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(C) The equation of the wave is $y = A \sin (\omega t - kx)$.To find the slope of the curve,we differentiate $y$ with respect to $x$ at a constant tim

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$kA$

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(C) The equation of the wave is $y = A \sin (\omega t - kx)$.To find the slope of the curve,we differentiate $y$ with respect to $x$ at a constant time $t$:$\frac{dy}{dx} = A \cos (\omega t - kx) \cdot (-k) = -kA \cos (\omega t - kx)$.At point $B$,the wave crosses the $x$-axis,meaning $y = 0$. Thus,$\sin (\omega t - kx) = 0$,which implies $\cos (\omega t - kx) = \pm 1$.For a wave traveling in the $+ve$ $x$-direction,at the point where the wave crosses the axis moving downwards (as seen in the figure),the slope is negative. Specifically,the magnitude of the slope is $|\frac{dy}{dx}| = |-kA \cos (\omega t - kx)| = kA$.Alternatively,using the relation between particle velocity $v_p$ and wave velocity $v$: $v_p = -v \cdot (	ext{slope})$. At point $B$,the particle velocity is at its maximum magnitude,$|v_p| = \omega A$. Since $v = \frac{\omega}{k}$,the magnitude of the slope is $|\frac{dy}{dx}| = \frac{|v_p|}{v} = \frac{\omega A}{\omega / k} = kA$.

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

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