Question
The radius of a sphere shrinks from 10 to 9.8cm. Find approximately the decrease in its volume.
✓
Answer
Let $\text{x}=10,\\\text{x}+\triangle\text{x}=9.8$
$\triangle\text{x}=9.8-\text{x}$
$=9.8-10$
$\triangle\text{x}=-0.2$
$\text{y}=\frac{4}{3}\pi\text{x}^3$
$\frac{\text{dy}}{\text{dx}}=4\pi\text{r}^2$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}\Rightarrow10}=4\pi(10)^2$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}\Rightarrow10}=400\pi\ \text{cm}^2$
$=400\pi\times(-0.2)$
$\triangle\text{y}=-80\pi\ \text{cm}^3$
so, approximate decrease in volume is $80\pi\ \text{cm}^3$
$\triangle\text{x}=9.8-\text{x}$
$=9.8-10$
$\triangle\text{x}=-0.2$
$\text{y}=\frac{4}{3}\pi\text{x}^3$
$\frac{\text{dy}}{\text{dx}}=4\pi\text{r}^2$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}\Rightarrow10}=4\pi(10)^2$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}\Rightarrow10}=400\pi\ \text{cm}^2$
$=400\pi\times(-0.2)$
$\triangle\text{y}=-80\pi\ \text{cm}^3$
so, approximate decrease in volume is $80\pi\ \text{cm}^3$
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