reciprocal cm?

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November 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 $\theta C$ measured at a distance $x cm$ is given by:

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

for distances $x\geq 1$

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

my answer was

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

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

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

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

$=-2.13C/cm$ to 3 sf

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

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

can someone please explain what the reciprocal cm is all about.

thanks.
November 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 $\theta C$ measured at a distance $x cm$ is given by:

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

for distances $x\geq 1$

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

my answer was

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

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

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

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

$=-2.13C/cm$ to 3 sf

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

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

can someone please explain what the reciprocal cm is all about.

thanks.

Your book has got it wrong. The unit is either $^0 \text{C/cm}$ or $^0 \text{C cm}^{-1}$ but not a mixture of both.
November 2nd 2009, 01:54 AM
sammy28

thanks for yor answer MF. it was me who got it wrong sorry. the book does say $^0\text{C cm}^{-1}$ now it makes sense. what an idiot!