1. ## Some help

The rate at which the temperaturex of a liquid decreases is inversely proportional ites temperature at time t. write down a differential eq and solve it. given that the temp of the liquid increases from 20 to 20.5 in 2 seconds find temp of body after 10 seconds...

Inversely proportional = 1/k

so eq would be

dx/dt = 1/kx

kx dx = dt

integrating

kx^2 = t + c

? 2 unknowns ? or what am i doing wrong here

2. $\displaystyle \frac {dx}{dt} = \frac 1 {kt}$ is more like it!

Mind, now I look at it I'm not sure why ...

3. Originally Posted by ramzel
The rate at which the temperaturex of a liquid decreases is inversely proportional ites temperature at time t. write down a differential eq and solve it. given that the temp of the liquid increases from 20 to 20.5 in 2 seconds find temp of body after 10 seconds...
$\displaystyle \frac{dx}{dt} = \frac{k}{x}$

4. \frac{dx}{dt} = \frac{k}{x}

separate variables ...

\frac{dx}{x} = k \, dt

huh when you cross multiply doesn't the x goes up ?

5. Originally Posted by ramzel
\frac{dx}{dt} = \frac{k}{x}

separate variables ...

\frac{dx}{x} = k \, dt

huh when you cross multiply doesn't the x goes up ?
you're right ... my error. was thinking of something else.

$\displaystyle \frac{dx}{dt} = \frac{k}{x}$

$\displaystyle x \, dx = k \, dt$

6. Thanks i got the right answer