Maya measures the radius and the height of a cone with 1% and 2% errors, respectively. She use these data to calculate the volume of the cone. What can Maya say about her percentage error in her volume calculation of the cone?

1 Answer

V_"actual"=V_"measured" pm4.05%, pm .03%, pm.05%Vactual=Vmeasured±4.05%,±.03%,±.05%

Explanation:

The volume of a cone is:

V=1/3 pir^2hV=13πr2h

Let's say we have a cone with #r=1, h=1. The volume is then:

V=1/3pi(1)^2(1)=pi/3V=13π(1)2(1)=π3

Let's now look at each error separately. An error in rr:

V_"w/ r error"=1/3pi(1.01)^2(1)Vw/ r error=13π(1.01)2(1)

leads to:

(pi/3(1.01)^2)/(pi/3)=1.01^2=1.0201=>2.01%π3(1.01)2π3=1.012=1.02012.01% error

And an error in hh is linear and so 2% of the volume.

If the errors go the same way (either too big or too small), we have a slightly bigger than 4% error:

1.0201xx1.02=1.040502~=4.05%1.0201×1.02=1.0405024.05% error

The error can go plus or minus, so the final result is:

V_"actual"=V_"measured" pm4.05%Vactual=Vmeasured±4.05%

We can go further and see that if the two errors go against each other (one is too big, the other too small), they will very nearly cancel each other out:

1.0201(0.98)~=.9997=>.03%1.0201(0.98).9997.03% error and

(1.02)(.9799)~=.9995=>.05%(1.02)(.9799).9995.05% error

And so, we can say that one of these values is correct:

V_"actual"=V_"measured" pm4.05%, pm .03%, pm.05%Vactual=Vmeasured±4.05%,±.03%,±.05%