Problems covered in recitation on Monday, Oct 6,2003October 8, 2003On Monday, we talked about Lagrange multipliers. We also did the followingproblems.Problem 1 Let w(x, y) = x3y2. Find a parametric equation for the line tangentto the curve w = 1 at the point (1, 1). Use the gradiant.Solution We compute∇w = h3x2y2, 2x3yiso we have ∇w(1, 1) = h3, 2i. Now, the curve w = 1 is a level curve for w, andwe know that ∇w is perpendicular to level curves. So if we rotate ∇w by 90◦,we get a vector tangent to the curve. This rotated vector is∇w(1, 1)⊥= h−2, 3i.A parametric equation for a line passing through (1, 1) in the direction of h−2, 3iis given byx(t) = −2t + 1y(t) = 3t + 1.Problem 2 Find the point of the plane x + 2y + 3z = 1 closest to the origin.Solution We use Lagrange multipliers. Our constraint equation is g(x, y, z) = 1whereg(x, y, z) = x + 2y + 3z.We want to minimize distance to the origin, which is really the same as min-imizing the s quare of the distance (get rid of that nasty square root). So thefunction we want to minimize isf(x, y, z) = x2+ y2+ z2.Now we do Lagrange multipliers. We look for solutions to∇f = λ∇g.1We compute∇f = h2x, 2y, 2ziλ∇g = λh1, 2, 3iComparing the components of these vectors, we see that we have a system ofequations2x = λ2y = 2λ2z = 3λOf course, we also have the constraint equation x + 2y + 3z = 1. Solving for x,y, and z in terms of λ, we get x = λ/2, y = λ, and z = 3λ/2. Plugging into theconstraint equation, we see thatλ/2 + 2λ + 9λ/2 = 7λ = 1so λ = 1/7. Therefore,x = 1/14y = 1/7z =
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