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a.) The maximal relative error of the volume of a ball is allowed within $1\%$. What is the maximal relative error measuring the radius of the ball?

b.) Given the equation $x^2-40x+1=0$, find its roots to five significant digits using $\sqrt{399} = 19.975$, correctly rounded to five digits.

What I did and know for a is that the volume of a sphere is $V = 4/3\pi * R^3$ and the only term that can have a relative error is the R term. So, I let $dV$ be the relative error of the volume and $dR$ the relative error of the radius then

$dV = dR + dR + dR = 3dR$

$dV/3 = dR$

$1\%/3 = dR$

and I am completely lost in b.

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I think a more useful result would be $\sqrt{15960025}=3995$. – copper.hat Jan 31 '13 at 5:44
You need $\frac{dV}{dR}=4\pi R^2$, so $dV=4\pi R^2\,dR$. This follows fom the formula for volume, $V=\frac{4}{3}\pi R^3$. Now calculate $\frac{dV}{V}$, express it in terms of $R$ and $\frac{dR}{R}$. – André Nicolas Jan 31 '13 at 5:57
Thank you @AndréNicolas ! – Lays Jan 31 '13 at 6:03
Everything will turn out fine, but it is probably expected that you will go through the reasoning. – André Nicolas Jan 31 '13 at 6:04
@Lays If you get stuck, I have included a detailed explanation of a) in my answer. Don't look until you have hit a road block! – Rustyn Jan 31 '13 at 6:13
up vote 1 down vote accepted

Solve for $x$ over the real numbers:
$x^2-40 x+1 = 0$
Solve the quadratic equation by completing the square.
Subtract $1$ from both sides:
$x^2-40 x = -1$
Take one half of the coefficient of $x$ and square it, then add it to both sides.
Add $400$ to both sides:
$x^2-40 x+400 = 399$
Factor the left hand side.
Write the left hand side as a square:
$(x-20)^2 = 399$
Eliminate the exponent on the left hand side.
Take the square root of both sides:
$x-20 = \sqrt{(399)} \text{ or } x-20 = -\sqrt{(399)}$
Look at the first equation: Solve for $x$.
Add $20$ to both sides:
$x = 20+\sqrt{(399)} \text{ or } x-20 = -\sqrt{(399)}$
Look at the second equation: Solve for x.
Add $20$ to both sides:
$$ x = 20+\sqrt{(399)} \text{ or } x = 20-\sqrt{(399)} $$

For question a): $$ V = \frac{4}{3} \pi R^3 \Rightarrow dV = 4\pi R^2 dR \Rightarrow \frac{dV}{V} = \frac{4\pi R^2dR}{\frac{4}{3} \pi R^3} \le .01 \Rightarrow \frac{3dR}{R} \le .01 \Rightarrow \frac{dR}{R} \le \frac{1}{300} \Rightarrow \max{\left(\frac{dR}{R}\right)} = \frac{1}{300} $$

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I believe OP said he was lost in b, so I'm not sure what you mean by he is incorrect... also, there is an extra $(399)$ in your 5th last line. – Calvin Lin Jan 31 '13 at 5:46
@CalvinLin I misread, it is fixed – Rustyn Jan 31 '13 at 5:47
Thank you very much for this great explanation for problem b! And do you know if I did question a correctly? – Lays Jan 31 '13 at 5:48
Rus, well I know the formula for the sphere and from that I knew only R will have a relative error. am I wrong on that? Should I start over? – Lays Jan 31 '13 at 5:57
@Lays you should come up with an expression for the relative error of volume, set up your inequality, and then try to determine something about relative error of the radius. – Rustyn Jan 31 '13 at 6:00

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