Math 149 Lab 5 Exercises

Math 149 Lab 5 Exercises - f on those intervals? Exercise 4...

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Math 149 Lab 5 - Exercises Name : < your name goes here > Section : < your section number goes here > Exercises ü Exercise 1 x 4 4 is an antiderivative of x 3 . Of course, you know that there are other antiderivatives of x 3 . (a) Name one other antiderivative of x 3 . (b) Find an antiderivative of 1 4 + x 2 . (c) Find an antiderivative of 1 2 + x 4 . (d) Find an antiderivative of 2 I x 2 M . ü Exercise 2 (a) Have Mathematica find the antiderivative of tan H x L 3 , and have Mathematica check that its derivative is what it should be. (b) Have Mathematica find the antiderivative of 1 1 + x 4 , and have Mathematica compute its derivative. (Do not simplify.) ü Exercise 3 Let f H x L = x 3 - 4 x 2 + 3 x , and @ a , b D = @ 0, 4 D . Also let F H x L = Ÿ a x f H t L t . è Create an animation similar to the one above in the example for the Fundamental Theorem and discuss what is happening. è Solve F ' H x L = 0 by hand and then with Mathematica and compare the results. What can you observe about the graphs of f and F ' at points where F ' H x L = 0? è On what intervals is the function F ' increasing or decreasing? What can you say about
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Unformatted text preview: f on those intervals? Exercise 4 The law of mass action states that the rate of a chemical reaction is proportional to the product of the concentrations of A and B , i.e., d @ C D d t = k @ A D@ B D , where @ A D , @ B D , and @ C D are all functions, and k is a constant. Let's assume the initial concentrations of A and B are equal, and write @ A D H L = @ B D H L = a . If we write c H t L = @ C D H t L for the concentration of C at time t , then the concentration of A (as well as that of B ) at time t is @ A D H t L = a-@ C D H t L . Therefore, in our case, the law of mass action can be stated as d c d t = k H a-c L 2 . Use Mathematica 's DSolve command to find c , the concentration of C , as a function of t . Hint: the reaction begins with an initial concentration @ C D H L = 0. Mathematica version 8.0, 3 April, 2012 2 Math 149 Lab 5 Exercises.nb...
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Math 149 Lab 5 Exercises - f on those intervals? Exercise 4...

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