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2 years ago

Find the second derivative y=1/5x^2+1/11x

Mathematics

1 answer:

Amiraneli [1.4K]2 years ago

8 0

$\frac{dy}{dx}=\frac{1}{5}.x^{2-1}.2 + \frac{1}{11}.x^{1-1}.1$

$\frac{dy}{dx}=\frac{2}{5}.x+\frac{1}{11}$

$\frac{d^2y}{dx^2}=\frac{2}{5}.x^{1-1}.1$

$\frac{d^2y}{dx^2}=\frac{2}{5}$

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Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

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d = distance in ly (light-years)

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Hence the apparent brightness of Alpha Centauri A is

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According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

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