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# If the radius of a sphere is measured as $70$ cm with an error of $0.03$cm, then find the approximate error in calculating its surface area

Toolbox:
• Let $y=f(x)$
• $\Delta x$ denote a small increment in $x$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)\ $$\Delta x • Surface area=S=4\pi r^2 Step 1: Radius of the sphere =70 cm Error in measurement =\Delta r = 0.03 cm Surface area=S=4\pi r^2 \Rightarrow \large\frac{dS}{dr}$$=8\pi r$ [Differentiating with respect to r]
Step 2:
$\Delta S=\large\frac{ds}{dr}$$\times \Delta r$
$\quad\;\;=8\pi r\times 0.03$
$\quad\;\;=8\pi \times 70 \times 0.03$
$\quad\;\;= 52.8\; cm^2$