anonymous
  • anonymous
The radius R of a spherical ball is measured as 14 in. with what accuracy must the radius R be measured to gaurantee an error of at most 2in^3 in the calculated volume I'm doing it as dV=4pir^2dr and I get 2/784pi but apparently that is incorrect. Can someone tell me how to do this type of calculation as well as propagated error and just accuracy in general... thanks
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1333375090395:dw|
anonymous
  • anonymous
nahh :\
anonymous
  • anonymous
The equation for the volume of a sphere is V = (4/3)*pi*r^3 so the radius R is being cubed, so if R = 14 +/- 3'sqrt(2) then R^3 would be 2744 +/- 2 in.. Am I misunderstanding the question? I don't understand the need for differentiation.

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anonymous
  • anonymous
its the linear approx section that wont fly lol I'm just trying to understand the section is all.
anonymous
  • anonymous
The approx. is d/dx f(x)*dx (dx being the change). Plug in the numbers into that formula.
anonymous
  • anonymous
can you type that with respect to the problem i dont understand what d/dx is in this problem my professor doesnt use d/dx
anonymous
  • anonymous
That just means the derivative. So, the derivative of f(x) (which is the formula for the sphere which @AsianDuck wrote out above) multiplied by the change = the error.
anonymous
  • anonymous
I dont get it. I dont have change in x and if I did I would be essentially plugging all of the unknowns in to the derivative it would be 4pi(14)^2(1/392pi) * (4/3pi*(14)^3*(1/392pi) is essentially waht your are saying
anonymous
  • anonymous
You have the radius, so you can find the area of the sphere and you have the error. All you are missing is the change that produces the error. You just need to plug in. Derivative of the area * change = 2
anonymous
  • anonymous
that is 1/392pi
anonymous
  • anonymous
I haven't actually done the problem.
anonymous
  • anonymous
gah this section is going to be the effing death of me.
anonymous
  • anonymous
isnt accuracy just dR? 1/392pi is dR by solving for 2=4pi(14)^2(dr)
anonymous
  • anonymous
I'm not sure where you got 4pi(14)^2, but if that is the derivative, then yes, that would be the correct equation, with dr being the change.
anonymous
  • anonymous
on the webwork solutions to type in the solution it says dR= right before the blank type in box
anonymous
  • anonymous
dV= 4/3 * 3r^2 (dr)
anonymous
  • anonymous
forgot a pi in there
anonymous
  • anonymous
Are you estimating the change in the area in class?
anonymous
  • anonymous
this is volume
anonymous
  • anonymous
For example if you were asked to find the change in the area as the radius of a circle increased from 3 to 3.04, then you would find the derivative of the area of a circle which is 2PiR and plug in. So, the equation would be 2Pi(3<=the original number)(.04 <=the change)= the error, which in this case is .754. You just need to do this with the volume.
anonymous
  • anonymous
so your example up there = dA i have dV its 2 I'm solving for the error in measurement of the radius. I must be doing something wrong. Thanks for your help I'm just not getting it.
anonymous
  • anonymous
dA and dV are sort of the same thing in this problem, you need to find the derivative of both.
anonymous
  • anonymous
essentially what I am saying is I did that thats what I typed in the question up top its just not giving me the correct answer either webwork is wrong or we're doing it wrong.
anonymous
  • anonymous
dV=4pir^2dr 2=784pidr 1/392pi=dr
anonymous
  • anonymous
dV/dr is only correct for infinitesimal changes in V and r, I think.
anonymous
  • anonymous
The derivative of the formula for a sphere is 4Pi*r^2, which you have, so just plug in your radius in order to get the volume. So... 4*Pi*(14^2)= 784Pi * X = 2 This is 1/392 Pi. It does look like you are correct, it must be an error in your book.
anonymous
  • anonymous
^^ that would be nice! Thanks for sticking through it haha
anonymous
  • anonymous
Sure thing =)
anonymous
  • anonymous
just an example of dy/dx being invalid.. y = x^2 dy/dx = 2x dy = 2x(dx) if we say we start at x=1, and induce a change to x=2 (dx=1) dy = 2(1)(1) = 2 but in fact, (2)^2 - (1)^2 = 3 So I don't think it is correct using dr for a change of 2inches because that is a relatively large change so dV/dr would not hold.
anonymous
  • anonymous
I'll probably just use V' and r' because I honestly dont have time to learn what these stupid symbols mean anyways.... I got 100% on the first exam and some how I am not getting this stuff and will probably bomb this one....
anonymous
  • anonymous
good luck
anonymous
  • anonymous
thank you
anonymous
  • anonymous
wups didnt give you a medal

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