Answer :
The area of a circle with radius r is given by the formula:
[tex]A=\pi r^2[/tex]If we change the value of r by a small amount Δr, the new value of the area will be:
[tex]\begin{gathered} A+\Delta A=\pi(r+\Delta r)^2 \\ =\pi(r^2+2r\Delta r+\Delta r^2) \end{gathered}[/tex]We can neglect the term Δr^2 since we are assuming that Δr is a very small quantity. Then:
[tex]A+\Delta A=\pi r^2+2\pi r\Delta r[/tex]Substitute A=πr^2 and isolate Δr from the equation:
[tex]\begin{gathered} \Rightarrow\pi r^2+\Delta A=\pi r^2+2\pi r\Delta r \\ \Rightarrow\Delta A=2\pi r\Delta r \\ \Rightarrow\Delta r=\frac{\Delta A}{2\pi r} \end{gathered}[/tex]Assuming that the ideal area of the disk is 1000in^2, calculate the ideal radius of the disk:
[tex]\begin{gathered} A=\pi r^2 \\ \Rightarrow r^2=\frac{A}{\pi} \\ \Rightarrow r=\sqrt[]{\frac{A}{\pi}} \\ \Rightarrow r=\sqrt[]{\frac{1000in^2}{\pi}} \\ \Rightarrow r=17.84124116\ldots in \end{gathered}[/tex]Substitute the value of r as well as the variation on the value of the area ΔA=4in^2 to find the variation in the value of the radius:
[tex]\begin{gathered} \Delta r=\frac{4in^2}{2\pi(17.84124116\ldots in)} \\ =0.03568248232\ldots in \end{gathered}[/tex]Up to 3 significant figures, the variation in the value of the radius must be less than:
[tex]0.0357in[/tex]