Question #99905

1 Answer
Dec 15, 2016

#e^-pi#

Explanation:

The gradient (slope) of a tangent to this curve is the derivative:
#f'(x) = e^x*(-sin x)+e^x*cos x = e^x(cosx-sinx)#
At point #x=pi# this slope equals to
#m = e^pi*(cos pi - sin pi)=-e^pi#
The gradient of a normal to a curve #n# at the same point is related with a gradient of a tangent in a simple equation:
#m*n = -1#.

Therefore, #n=-1/(-e^pi) = e^-pi#