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Sin(theta) = dy/dx ? 
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#1
Jun2813, 01:49 PM

P: 31

In preparing for an acoustics course, I ran across the following sentence which confused me:
"If (theta) is small, sin(theta) may be replaced by [partial]dy/dx." I expected to see sin(theta) = (theta) so this threw me off. This came up in the derevation of the one dimensional wave equation after approximating (by Taylor series) the transverse force on a mass element of a tensioned string with [partial]d(Tsin(theta))/dx. The approximation in question thus gave T*([partial]d2y/dx2)*dx. In the original setup, x and y are cartesian axis in physical 2D space and (theta) is the angle the string (with tension T) makes from the xaxis after displacement from equalibrium. I've never seen sine approximated by dy/dx before and was hoping somebody might shed some light for me :) 


#2
Jun2813, 03:12 PM

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PF Gold
P: 7,581

When ##\theta## is small, ##sin\theta \approx \tan\theta = \frac{dy}{dx}##. In the derivation for the vibrating string, ##\theta## is the slope angle of the string.



#3
Jun2913, 09:43 AM

P: 31

Thank you! :)



#4
Jun2913, 11:51 AM

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P: 7,034

Sin(theta) = dy/dx ?
sin(θ) = Δy / sqrt(Δy^2 + Δx^2). For small θ, Δy is small compared to Δx, so
sin(θ) ≈ Δy / sqrt(0 + Δx^2) = Δy / Δx 


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