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Michael Sullivan, Kathleen Miranda

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Chapter 3, Problem 59

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The radius of a spherical ball is found by measuring the volume of the sphere (by finding how much water it displaces). It is determined that the volume is 40 cubic centimeters $\left(\mathrm{cm}^{3}\right),$ with a tolerance of $1 \% .$ Find the percentage error in the radius of the sphere caused by the error in measuring the volume.

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Transcript

we’re measuring a sphere the radius of the sphere by measuring its volume and we’ll see that. That’s actually a good way of measuring the radius. If we can do it that way, measuring the radius with a tape measure is always kind of hard because you need to get it right around the circumference of the great circle of this, not like up here, somewhere off. Um so that can be a little bit challenging, but this is probably a good way to do it. And so we can measure the volume and we say that they get 40 cubic centimeters plus or minus 1% which is 40 plus or minus zero point for cubic centimeters. And the volume is of course 4/3 pi r cubed and the nominal the nominal radius then, given the center of our best estimate here is 2.12 cm now, um we can again taking the equation for the volume differentiating, we get DV equals four pi r squared er and so D r is DV over four pi r squared. And so delta are, is going to be approximately delta v over four pi uh are square. And so we know delta v is this value here and we know are not this is we are not…