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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
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you just take the +0.2 to the cylinder height 0.2+4, find the volume and find the volume again for 40.2 and find the volume again and get the difference

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[Error=\pi r^2(h+0.2)  \pi r^2(h0.2)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh another of those sneaky questions...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thisis prolly what it's asking the volume different between thisdw:1438420033333:dw

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0yes it wants the difference between the volume when r = 4.2 and volume when r = 4

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0pi*4.2^2*12  pi*4^2*12 = 665  603.2

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3\(V = \pi \ r^2 \ h\) \(dV = 2 \pi r \ dr \ h\) \(dr = ±0.2\)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0oh I see then that will give +/ 60.32 cu ins

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3that 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.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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 \[V_{2}=(1 \pm 0.1) V_{1}\]\[V_{2}=V_{1} \pm 0.1V_{1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i fell asleep, sorry. but thank you for your help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its an open ended question and i need to do it using calculus, does anyone know how to do that that way?

phi
 one year ago
Best ResponseYou've already chosen the best response.1See 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

phi
 one year ago
Best ResponseYou've already chosen the best response.1**perhaps it is better to say \[ \Delta V \approx 2 \pi r h\ \Delta r \]
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