Solving for Time: Differentiating Equation for Δt

[DM1984]

Homework Statement

How do you differentiate this to get a change in time equation?

we have:
(dm/dt) = (pi/4)(D^2)(V(D))(LWC)(E)

LWC = 2E-6
E=1

these two values aren't important until the very end when finding the actual time, just a plug and chug then. I'm having trouble solving for Δt!
need to get (delta t) so find the time.

Homework Equations

this is what we're looking for : Δt = ?

The Attempt at a Solution

(dm/dt)= (row)(pi/6)(3D^2)(dD/dt) = 343(pi/4)(D^2.6)(LWC)(E)

 

[Mark44]

Homework Statement

How do you differentiate this to get a change in time equation?

we have:
(dm/dt) = (pi/4)(D^2)(V(D))(LWC)(E)

LWC = 2E-6
E=1

This is very unclear. In fact, I am completely mystified by it.
For starters, you wouldn't differentiate this, since what you have already is a derivative.
To continue, you have a bunch of letters whose meaning you don't give. Are they variables? Do D and V(D) depend on t?

If E = 1 does it mean that LWC = 2*1 - 6 = -4?
Or are you writing 2 X 10^-6?

these two values aren't important until the very end when finding the actual time, just a plug and chug then. I'm having trouble solving for Δt!
need to get (delta t) so find the time.

Homework Equations

this is what we're looking for : Δt = ?

The Attempt at a Solution

(dm/dt)= (row)(pi/6)(3D^2)(dD/dt) = 343(pi/4)(D^2.6)(LWC)(E)

What is "row" and where did it come from? Do you mean the Greek letter "rho" ($\rho$)?

How did (dm/dt) = (pi/4)(D^2)(V(D))(LWC)(E) change to (dm/dt)= (row)(pi/6)(3D^2)(dD/dt) and how is this equal to 343(pi/4)(D^2.6)(LWC)(E)?

Finally, if E = 1, why is it still being dragged along?

 

[DM1984]

"(dm/dt)= (row)(pi/6)(3D^2)(dD/dt)=343(pi/4)(D^2.6)(LWC)(E)"

is directly from my instructor.

yes, I meant "rho".

this is for graupel growth. trying to find out how long it takes for a 1mm piece of graupel to grow to 5mm.

The original equation is (dm/dt)=(pi/4)((D)^2)(V(D))(LWC)(E)

D= diameter
LWC = liquid water content = 2 x 10^-6 g/cm^3
E = 1.0
rho = 0.6 g/cm^3

 

[micromass]

"(dm/dt)= (row)(pi/6)(3D^2)(dD/dt)=343(pi/4)(D^2.6)(LWC)(E)"

is directly from my instructor.

yes, I meant "rho".

this is for graupel growth. trying to find out how long it takes for a 1mm piece of graupel to grow to 5mm.

The original equation is (dm/dt)=(pi/4)((D)^2)(V(D))(LWC)(E)

D= diameter
LWC = liquid water content = 2 x 10^-6 g/cm^3
E = 1.0
rho = 0.6 g/cm^3

You never really answered Mark his questions, so I'll ask them again:

How does (dm/dt)= (row)(pi/6)(3D^2)(dD/dt)=343(pi/4)(D^2.6)(LWC)(E) change to (dm/dt)=(pi/4)((D)^2)(V(D))(LWC)(E)?

Do any of the variable D, LWC, V and other depend on t?

Why would you differentiate something if you're already given a derivative??

Why drag E along if E=1?

Can you present us with the exact problem description as it is given to you??