hel Posted April 13, 2011 Report Share Posted April 13, 2011 Hi,I was working through an exam problem(May 2009 P1) and I have a question....Basically the problem is worth 6 marks. You're given a function, f(x)=e^x cosx and need to find the gradient of the normal to the curve of f at x=π (pi).So I used the product rule to differentiate and then substituted π. I simplified the answer which gave -e^π and I left that as an answer. However it says that you should take negative reciprocal -1/f'(π) which results in the gradient being 1/e^π. Why do we need to take the negative reciprocal? Thanks in advance. Reply Link to post Share on other sites More sharing options...
Keel Posted April 13, 2011 Report Share Posted April 13, 2011 Yes that is correct, you must take the negative reciprocal of the gradient to find the gradient of the normal. Remember just diferentiating it will give you the gradient of the tangent. Gradient of tangent = negative reciprocal gradient of normal and vice versa. 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted April 13, 2011 Report Share Posted April 13, 2011 (edited) what you got (-e^ π) is the gradient of the tangent of the curve. remember, dy/dx at a point is the gradient of the tangent of y the normal is perpendicular to the tangent. thus m(normal)*m(tangent)=-1 EDIT: meh, Keel beats me to it Edited April 13, 2011 by Desy ♫ 1 Reply Link to post Share on other sites More sharing options...
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