Question

Question #99905

Answer 1

`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`