BU EECE 301 - How to Maximize Your Partial Credit (2 pages)

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How to Maximize Your Partial Credit



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How to Maximize Your Partial Credit

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Pages:
2
School:
Binghamton University
Course:
Eece 301 - Signals and Systems (DIS)
Unformatted text preview:

How to Maximize Your Partial Credit The key to getting maximum credit on your work is to make sure that you make it easy for me to figure out what you have done I can only do that if you present your ideas in a readable logical and understandable way When writing your answers you will increase your chances of getting partial credit if you 1 Use drawings and diagrams to explain what you are doing these don t have to be elaborate hand drawn sketches are fine 2 Write legibly if I can t read it I can t grade it 3 Clearly state your assumptions 4 Indicate why you are doing something a Tell what you are doing and why you are doing it b E g before you take the derivative of something state why you are taking the derivative 5 Use words to transition between your equations believe it or not a string of equations left unbroken by commentary is virtually impossible to decipher try blacking out all the words in your text book and read only the equations to see what I mean a If you only write down equations you will get very little credit 6 Use words to define symbols that are used 7 Before you begin your detailed development provide a brief overview or plan of what you are going to do 8 At the end of your development comment on what the result tells you This is what engineering is all about Here s an example Problem Find the maximum height of a projectile following a parabolic trajectory Well Written Solution Plan Find an expression for the trajectory height vs time Peak occurs when derivative is zero so differentiate trajectory and set equal to zero Solve resulting equation for time value to find the location of the peak tp Substitute time value into trajectory expression to find the maximum height hp height dh dt 0 hp tp t 2 Details A general parabolic trajectory is given by h t at2 bt c Since the peak occurs when the derivative is zero we differentiate h t with respect to time to give dh dt 2at b Now setting this to zero gives dh dt 0 gives 2atp b 0 Note that tp is used here not t Now solving for the value of t that solves this gives tp b 2a Notice that a larger b gives a later peak a smaller a gives a later peak Notice also that the case when the trajectory starts at t 0 and the peak is at some positive time i e tp 0 requires that b 0 is negative and that a 0 Poorly Written Solution Equation to work with is h t at2 bt c dh dt 2at b 2atp b 0 tp b 2a


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