## anonymous one year ago The radius of a 12 inch right circular cylinder is measured to be 4 inches, but with a possible error of ±0.2 inch. What is the resulting possible error in the volume of the cylinder? Include units in your answer

1. anonymous

you just take the +0.2 to the cylinder height 0.2+4, find the volume and find the volume again for 4-0.2 and find the volume again and get the difference

2. anonymous

$Error=\pi r^2(h+0.2) - \pi r^2(h-0.2)$

3. anonymous

i don't know it's calculus and i'm pretty sure it some how involves the derivative of the volume just not too sure in what way

4. anonymous

oh another of those sneaky questions...

5. anonymous

indeed

6. anonymous

Thisis prolly what it's asking the volume different between this|dw:1438420033333:dw|

7. welshfella

yes it wants the difference between the volume when r = 4.2 and volume when r = 4

8. welshfella

pi*4.2^2*12 - pi*4^2*12 = 665 - 603.2

9. IrishBoy123

$$V = \pi \ r^2 \ h$$ $$dV = 2 \pi r \ dr \ h$$ $$dr = ±0.2$$

10. welshfella

oh I see then that will give +/- 60.32 cu ins

11. IrishBoy123

that might be how they want it done [depends on what is being studied by the OP, i suppose]; but there clearly is nothing wrong in plugging in to the volume equation itself as that will give the exact answer.....

12. anonymous

The volume of a right circular cylinder is proportional to the square of it's radius $V \alpha r^2$ This means any change in the radius will a change the volume by the square of the radius Suppose for the radius of 4 inches, we had a volume V1 before $V_{1}=16k$ Now when we change the radius by plus minus 2, we get $V_{2}=(4 \pm 0.2)^2k=(16+0.04 \pm 1.6)k = (16 \pm 1.6)k$ We can ignore the 0.04 as it is quite small Dividng V2 by V1 we get $\frac{V_{2}}{V_{1}}=\frac{16 \pm 1.6}{16}$ Therefore $V_{2}=(1 \pm 0.1) V_{1}$$V_{2}=V_{1} \pm 0.1V_{1}$

13. welshfella

good result

14. anonymous

i fell asleep, sorry. but thank you for your help

15. anonymous

its an open ended question and i need to do it using calculus, does anyone know how to do that that way?

16. phi

See Irish boy's post. $dV = 2 \pi r \ dr \ h$ or if we replace the infinitesimals we have $\Delta V= 2 \pi r h\ \Delta r$ replace the $$\Delta r$$ with the uncertainty $$\pm 0.2$$ and calculate the uncertainty in V

17. phi

**perhaps it is better to say $\Delta V \approx 2 \pi r h\ \Delta r$