The radius and the height of a right circular | vedclass

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(B) The total surface area $S$ of a right circular cone is given by $S = \pi r^2 + \pi r l$,where $l = \sqrt{r^2 + h^2}$.Given $r = 7$,$h = 7$,then $l

Vedclass · Accepted Answer

$(0.616)(\sqrt{2}+1)$

Vedclass · Answer

(B) The total surface area $S$ of a right circular cone is given by $S = \pi r^2 + \pi r l$,where $l = \sqrt{r^2 + h^2}$.Given $r = 7$,$h = 7$,then $l = \sqrt{7^2 + 7^2} = 7\sqrt{2}$.The error in measurement is $\Delta r = \Delta h = 0.002 	imes 7 = 0.014$.$S = \pi r^2 + \pi r \sqrt{r^2 + h^2}$.Taking the differential $dS = \frac{\partial S}{\partial r} dr + \frac{\partial S}{\partial h} dh$.$\frac{\partial S}{\partial r} = 2\pi r + \pi \sqrt{r^2 + h^2} + \pi r \frac{r}{\sqrt{r^2 + h^2}} = 2\pi r + \pi l + \frac{\pi r^2}{l}$.$\frac{\partial S}{\partial h} = \pi r \frac{h}{\sqrt{r^2 + h^2}} = \frac{\pi r h}{l}$.Substituting $r=7, h=7, l=7\sqrt{2}$ and $dr=dh=0.014$:$\frac{\partial S}{\partial r} = 14\pi + 7\sqrt{2}\pi + \frac{49\pi}{7\sqrt{2}} = 14\pi + 7\sqrt{2}\pi + \frac{7\pi}{\sqrt{2}} = 14\pi + 7\sqrt{2}\pi + 3.5\sqrt{2}\pi = 14\pi + 10.5\sqrt{2}\pi$.$\frac{\partial S}{\partial h} = \frac{49\pi}{7\sqrt{2}} = 3.5\sqrt{2}\pi$.$dS = (14\pi + 10.5\sqrt{2}\pi)(0.014) + (3.5\sqrt{2}\pi)(0.014) = (14\pi + 14\sqrt{2}\pi)(0.014) = 14\pi(1+\sqrt{2})(0.014) = 0.196\pi(1+\sqrt{2})$.Using $\pi \approx 3.14$,$0.196 	imes 3.14 \approx 0.61544 \approx 0.616$.Thus,the error is $(0.616)(\sqrt{2}+1)$.

The radius and the height of a right circular solid cone are measured as $7 \text{ ft}$ each. If there is an error of $0.002 \text{ ft}$ for every foot in measuring them,then the error in the total surface area of the cone (in $\text{sq. ft}$) is:

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