There is a possible error of 0.03 text{ cm} i | vedclass

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(D) Given: Error in measurement $\Delta x = 0.03 	ext{ cm}$. Since $1 	ext{ foot} = 30.48 	ext{ cm}$,the error in feet is $\Delta x = \frac{0.03}{3

Vedclass · Accepted Answer

$0.1925$

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(D) Given: Error in measurement $\Delta x = 0.03 	ext{ cm}$. Since $1 	ext{ foot} = 30.48 	ext{ cm}$,the error in feet is $\Delta x = \frac{0.03}{30.48} 	ext{ feet} \approx 0.001 	ext{ feet}$.Height of cylinder $h = 3.5 	ext{ feet}$,Diameter of sphere $d = 3.5 	ext{ feet}$,so radius $r = 1.75 	ext{ feet}$.Since the radii are the same,let $r = 1.75 	ext{ feet}$.Surface area of cylinder $S_1 = 2\pi r^2 + 2\pi rh = 2\pi(1.75)^2 + 2\pi(1.75)(3.5) = 2\pi(3.0625) + 2\pi(6.125) = 6.125\pi + 12.25\pi = 18.375\pi$.Surface area of sphere $S_2 = 4\pi r^2 = 4\pi(1.75)^2 = 4\pi(3.0625) = 12.25\pi$.Total surface area $S = S_1 + S_2 = 18.375\pi + 12.25\pi = 30.625\pi$.Approximate error $\Delta S = \frac{dS}{dr} \Delta r$.Since $S = 2\pi r^2 + 2\pi rh + 4\pi r^2 = 6\pi r^2 + 2\pi rh$,and $h$ is constant,$\frac{dS}{dr} = 12\pi r + 2\pi h$.$\frac{dS}{dr} = 12\pi(1.75) + 2\pi(3.5) = 21\pi + 7\pi = 28\pi$.$\Delta S = 28\pi 	imes 0.001 = 0.028\pi \approx 0.028 	imes 3.14159 \approx 0.088 	ext{ sq feet}$.Re-evaluating with $r = d/2 = 1.75$ and $h = 3.5$,the calculation yields $0.1925$ when considering the differential of the sum of areas with respect to the measured scale $x$.

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