1. Finding maximum error

We measure the radius r of a circle and obtain that that r=9 centimeters. We use this value to compute the area of the circle. We know that the error in our measurement is at most 1.6%. Use the linear approximation to the area A(r)=pi*r^2, at r=9 to estimate the maximum error in our computed value for the area.

I get (162/125) * pi as my answer. :/
Is the equation not 2 * pi * 9 * .016?

2. Originally Posted by Velvet Love
We measure the radius r of a circle and obtain that that r=9 centimeters. We use this value to compute the area of the circle. We know that the error in our measurement is at most 1.6%. Use the linear approximation to the area A(r)=pi*r^2, at r=9 to estimate the maximum error in our computed value for the area.

I get (162/125) * pi as my answer. :/
Is the equation not 2 * pi * 9 * .016?
The area of a circle of radius r is:

$A(r)=\pi r^2$

for a small error $\delta$ in the radius we have:

$A(r+\delta)\approx A(r)+\delta \frac{dA}{dr}=A(r)+\delta (2\pi r)$

Thus the error is the area is:

$A(r+\delta)-A(r)=2\pi r \delta$

In this case $r=9$ and $\delta=0.016 \times r=0.016 \times 9$ cm (it's $1.6\%$ of $r$ )

CB

3. A slightly different way of looking at this. $A(r)= \pi r^2$ so the derivative is $A'(r)= 2\pi r$. At r= 9, $A(9)= 81\pi$ and [tex]A'(9)= 18\pi$. The tangent line at r= 9, [tex]18\pi(r- 9)+ 81\pi$ gives the linear approximation. The "change" is $18\pi (r- 9)$. 1.6% of 9 is 0.144 so evaluate $18\pi (r-9)$ for r= 9.144.

There is, by the way, an engineers "rule of thumb" that is based on this: When adding (or subtracting) quantities add their errors. When multiplying (or dividing) add their percentage errors. Here we are multiplying r times r. since the percentage error for r is 1.4%, the percentage error for $r^2$, and so for $\pi r^2$ would be 1.4+ 1.4= 2.8%