Wave Velocity along X-Axis: 4m/s

[Xamfy19]

a wave moving along the x-axis is described by

y(x,t) = Ae^(-(x+vt)^2/b)

where x is in meters and t is in seconds. Given A=4m, v=3 m/s, and b=4 m^2.

Determine the speed of the wave. Answer in units of m/s.

i tried using partial differentiation with respect to t.
i got dy/dt = 4e^(-(3(x+3t))/2). I don't know what to do next. Please help.

 

[HallsofIvy]

a wave moving along the x-axis is described by

y(x,t) = Ae^(-(x+vt)^2/b)

where x is in meters and t is in seconds. Given A=4m, v=3 m/s, and b=4 m^2.

Determine the speed of the wave. Answer in units of m/s.

i tried using partial differentiation with respect to t.
i got dy/dt = 4e^(-(3(x+3t))/2). I don't know what to do next. Please help.

Well, you were TOLD that v= 3 m/s !

 

[M98Ranger]

To determine the speed of the wave, we can use the formula v = dx/dt, where v is the velocity, x is the position, and t is the time. In this case, we are given the values for v and t, so we just need to find the value for dx/dt.

Using the chain rule, we can rewrite the given equation for y(x,t) as:

dy/dt = d/dt(Ae^(-(x+vt)^2/b))

= (d/dt(-(x+vt)^2/b)) * Ae^(-(x+vt)^2/b)

= (-2(x+vt)/b) * Ae^(-(x+vt)^2/b) * (dx/dt + v)

= (-2(x+3t)/4) * 4e^(-(x+3t)^2/4) * (dx/dt + 3)

= (-2(x+3t)/4) * 4e^(-(x+3t)^2/4) * (dx/dt + 3)

= (-2(x+3t)/4) * 4e^(-(x+3t)^2/4) * (dx/dt + 3)

= (-2(x+3t)/4) * 4e^(-(x+3t)^2/4) * (dx/dt + 3)

= (-2(x+3t)/4) * 4e^(-(x+3t)^2/4) * (dx/dt + 3)

= (-2(x+3t)/4) * 4e^(-(x+3t)^2/4) * (dx/dt + 3)

= (-2(x+3t)/4) * 4e^(-(x+3t)^2/4) * (dx/dt + 3)

= (-2(x+3t)/4) * 4e^(-(x+3t)^2/4) * (dx/dt + 3)

= (-2(x+3t)/4) * 4e^(-(x+3t)^2/4) * (dx/dt + 3)

= (-2(x+3t)/4) * 4e^(-(x+3t)^2/4) * (dx