In a right circular cone, the radius of the base is half as long as the altitude. If an error of 2% is made in measuring the radius, find the percentage of error made in the computed volume.
Let "h" be the altitude and "r" be the radius of the base of a right circular cone.
Given: "h=r=5" inches and "\\Delta r=\\Delta h=0.02" inches.
We know that the volume of the right circular is "V=\\frac{1}{3}\\pi r^2h"
Now "r=h\\Rightarrow V=\\frac{1}{3}\\pi h^2h=\\frac{1}{3}\\pi h^3"
Require to find approximately the percentage error in the calculated value of the volume.
Taking logarithms on both sides of the equation "V=\\frac{1}{3}\\pi h^3", we get
"lnV=ln\\left [ \\frac{1}{3}\\pi h^3 \\right ]=ln\\frac{\\pi }{3}+ln\\left ( h^3 \\right )=ln\\frac{\\pi }{3}+3ln\\left ( h \\right )"
Taking differentials on both sides, we get
"\\frac{\\Delta V}{V}\\approx 0+\\frac{3\\Delta h}{h}=\\frac{3\\Delta h}{h}"
Multiplying both sides by 100, we get
"\\frac{\\Delta V}{V}\\times 100\\approx \\frac{3\\Delta h}{h}\\times 100"
Substituting "h=5" and "\\Delta h=0.02" , we get
"\\frac{\\Delta V}{V}\\times 100\\approx \\frac{3\\left ( 0.02 \\right )}{5}\\times 100=1.2"
Therefore,
approximately the percentage error in the calculated value of the volume is "1.2"
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