The central angle of a cirular sector is $80°$. It is desired to reduce it by $1°$. By how much should the radius of the sector be increased so that the area will remain unchanged if the original length of the radius is $20cm$?
Let $A$ be the area of the sector if function of the angle $\theta rad$ and the radius $R$. Then, we have that $A(\theta, R) = \frac{\theta R^2}{2} $. I want to aproximate my $\triangle A$, that is, the variation of the area, by
$$ dA = \frac{\partial A}{\partial \theta} d\theta + \frac{\partial A}{\partial R}dR$$
We have that $dA = 0$ and we want to find $dR$
I've found that:
• $80° = \frac{4 \pi}{9} rad$;
• $R = \frac{45}{\pi} cm$
•$1° = \frac{\pi}{180} rad$
Using this, I couldn't get the answer given ($1/8 cm$). Maybe I am thinking wrong.
Can you help me?
Thanks in advance!