Question
Using differentials, find the approximate values of the following:
$\sqrt{49.5}$
$\sqrt{49.5}$
✓
Answer
Define a function $\text{y}=\text{x}^{\frac{3}{2}}$
For x = 4, y = 8
$\text{x}+\triangle\text{x}=3.968\\\Rightarrow\triangle\text{x}=3.968-4=-0.032$
$\frac{\text{dy}}{\text{dx}}=\frac{3}{2}\text{x}^{\frac{1}{2}}$
$\text{dy}=\Big(\frac{3}{2}\text{x}^{\frac{1}{2}}\Big)\text{dx}$
$\Rightarrow\triangle\text{y}\mid_{\text{x}=4}\simeq(3)\triangle\text{x}$
$\Rightarrow\triangle\text{y}\mid_{\text{x}=4}\simeq(3)\times(-0.032)=-0.096$
$(3.968)^{\frac{3}{2}}=\text{y}+\triangle\text{y}=8-0.096$
$=7.904$
For x = 4, y = 8
$\text{x}+\triangle\text{x}=3.968\\\Rightarrow\triangle\text{x}=3.968-4=-0.032$
$\frac{\text{dy}}{\text{dx}}=\frac{3}{2}\text{x}^{\frac{1}{2}}$
$\text{dy}=\Big(\frac{3}{2}\text{x}^{\frac{1}{2}}\Big)\text{dx}$
$\Rightarrow\triangle\text{y}\mid_{\text{x}=4}\simeq(3)\triangle\text{x}$
$\Rightarrow\triangle\text{y}\mid_{\text{x}=4}\simeq(3)\times(-0.032)=-0.096$
$(3.968)^{\frac{3}{2}}=\text{y}+\triangle\text{y}=8-0.096$
$=7.904$
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