quick calculus question

TX123

ok just want to get my head around this problem. say u have a glass with volume 1m^3. you want to fill the glass with water from a tap. usuing calculus how to we find the rate at which the glass is filled. what variables are needed and with thos variables how do we show the derivative or integral. you can oput up ur own scenario for urselves using your own variables.

Fremen

I'm not really sure what it is you're asking, but I guess you were thinking about what rate the water level rises at.
It's been a few years since I've done maths like that, but if you can find an expression for the volume of water in the glass at time t, and the water level once you've put a certain amount of water in, you're pretty much finished.

In mathematical terms:
Let H(v) be the water level once you've poured in a specific volume of water, v, in.

Let V(t) be the amount of water that has been put in after time t.

Now (dH/dV)*(dV/dT) = dH/dT or the rate at which the water rises.

H(v) could be tricky to work out for a lot of shapes like spheres, and in many cases it will require a double integral.
For cubes and pyramids and such, it should be fairly straightforward, though.

Feel free to jump in on this if I've made a mistake, like I said, my knowledge of this area is a bit rusty.

TX123

ok so im a bit confues now. if we take h(v) to be 1000ml^3 and V(t) to be 10s then is the rate just 100ml/s. when would you involve derivatives . i spent years doing derivates nand want to apply them but am just unsure how to go about it using complex methods.

what is the point of using e.g 2x^2 = 4x in this type of calculations.

IconoclastCynic

TX123 wrote:

ok just want to get my head around this problem. say u have a glass with volume 1m^3. you want to fill the glass with water from a tap. usuing calculus how to we find the rate at which the glass is filled. what variables are needed and with thos variables how do we show the derivative or integral. you can oput up ur own scenario for urselves using your own variables.

Okay, unless you change the knobs or something, the rate at which the glass fills is constant, so:

dV/dt = c

where V = volume, t = time, and c = whatever the rate is and dV/dt = the instantaneous rate of change of the volume.

If you want to find the volume at any specific time, you'd integrate yielding:

V = c*t + k

where k = the initial volume.