Question

1.     The diameter of a right circular cylinder is equal to its altitude. Letting V and S be the respective volume and surface area of the cylinder and 2r the diameter, use differentials to estimate the maximum possible error, relative error, and percentage error in computing

(a)   the volume of this cylinder and

(b)  the surface area of the cylinder

when the measured diameter is 20 cm and the maximum possible error in the measurement is 0 1 cm.

Answer 1

(a) Maximum possible error of the volume

r=10cm

d=h=20cm

$V= \pi r^2h$

$\frac{dV}{dr}=2\pi rh$

$dV=2\pi rhdr$

$dV=(2\pi\times10\times20)0.1$

$dV=40\pi=125.66cm^3$

Relative error of the volume

$=\frac{dV}{V}$

$=\frac{40\pi}{2000\pi}$

$=0.02$

Percentage error of the volume

$= Relative\ error \times 100$

$=0.02\times100\\=2\%$

(b)Maximum possible error of surface area

$S=2\pi rh+2\pi r^2$

$\frac{dS}{dr}=\frac{d}{dr}(2\pi rh)+\frac{d}{dr}(2\pi r^2)$

$dS=(2\pi h+4\pi r)dr$

$dS=(2\pi\times 20+4\pi\times10)0.1$

$dS=80\pi\times0.1=8\pi=25.13cm^2$

$=25.13cm^2$

Relative error of surface area

$\frac{dS}{S}=\frac{8\pi cm^2}{2000\pi cm^2}$

$=0.0133$

percentage error of surface area

$=Relative \ error \times100\\=0.0133\times100\\=1.33\%$