Question

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?

Answer 1

`V_"actual"=V_"measured" pm4.05%, pm .03%, pm.05%`

Explanation:

The volume of a cone is:

`V=1/3 pir^2h`

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

`V=1/3pi(1)^2(1)=pi/3`

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

`V_"w/ r error"=1/3pi(1.01)^2(1)`

leads to:

`(pi/3(1.01)^2)/(pi/3)=1.01^2=1.0201=>2.01%` error

And an error in `h` 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%` error

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

`V_"actual"=V_"measured" pm4.05%`

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%` error and

`(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%`