SolutionsMT

# SolutionsMT - Mathematics 1c Solutions Midterm Examination...

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Mathematics 1c: Solutions, Midterm Examination Due: Monday, May 3, at 10am 1. Do each of the following calculations. (a) If a particle follows the curve c ( t ) = e t - 1 i - ( t - 1) j + sin( πt ) k and ﬂies oﬀ on a tangent at t = 1, where is it at t = 2? Solution. First note that c (1) = i . We compute that c 0 ( t ) = e t - 1 i - j + π cos( πt ) k . Therefore c 0 (1) = i - j - π k . Hence the position of the particle at time t = 2 is i + (2 - 1)( i - j - π k ) = 2 i - j - π k . (b) Find the equation of the tangent plane to the surface x 2 - e xy + z 2 = 1 at the point (1 , 0 , 1). Solution. The tangent plane consists of vectors based at (1 , 0 , 1) that are perpen- dicular to the gradient of f ( x,y,z ) = x 2 - e xy + z 2 . We compute that f ( x,y,z ) = (2 x - ye xy , - xe xy , 2 z ) . Evaluating at (1 , 0 , 1), we obtain f (1 , 0 , 1) = (2 , - 1 , 2) . Therefore, the tangent plane at (1 , 0 , 1) is deﬁned by (2 , - 1 , 2) · ( x - 1 ,y,z - 1) = 0 , namely by z = - x + y 2 + 2 . (c) Let f ( x,y,z ) = 5 + xy - zx be the concentration of chemical X. Find the direction at (1 , 1 , 1) in which X is decreasing the fastest. In which directions is it decreasing at 30% of its maximum rate? Give your answer in terms of the angle made with the direction of fastest decrease. Solution. The concentration of X is increasing the fastest in the direction of the gradient of f , and hence decreasing the fastest in the opposite of this direction. We compute that f ( x,y,z ) = ( y - z,x, - x ) , and hence that f (1 , 1 , 1) = (0 , 1 , - 1) . 1

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Therefore the concentration of X is decreasing fastest in the direction -∇ f (1 , 1 , 1) = (0 , - 1 , 1) . Let n be a unit vector, and let θ be the angle between n and -∇ f (1 , 1 , 1). Then the directional derivative of f in the direction n is given by f (1 , 1 , 1) · n = k∇ f (1 , 1 , 1) kk n k cos( π - θ ). This is 30% of its most negative value when cos( π - θ ) = - 0 . 3, which is equivalent to cos( θ ) = 0 . 3. This means that θ = cos - 1 (0 . 3). 2. Answer each of the following questions. (a) Let f ( r,s ) be a (smooth) function of r and s and, let r = x + 2 y and s = x - 2 y . Deﬁne the function h by h ( x,y ) = f ( x + 2 y,x - 2 y ). Calculate 2 h ∂x∂y in terms of the partial derivatives of f . Solution. Let r ( x,y ) = x + 2 y and s ( x,y ) = x - 2 y . Then we have that ∂r ∂x = 1 , ∂r ∂y = 2 , ∂s ∂x = 1 , ∂s ∂y = - 2 . Applying the chain rule to h ( x,y ) = f ( r ( x,y ) ,s ( x,y )), we ﬁnd that 2 h ∂x∂y = ∂x ± ∂h ∂y ² = ∂x ± ∂f ∂r ∂r ∂y + ∂f ∂s ∂s ∂y ² = ∂x ± 2 ∂f ∂r - 2 ∂f ∂s ² = 2 ∂x ± ∂f ∂r ² - 2 ∂x ± ∂f ∂s ² = 2 ± 2 f ∂r 2 ∂r ∂x + 2 f ∂s∂r ∂s ∂x ² - 2 ± 2 f ∂r∂s ∂r ∂x + 2 f ∂s 2 ∂s ∂x ² = 2 ± 2 f ∂r 2 + 2 f ∂s∂r ² - 2 ± 2 f ∂r∂s + 2 f ∂s 2 ² . Because
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## This note was uploaded on 07/19/2010 for the course MA 1C taught by Professor Ramakrishnan during the Spring '08 term at Caltech.

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SolutionsMT - Mathematics 1c Solutions Midterm Examination...

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