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Unformatted text preview: 1 . Suppose that the temperature z at the point ( x,y ) is z = f ( x,y ) = 40 3 x 2 + 2 y 2 . a) In what direction should a bug at (1 , 2) to move to get warmer the most quickly? What is the value of the maximal temperature increase rate at (1 , 2)? What is the temperature increase rate in the direction of ~v = h 3 , 4 i ? Answer . In the direction of ∇ f (1 , 2). Since ∇ f ( x,y ) = h 6 x, 4 y i , ∇ f (1 , 2) = h 6 , 8 i . The maximal rate is ∇ f (1 , 2)  = 10 . The unit vector in the direction of ~v is ~u = D 3 5 , 4 5 E , and the increase rate is D ~u f (1 , 2) = ∇ f (1 , 2) · D 3 5 , 4 5 E = 18 5 + 32 5 = 14 5 . b) Assume the position of the bug at time t is ~ r ( t ) = D sin π 2 t , 2 t E . Find at what rate the temperature increases on the bug’s path at t = 1 . Answer . The temperature on the bug’s path at t is z = f ( x ( t ) ,y ( t )) . We have ~ r (1) = h x (1) ,y (1) i = h 1 , 2 i , d~ r dt = D π 2 cos π 2 t , 2 E and by chain rule, dz dt  t =1 = f x ( x (1) ,y (1)) dx dt  t =1 + f y ( x (1) ,y (1)) dy dt  t =1 = f x (1 , 2) dx dt  t =1 + f y (1 , 2) dy dt  t =1 = ( 6)(0) + 8(2) = 16 . 2 . Consider the surface S : z = y + e xy . a) Write an equation of the tangent plane to S at A (0 , 1 , 2) . Find the distance between that plane and the point B (1 , 2 , 3) . Answer . The vector ~n orthogonal to S at (0 , 1 , 2) is the gradient ∇ ( z y e xy ) = h ye xy , 1 xe xy , 1 i at (0 , 1 , 2) : ~n = h 1 , 1 , 1 i and the equation of the tangent plane is x ( y 1) + z 2 = 0 or x y + z 1 = 0 . The distance is...
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This note was uploaded on 12/16/2011 for the course MATH 226 at USC.
 '07
 KAMIENNY
 Math, Calculus

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