# reciprocal cm?

• Nov 2nd 2009, 01:17 AM
sammy28
reciprocal cm?
hi all,

im confused with the way an answer is given in reciprocal cm format the questions was:

The temperature $\displaystyle \theta C$ measured at a distance $\displaystyle x cm$ is given by:

$\displaystyle \theta=16+\frac{450}{x^\frac{3}{2}}$

for distances $\displaystyle x\geq 1$

Find the temeparture gradient (i.e. the rate at which temperature decreases with distance) 10cm from the flame.

$\displaystyle t=f(d)=16+450x^{-\frac{3}{2}}$, where t(emp), d(istance)

$\displaystyle f'(d)=-675x^{-\frac{5}{2}}$

$\displaystyle f'(10)=-675.\frac{1}{100\sqrt{x}}$

$\displaystyle =-\frac{6.75}{\sqrt{10}} \equiv -0.675\sqrt{10}$

$\displaystyle =-2.13C/cm$ to 3 sf

but the book (Edexcel A level maths} gives the answer as

$\displaystyle =-2.13C/cm^{-1}$

thanks.
• Nov 2nd 2009, 01:39 AM
mr fantastic
Quote:

Originally Posted by sammy28
hi all,

im confused with the way an answer is given in reciprocal cm format the questions was:

The temperature $\displaystyle \theta C$ measured at a distance $\displaystyle x cm$ is given by:

$\displaystyle \theta=16+\frac{450}{x^\frac{3}{2}}$

for distances $\displaystyle x\geq 1$

Find the temeparture gradient (i.e. the rate at which temperature decreases with distance) 10cm from the flame.

$\displaystyle t=f(d)=16+450x^{-\frac{3}{2}}$, where t(emp), d(istance)

$\displaystyle f'(d)=-675x^{-\frac{5}{2}}$

$\displaystyle f'(10)=-675.\frac{1}{100\sqrt{x}}$

$\displaystyle =-\frac{6.75}{\sqrt{10}} \equiv -0.675\sqrt{10}$

$\displaystyle =-2.13C/cm$ to 3 sf

but the book (Edexcel A level maths} gives the answer as

$\displaystyle =-2.13C/cm^{-1}$

Your book has got it wrong. The unit is either $\displaystyle ^0 \text{C/cm}$ or $\displaystyle ^0 \text{C cm}^{-1}$ but not a mixture of both.
thanks for yor answer MF. it was me who got it wrong sorry. the book does say $\displaystyle ^0\text{C cm}^{-1}$ now it makes sense. what an idiot!