Damien R. Posted January 28, 2012 Report Share Posted January 28, 2012 (edited) Let f(x) = excosx. Find the gradient of the normal to the curve of 'f' at x=π.How would you go about answering this? Any help would be greatly appreciated! Edited January 28, 2012 by Damien R. Reply Link to post Share on other sites More sharing options...
dessskris Posted January 28, 2012 Report Share Posted January 28, 2012 f'(x) at that point gives the gradient of the tangent.gradient of tangent * gradient of normal = -1 Reply Link to post Share on other sites More sharing options...
Damien R. Posted January 28, 2012 Author Report Share Posted January 28, 2012 So the gradient of the normal is -1/excosx ..? If this is correct, am I to then plug it into the equation "y-y1=m(x-x1)" as 'm'? However, how am I supposed to calculate with 'e' in the equation - is it being used as a costant or just an algebraic value? I am stuck on this past paper question, I will admit. Reply Link to post Share on other sites More sharing options...
dessskris Posted January 28, 2012 Report Share Posted January 28, 2012 e is the euler number whose value you can find from your calculator just press 'e' when asked for the normal, you always have to find the gradient of the tangent first. so you differentiate f(x) to find f'(x), and find its value at x=π i.e. find f'(π). after that, since normals are always perpendicular to tangents, gradient of tangent * gradient of normal = -1 so you can find the gradient of the normal from there. Reply Link to post Share on other sites More sharing options...
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