math226finalspring08s

math226finalspring08s - 1 Suppose that the temperature z at...

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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 d = | A ~ B · ~n | | ~n | = 1 q ( - 1) 2 + ( - 1) 2 + 1 2 = 1 3 .
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