rate of change

**mark** · Nov 17th 2009, 07:39 AM

hi, i'm starting to look at work on rates of change and was wondering if someone could explain something to me.

the example question goes like this:

the volume, $Vm^3$ of water is given by $V = t^3 - 5t^2 + 6t + 2$ . find the rate of change of volume with respect to time after 0.5 hours

it then goes on to $\frac{dV}{dt} = 3t^2 - 10t + 6$ and explains that:

when t = 0.5 $\frac{dV}{dt} = 3(0.5)^2 - 10(0.5) + 6 = 1.75 \implies$ after 0.5 hours the volume is increasing at a rate of $1.75m^3h^-1$

i don't understand where the power of minus 1 $(h^-1)$ comes from?

can someone please explain this? thankyou

**tom@ballooncalculus** · Nov 17th 2009, 08:09 AM

per hour! (same as /h)

**mark** · Nov 17th 2009, 08:13 AM

i understand that h stands for per hour. i meant the "to the power of minus 1" part. why is it to the power of minus one?

also how do you write it in latex so that the 1 appears small and at the top like in $t^3$ . it comes out big with minus 1 $h^-1$
cheers

**tom@ballooncalculus** · Nov 17th 2009, 08:20 AM

use the 'hat' symbol (shift 6?) like in excel etc. i.e. h^{-1} oh yes and use curly braces if the power contains more than one character

h means hour, power minus one means per

**mark** · Nov 17th 2009, 08:23 AM

ah i see. thanks

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