MT171 - FEEDBACK on QUESTION SHEET 6 - 2018/19 MARKING COMMENTS Questions marked are: Q1(c)/Q2(b) and (d)/ Q3 (a) or (b)/ Q4(b) and (c)/Q5(a) Each question will be classified as 0-if it is mostly incorrect; 1-if it is about half correct; 2-if it is mostly correct. The maximum you can get is 10. For continuous assessment: if you didn't do enough work or gave in your work after the deadline, there will be a box encircling a zero at the top of page 1 of your work. Otherwise your work will have gained a 'one' and nothing will appear written. Comment 1 (C1) You need to improve the clarity of your presentation. Comment 2 (C2) Explain what you are doing. Comment 3 (C3) Use correct notation! An equation needs to be written here. This needs a LHS (left hand side) and a RHS (right hand side)! Comment 4 (C4) Use correct mathematical language. Comment 5 (C5) State your conclusion! Q1 (c) Comment 1.1 (C1.1) Explain what you are doing: you differentiate the equation implicitly with respect to x. Comment 1.2 (C1.2) You need to use the chain rule to differentiate implicitly both sides of the equation. For example in the RHS you have darctan(y /x) _darctan(u) du for u= >. See solutions for more details. dx du dx x Comment 1.3 (C1.3) Learn the derivatives of standard forms like (arctan(u))! Q2 (b) and (d) Comment 2.1 (C2.1) To prove the expressions given, start in one of the sides of the equation and finish by getting the other side of the equation. Comment 2.2 (C2.2) eix-e" -ix Use Euler's formula to define sin x as sin(x) = and then the proof is straightforward. ! 2i See solutions.
Q3 (a or b) Comment 3.1 (C3.1) Add and subtract the two given equations so that you have a new system of two equations from where it is easy to determine the two unknowns. Comment 3.2 (C3.2) Exp(x) is always positive so if you get a negative value for it, as exp(x) =- 1, this means that there is not a solution satisfying such equation. In case of part (a) you only have one possible solution but you need to give the argument of why you discard the other one. Comment 3.3 (C3.3) Use the result from Question 2(a) to simplify your system, when adding and subtracting the two equations. Q4 (b) and (d) Comment 4.1 (C4.1) In part (b) you have to use the chain rule, dx darcsech(u) = darcsech(u) du , du dx for u= ^. ! a darcsech(u) Unfortunately = du is not a standard form: this means you need to deduce it even if not asked. To do that, proceed as in lectures for the inverse hyperbolic functions. From the definition of arcsech, you have „y = arcsech(u) - sechy ) =u. Then you need to differentiate implicitly with respect to x the equation to get the derivative you want dy . See solutions for all details. ?du „sech(y) =u