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So I've been trying to figure out the problem for about an hour and I cannot figure it out. Question: To study the effect an earthquake has on a structure, engineers look at the way a beam bends when subjected to an earth tremor. The equation $$D = a − a\cos(\frac{\pi h}{2L} )$$$$ 0 ≤ h ≤ L$$ where L is the length of a beam and a is the maximum deflection from the vertical, has been used by engineers to calculate the deflection D at a point on the beam h ft from the ground. Suppose that a 14-ft vertical beam has a maximum deflection of $\frac{1}{3}$ft when subjected to an external force. Using differentials, estimate the difference in the deflection between the point midway on the beam and the point $\frac{1}{10}$ ft above it. (Round your answer to four decimal places.) So what I hav so far is $$h=7$$$$L=14$$$$a=\frac{1}{3}$$$$dD=.1$$ So the way I thought I would get the difference in deflection is by $D'*dD$ So for $D'$ I got $$D'=1-\cos(\frac{\pi h}{2L})+a(-\sin(\frac{\pi h}{2L})*\frac{2L\pi-2h\pi}{(2L)^2})$$ After plugging in each of the values I have I ended up with $.0267$ rounded to four places but the answer is $.0026$ rounded off to four places, am I going on the right way to solve this problem? A step by step solution would be greatly appreciated! (I know I can just plug in the values for 7.1 and 7, but I'm being tested on the question) Thanks In advance!

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You want to find $\delta D$ and you're given $\delta h=0.1$. So differentiate $D$ with respect to $h$,

$$ \frac{dD}{dh}=\frac{d}{dh} a(1-\cos\frac{\pi h}{2L})=\frac{a\pi}{2L}\sin\frac{\pi h}{2L} $$

and plug $a=1/3$, $L=14$, $h=7$, and $\delta h=0.1$ into

$$ \delta D\approx \frac{dD}{dh}\delta h $$

and you will get the correct answer.

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