2011_Solution_7 - MA 1505 Mathematics I Tutorial 7...

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Unformatted text preview: MA 1505 Mathematics I Tutorial 7 Solutions 1. (i) As the temperature function is only valid within the hotel room, its domain is { ( x,y ) : 0 ≤ x ≤ 10 , ≤ y ≤ 10 } . (ii) The heater is at the location where the temperature is highest. It is clear that the largest value of T ( x,y ) is 36 at (0 , 5). (iii) The level curve of 20 is x 2 + ( y- 5) 2 = 80 which is an arc of the circle centered at (0 , 5) with radius √ 80 ≈ 8 . 9. The level curve of 25 is x 2 + ( y- 5) 2 = 55 which is an arc of the circle centered at (0 , 5) with radius √ 55 ≈ 7 . 4. So the bed should be placed somewhere between the two arcs. (iv) The level curves of c is x 2 + ( y- 5) 2 = 5(36- c ). These are circles centered at (0 , 5), the values of c decreasing as the radius increases. The largest circle intersecting the domain intersects the domain at (10 , 0) and (10 , 10), so these points have the lowest temperature which is 11. 2. V = I × R = ⇒ I = V R . (i) ∂I ∂V = 1 R . If R = 15, then ∂I ∂V = 1 15 ≈ . 0667 A/V....
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This note was uploaded on 01/05/2012 for the course MATHEMATIC MA1505 taught by Professor Freudleong during the Spring '10 term at National University of Singapore.

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2011_Solution_7 - MA 1505 Mathematics I Tutorial 7...

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