A candle of diameter d is floating on a liquid in a cylindrical container of diameter D (D>>>d). If it is burning at a rate of 2cm/hr, then the top of the candle will? (options)
A) Fall at the rate of 1cm/hr
B) Go up at the rate of 1cm/hr
Assume density of wax to be half that of liquid
Initially, half of the candle is submerged in liquid.
In time $t$, the length of remaing candle will be $2l-2t$, where $l$ is the length of half the initial candle.
Balancing weight of the candle with upthrust
$$A(2l-2t)\rho g = 2A y\rho g$$
Here $y$ is the length of candle in the liquid after time $t$
So$$y=l-t$$
then
$$\frac{dy}{dt}=-1$$
Now the part am really confused about is what does the negative sign imply? It should be easy to tell intuitively, but what i want to know the mathematical significance.
Best Answer
Mathematically the derivative being negative at $t_0$ means that the function is decreasing at $t_0$. Being negative for all $t\in[0,l]$ means that the function is strictly decreasing in this interval. Physically means that the part of the candle under the surface is diminishing, it so happens in the other part and then, the top is falling. But the rate is -1 cm/hr, so, none of the options seems correct.