Differential - Aproximation to $\triangle A$ (area variation)

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?

Guess it's just a little error in your calculation. Start with keeping everything in cm and degrees; they are just units. $$0 = \frac{20^2}{2} \times (-1) + 20 \times 80 \times dR$$ Giving the outcome as required.