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Apply Taylor series expansion with f(x) = (Tsin 0)x so that (Tsin 0)x+dx = (Tsin0)x+ a(T sine) -.dx + ... ax and df, = [(T sine), + (T sine) .dx + ...- (T sine), дх ⇒df, y = a(T sino) ax .dx, from the first non-vanishing term. When 0 is small, sin 0 dy/dx and the net transverse force on the element is . ду df₁ = x (² 2 (Tox) dx əx 1dx = To²y 2x² .dx (as T is constant) (1.1)

Apply Taylor series expansion with f(x) = (Tsin 0)x so that (Tsin 0)x+dx = (Tsin0)x+ a(T sine) -.dx + ... ax and df, = [(T sine), + (T sine) .dx + ...- (T sine), дх ⇒df, y = a(T sino) ax .dx, from the first non-vanishing term. When 0 is small, sin 0 dy/dx and the net transverse force on the element is . ду df₁ = x (² 2 (Tox) dx əx 1dx = To²y 2x² .dx (as T is constant) (1.1)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Apply Taylor series expansion with f(x) = (Tsin 0)x so that (Tsin 0)x+dx = (Tsin0)x+ a(T sine) -.dx + ... ax and df, = [(T sino), + (Tsine) .dx + ...- (T sine), дх ⇒df, y = a(T sine) ax .dx, from the first non-vanishing term. When 0 is small, sin 0 dy/dx and the net transverse force on the element is . ду ax df₁ = x (₁ T 1.dx = 70²y Əx² .dx (as T is constant) (1.1)
Transcribed Image Text:Apply Taylor series expansion with f(x) = (Tsin 0)x so that (Tsin 0)x+dx = (Tsin0)x+ a(T sine) -.dx + ... ax and df, = [(T sino), + (Tsine) .dx + ...- (T sine), дх ⇒df, y = a(T sine) ax .dx, from the first non-vanishing term. When 0 is small, sin 0 dy/dx and the net transverse force on the element is . ду ax df₁ = x (₁ T 1.dx = 70²y Əx² .dx (as T is constant) (1.1)
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