Liu_ 30171369_Assessment 1 (1).pdf

# Liu_ 30171369_Assessment 1 (1).pdf - MAT9004-TP1 2019 Due...

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MAT9004-TP1 2019 Due: 21 January 2019 Assignment 1 Functions, Basic Calculus Answer 1 . The mass of bacteria (in mg) at the start of the n-th every hour, starting from n = 0, is found to be (a) Each hour, 1mg of bacteria excrete 0:1ml of a toxic substance. Using sigma notation, write an expression for S(m), the total volume of the toxic substance excreted until the end of the m-th hour. (b) formula for sum of geometric series based on above sigma notation Let : The formula of sum of volume of toxic substance Calculate how many hours does the volume reach 20ml m 29 f ( x ) = e 0.1 n S ( m ) = m n =0 0.1 * e 0.1 n S ( m ) = 0.1 * ( e 0.1*0 + e 0.1*1 + e 0.1*2 + e 0.1*3 + . . . + e 0.1* m ) e 0.1 = a S ( m ) = 0.1 * ( a 0 + a 1 + a 2 + a 3 + . . . + a m ) = 0.1 * (1 a )( a 0 + a 1 + a 2 + a 3 + . . . + a n )/(1 a ) = 0.1 × (1 a ( m +1) ) (1 a ) S ( m ) = 0.1 × ( 1 e 0.1 × ( m +1) 1 e 0.1 ) 0.1 * ( 1 e 0.1*( m +1) 1 e 0.1 ) 20 1 e 0.1*( m +1) 200 * (1 e 0.1 ) e 0.1*( m +1) 200 * (1 e 0.1 ) 1 ln ( e 0.1*( m +1) ) ln ( 200 * (1 e 0.1 ) + 1) ( m + 1) ln ( 200 * (1 e 0.1 ) + 1) * 10 1 Liu_ 30171369_Assessment 1.pdf

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MAT9004-TP1 2019 Due: 21 January 2019 After 30 hours does the volume reach 20ml (c) We now allow the function f to take any real number as input. Using the computation rules for exponentials, compute for arbitrary t. Interpret the result in words. Answer: The equation can be use to computer at arbitrary point of time t , after 10*ln(2) ( about 7 hour), the mass of bacteria doubled based on the mass on time t . 2 . accurate formula for the mass of bacteria at time t . (a) Plot the function g on the interval [0; 120] and the function f on [0; 20] on the same sketch. f ( t + (10 ln 2)) f ( t ) f ( x ) = e 0.1 x f (2) = e 0.1*2 10 * ln 2 = f 1 (2) f ( t + f 1 (2)) f ( t ) = 2 g ( t ) = e 0.1 t 1 + 10 3 * e 0.1 t 2 Liu_ 30171369_Assessment 1.pdf
MAT9004-TP1 2019 Due: 21 January 2019 (b) Let start with find derivative of f(t) and g(t) => f’(0) = 0.1 ~ f(10) = 0.27 => g’(0) = 0.1 ~ g’(10) = 0.27 When the rate of the change of the f(t) and g(t) is almost identical, therefore the function f(t) and g(t) if is changing in same rate. (c) Function g(t) with in its domain [0, ) is one to one function , therefore g : ℝ→ [0, ) is invertible (d) find inverse , : ( 0 < t < 1000) f ( t ) = 0.1 × e 0.1 t g ( t ) = 10 5 × e 0.1 t ( e 0.1 t + 10 3 ) 2 t [0,10]; g 1 ( t ) y = e 0.1 t 1 + 10 3 * e 0.1 t y (1 + 10 3 * e 0.1 t ) = e 0.1 t y + 10 3 y * e 0.1 t = e 0.1 t y = e 0.1 t 10 3 y * e 0.1

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• Fall '19
• Derivative, Mathematical analysis, Convex function, Concave function, Wolfram

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