal = ∇w(1, e) = h−e, 1i.proof: Ifbu is tangent to the level curve at P0thendwdsP0,bu= 0since w is constant along the level curve, i.e.,dwdsP0,bu= ∇w(P0) ·bu = 0. QED(continued)18.02A topic 26 3Example: Find the tangent plane to the surface x2+ 2y2+ 3z2= 6 at the pointP = (1, 1, 1).Introduce a new variable w = x2+2y2+3z2= 6. Our surface is the level surface w = 6⇒ normal to surface is ∇w = h2x, 4y, 6zi. At the point P we have ∇w|P= h2, 4, 6i.Using point normal form the equation of the tangent plane is 2(x − 1) + 4(y − 1) +6(z − 1) = 0 ⇔ 2x + 4y + 6z =
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