-z.57 dx formula that estimates the change

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SUMMARY

The discussion focuses on deriving the differential formula for estimating the change in the volume of a sphere, represented by the equation $V=\frac{4}{3}\pi{r}^{3}$. The differential change in volume, denoted as $dV$, is calculated using the formula $dV=4\pi{r}_{0}^{2}dr$, where $r_0$ is the initial radius and $dr$ is the small change in radius. The discussion also references the concept of derivatives, emphasizing that the formula approximates the change in volume as the radius changes from $r_0$ to $r_0 + dr$. This approach is grounded in the principles of calculus.

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karush
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Write a differential formula that estimates the change in the volume
$V=\frac{4}{3}\pi{r}^{3}$ of a sphere when the radius changes from $r_0$ to $r_0+dr$

$$dV=4\pi{r}_{0}^{2}dr$$

this was one of the selections but I didn't know how to account for the
$r_0$ to $r_0+dr$
 
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karush said:
Write a differential formula that estimates the change in the volume
$V=\frac{4}{3}\pi{r}^{3}$ of a sphere when the radius changes from $r_0$ to $r_0+dr$

$$dV=4\pi{r}_{0}^{2}dr$$

this was one of the selections but I didn't know how to account for the
$r_0$ to $r_0+dr$

Recall the formula for a derivative (instantaneous rate of change) from first principles:

$\displaystyle \begin{align*} f'(x) = \lim_{\Delta x \to 0} \frac{f \left( x + \Delta x \right) - f(x)}{\Delta x} \end{align*}$

so this means that as long as $\displaystyle \begin{align*} \Delta x \end{align*}$ is small

$\displaystyle \begin{align*} f'(x) &\approx \frac{f\left( x + \Delta x \right) - f(x)}{\Delta x} \\ f \left( x + \Delta x \right) - f(x) &\approx \Delta x \, f'(x) \\ \Delta f &\approx \Delta x \, f'(x) \end{align*}$

Here treat V as f and r as x. "dr" is just a small change in r.
 

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