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I'm stumped with an answer to a question regarding Differentation (in a book for Edexcel Core Mathematics 2):

A sector of a circle has area $100 \text{cm}^2$. Show that the permimeter of this sector is given by the formula $P=2r+(200/r)$, $r > \sqrt{100/\pi}$.

Find the minimum value for the perimeter of such a sector.

The answer is shown as:

Perimeter $$P = 2r+r\theta \label{1}\tag{1}$$

Area $A = (1/2)r^2\theta$

But area is $100\text{cm}^2$, so

$(1/2)r^2\theta = 100$

$r\theta = 200/r$

Substitute into equation \eqref{1} to give $$P = 2r + 200/r\label{2}\tag{2}$$

and since area of circle > area of sector, $\pi r^2 > 100$, so $r > \sqrt{100/\pi}$.

$dP/dr = 2 - 200/r^2$

So, for $dP/dr = 0$, $r = 10$

Substitute into equation \eqref{2} to give $P = 20 + 20 = 40\text{cm}$.

I understand the answer up until the derivative function of the perimeter is given.

The original equation of the perimeter is $P = 2r + 200/r$, as shown in the question and proved in the workings.

I do not understand how the derivative function of $P = 2r + 200/r$ is $p'(r) = 2 - 200/r^2$. Can anyone help by walking through how the derivative equation is derived, please? I think I must be missing something obvious as I haven't had issues with other questions in the chapter. Thanks

2 Answers 2

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$$P(r) = 2r +\frac{200}{r}=2r^1+200r^{-1}$$ $$\frac{dP}{dr} =2\cdot 1r^{1-1}+200\cdot(-1)r^{-1-1}$$ $$=2r^0-200r^{-2}$$ $$=2-\frac{200}{r^2}$$

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    I'm guessing the second term is what tripped you up. Generally, for $p>0$, $$\frac{d}{dx}\left\{\frac a{x^p}\right\} = -\frac{ap}{x^{p+1}}$$ by similar reasoning (work it out!). – 2017-01-02
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    Yeah it was the second term, silly for me not to see it in retrospect - I'll blame it on being rusty after festive period! Thanks again. – 2017-01-02
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your function isn $$P(r)=2r+200r^{-1}$$ for the derivative we need the rule $$(x^n)'=nx^{n-1}$$ thus we have $$P'(r)=2+200(-1)r^{-2}$$

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    Thanks, both this and MPW's answer explains it. – 2017-01-02