Question #294693

A knuckleball thrown with a different grip than that of the problem above has left/right

position as it crosses the plate given by

f(W) =

(0.625/w2)(1-sin (2.72w+π/2))

Again, use graphical and tabular evidence to make a conjecture about the lim f(w) as w approaches 0 from the right.


1
Expert's answer
2022-02-08T07:19:35-0500

Solution:


Graph shows that when ω0+,f(w)=2.312\omega\rightarrow0^+, f(w)=2.312

f(w)=0.625ω2[1sin(2.72ω+π2)]f(w)=0.625ω2[1cos(2.72ω)]limω0+f(w)=limω0+0.625ω2[1cos(2.72ω)]=limω0+0.625(1cos(2.72ω)ω2)   [0/0 form]=limω0+0.625(2.72sin(2.72ω)2ω)  [L Hopital Rule]f(w)=\dfrac{0.625}{\omega^2}[1-\sin(2.72\omega+\dfrac {\pi}2)] \\\Rightarrow f(w)=\dfrac{0.625}{\omega^2}[1-\cos(2.72\omega)] \\\Rightarrow \lim_{ \omega\rightarrow0^+}f(w)=\lim_{ \omega\rightarrow0^+}\dfrac{0.625}{\omega^2}[1-\cos(2.72\omega)] \\ =\lim_{ \omega\rightarrow0^+}0.625(\dfrac{1-\cos(2.72\omega)}{\omega^2}) \ \ \ [0/0\ form] \\=\lim_{ \omega\rightarrow0^+}0.625(\dfrac{2.72\sin(2.72\omega)}{2\omega}) \ \ [L\ 'Hopital \ Rule]

=limω0+0.6252×2.722(sin(2.72ω)2.72ω)=0.6252×2.722(1)=2.312\\=\lim_{ \omega\rightarrow0^+}\dfrac{0.625}2\times2.72^2(\dfrac{\sin(2.72\omega)}{2.72\omega}) \\=\dfrac{0.625}2\times2.72^2(1) \\=2.312

Thus, we get same value, i.e., 2.312 from both methods.


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