Figure below shows the wave $\mathrm{y} = \mathrm{A} \sin (\mathrm{\omega t}-\mathrm{kx})$ at any instant travelling in the +ve x-direction. What is the slope of the curve at B?

 

The particle velocity is maximum at B and is given by

$\frac{\mathrm{dv}}{\mathrm{dt}} = {\left({\mathrm{v}}_{\mathrm{p}}\right)}_{\max } = \mathrm{\omega A}$

Also wave velocity is $\frac{\mathrm{dx}}{\mathrm{dt}} = \mathrm{v} = \frac{\mathrm{\omega }}{\mathrm{k}}$

So slope $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\left({\mathrm{v}}_{\mathrm{p}}\right)}_{\max }}{\mathrm{v}} = \mathrm{kA}$