Week 5 Summary

18.02 Lecture 11. – Tue, Oct 2, 2007 Differentials. Recall in single variable calculus: y =√f (x) ⇒ dy = f � (x) dx. Example: y = sin−1 (x) ⇒ x = sin y, dx = cos y dy, so dy/dx = 1/ cos y = 1/ 1 − x2 . Total differential: f = f (x, y, z) ⇒ df = fx dx + fy dy + fz dz. This is a new type of object, with its own rules for manipulating it (df is not the same as Δf ! The textbook has it wrong.) It encodes how variations of f are related to variations of x, y, z. We can use it in two ways: 1. as a placeholder for approximation formulas: Δf ≈ fx Δx + fy Δy + fz Δz. 2. divide by dt to get the chain rule: if x = x(t), y = y(t), z = z(t), then f becomes a function df dx dy dz of t and = fx + fy + fz dt dt dt dt Example: w = x2 y + z, dw = 2xy dx + x2 dy + dz. If x = t, y = et , z = sin t then the chain rule gives dw/dt = (2tet ) 1 + (t2 ) et + cos t, same as what we obtain by substitution into formula for w and one-variable differentiation. Can justify the chain rule in 2 ways: 1. dx = x� (t) dt, dy = y � (t) dt, dz = z � (t) dt, so substituting we get dw = fx dx + fy dy + fz dz = fx x� (t) dt + fy y � (t) dt + fz z � (t) dt, hence dw/dt. 2. (more rigorous): Δw � fx Δx + fy Δy + fz Δz, divide both sides by Δt and take limit as Δt → 0. Applications of chain rule: Product and quotient formulas for derivatives: f = uv, u = u(t), v = v(t), then d(uv)/dt = fu u� + fv v � = vu� + uv � . Similarly with g = u/v, d(u/v)/dt = gu u� + gv v � = (1/v) u� + (−u/v 2 ) v � = (u� v − uv � )/v 2 . Chain rule with more variables: for example w = f (x, y), x = x(u, v), y = y(u, v). Then dw = fx dx + fy dy = fx (xu du + xv dv) + fy (yu du + yv dv) = (fx xu + fy yu ) du + (fx xv + fy yv ) dv. Identifying coefficients of du and dv we get ∂f /∂u = fx xu + fy yu and similarly for ∂f /∂v. It's not legal to “simplify by ∂x”. Example: polar coordinates: x = r cos θ, y = r sin θ. Then fr = fx xr + fy yr = cos θ fx + sin θ fy , and similarly fθ . 18.02 Lecture 12. – Thu, Oct 4, 2007 Handouts: PS4 solutions, PS5. Gradient. dw dx dy dz dw d�r Recall chain rule: = wx + wy + wz . In vector notation: = �w · . dt dt dt dt dt dt Definition: �w = �wx , wy , wz � – GRADIENT VECTOR. Theorem: �w is perpendicular to the level surfaces w = c. Example 1: w = ax + by + cz, then w = d is a plane with normal vector �w = �a, b, c�. Example 2: w = x2 + y 2 , then w = c are circles, �w = �2x, 2y� points radially out so ⊥ circles. Example 3: w = x2 − y 2 , shown on applet (Lagrange multipliers applet with g disabled). 1 2 �w is a vector whose value depends on the point (x, y) where we evaluate w. Proof: take a curve �r = �r(t) contained inside level surface w = c. Then velocity �v = d�r/dt is in the tangent plane, and by chain rule, dw/dt = �w · d�r/dt = 0, so �v ⊥ �w. This is true for every �v in the tangent plane. Application: tangent plane to a surface. Example: tangent plane to x2 + y 2 − z 2 = 4 at (2, 1, 1): gradient�is �2x, 2y, −2z� = �4, 2, −2�; tangent plane is 4x + 2y − 2z = 8. (Here we could also solve for z = x2 + y 2 − 4 and use linear approximation formula, but in general we can’t.) (Another way to get the tangent plane: dw = 2x dx + 2y dy − 2z dz = 4dx + 2dy − 2dz. So Δw ≈ 4Δx + 2Δy − 2Δz. The level surface is Δw = 0, its tangent plane approximation is 4Δx + 2Δy − 2Δz = 0, i.e. 4(x − 2) + 2(y − 1) − 2(z − 1) = 0, same as above). Directional derivative. Rate of change of w as we move (x, y) in an arbitrary direction. Take a unit vector û = �a, b�, and look at straight line trajectory �r(s) with velocity û, given by x(s) = x0 + as, y(s) = y0 + bs. (unit speed, so s is arclength!) dw Notation: . ds |û Geometrically: slice of graph by a vertical plane (not parallel to x or y axes anymore). Directional derivative is the slope. Shown on applet (Functions of two variables), with w = x2 + y 2 + 1, and rotating slices through a point of the graph. dw d�r Know how to calculate dw/ds by chain rule: = �w · = �w · û. ds |û ds Geometric interpretation: dw/ds = �w · û = |�w| cos θ. Maximal for cos θ = 1, when û is in direction of �w. Hence: direction of �w is that of fastest increase of w, and |�w| is the directional derivative in that direction. We have dw/ds = 0 when û ⊥ �w, i.e. when û is tangent to direction of level surface. 18.02 Lecture 13. – Fri, Oct 5, 2007 (estimated – written before lecture) Practice exams 2A and 2B are on course web page. Lagrange multipliers. Problem: min/max when variables are constrained by an equation g(x, y, z) = c. � Example: find point of xy = 3 closest to origin ? I.e. minimize x2 + y 2 , or better f (x, y) = x2 + y 2 , subject to g(x, y) = xy = 3. Illustrated using Lagrange multipliers applet. Observe on picture: at the minimum, the level curves are tangent to each other, so the normal vectors �f and �g are parallel. So: there exists λ (“multiplier”) such that �f = λ�g. We replace the constrained min/max problem in 2 variables with equations involving 3 variables x, y, λ: ⎧ ⎧ ⎪ ⎪ ⎨fx = λgx ⎨2x = λy i.e. here fy = λgy 2y = λx ⎪ ⎪ ⎩ ⎩ g=c xy = 3. 3 � 2x − λy = 0 In general solving may be hard and require a computer. Here, linear algebra: −λx + 2y = 0 2 requires either x = y = 0 (impossible, √ det = 4 − λ = 0. So λ = ±2. No solutions √ √ since xy =√3), or for λ = −2, while λ = 2 gives ( 3, 3) and (− 3, − 3). (Checked on applet that �f = 2�g at minimum). Why the method works: at constrained min/max, moving in any direction along the constraint df = �f · û = 0, i.e. surface g = c should give df /ds = 0. So, for any û tangent to {g = c}, ds |û û ⊥ �f . Therefore �f is normal to tangent plane to g = c, and so is �g, hence the gradient vectors are parallel. Warning: method doesn’t say whether we have a min or a max, and second derivative test doesn’t apply with constrained variables. Need to answer using geometric argument or by comparing values of f . Advanced example: surface-minimizing pyramid. Triangular-based pyramid with given triangle as base and given volume V , using as little surface area as possible. Note: V = 13 Abase h, so height h is fixed, top vertex moves in a plane z = h. We can set up problem in coordinates: base vertices P1 = (x1 , y1 , 0), P2 , P3 , and top vertex P = (x, y, h). Then areas of faces = 12 |P�P1 × P�P2 |, etc. Calculations to find critical point of function of (x, y) are very hard. Key idea: use variables adapted to the geometry, instead of (x, y): let a1 , a2 , a3 = lengths of sides of the base triangle; u1 , u2 , u3 = distances in the xy-plane from the projection of� P to the sides of the base triangle. Then each face is a triangle with base length ai and height u2i + h2 (using Pythagorean theorem). � � � So we must minimize f (u1 , u2 , u3 ) = 12 a1 u21 + h2 + 12 a2 u22 + h2 + 12 a3 u23 + h2 . Constraint? (asked using flashcards; this was a bad choice, very few students responded at all.) Decomposing base into 3 smaller triangles with heights ui , we must have g(u1 , u2 , u3 ) = 1 1 1 2 a1 u1 + 2 a2 u2 + 2 a3 u3 = Abase . Lagrange multiplier method: �f = λ�g gives a1 u a1 � 1 = λ , similarly for u2 and u3 . 2 u21 + h2 2 u1 u2 u3 We conclude λ = � 2 = � 2 = � 2 , hence u1 = u2 = u3 , so P lies above the 2 2 u1 + h u2 + h u 3 + h2 incenter.