Question

A spherical raindrop evaporates at a rate proportional to ists surface area.Write a differential equation for the volume of raindrop as a function of time

Answer 1

dV/dt = -kS , where S is surface area at time t

Answer 2

S = 4Pi * r^2
V = 4/3 Pi * r^3

Answer 3

so you can solve S in terms of V

Answer 4

in V , solve for r

V = 4/3 Pi * r^3

((3/4)* V /Pi)^(1/3) = r

Answer 5

now plug this into S , and you will have a differential equation in terms of V or t only

Answer 6

V(t)=4*pi*((3/4)/V/pi)^(2/3) is that true

Answer 7

The type of differential equation you are solving here has one dependent variable (time) and one independent variable (volume), and derivatives of the dependent variable with respect to the independent variable

Answer 8

wait

Answer 9

you need to plug that into S, not V

Answer 10

V = 4/3 Pi * r^3
r = (3V/(4*Pi) ) ^(1/3)

plug into S

S = 4Pi * [(3V/(4Pi))^(1/3)]^2

Answer 11

Here check this out why not: \[\LARGE \frac{\partial V}{\partial t}=-k \frac{\partial V}{\partial r}\] Chain rule\[\LARGE \frac{\partial V}{\partial r}\frac{\partial r}{\partial t}=-k \frac{\partial V}{\partial r}\] divide out dV/dr\[\LARGE \frac{dr}{dt}=-k\] \[\LARGE r=-kt+C\] Now you just plug into the volume formula \[\LARGE V(t)=\frac{4}{3} \pi (-kt+C)^3\]

Answer 12

now you have

dV/dt = -k*S

dV/dt = -k * 4Pi * [(3V/(4Pi))^(1/3)]^2

Answer 13

I concur

Answer 14

well done

Answer 15

dV/dt = -k*S

dV/dt = -k * 4Pi * [(3V/(4Pi))^(1/3)]^2

dV/dt = -k * 4Pi * [ (3V)/(4pi)] ^2/3

dV/dt = -k * 4Pi * [3/(4pi)] ^(2/3) * V^(2/3)

dV/dt = - c * V^(2/3)

Answer 16

I think you're wasting time, \[\LARGE S=\frac{dV}{dr}\] as your substitution makes it easier to just apply the chain rule and solve.

Answer 17

S = V' makes sense right ?

Answer 18

with.respect.to r

Answer 19

Yeah @ganeshi8 and in this case our function looks explicitly like this: \[\LARGE V(r(t))\]

Answer 20

looks like you solved the differential equation :)

Answer 21

then you can take dV/dt , that is what the question asked, if i read correctly

Answer 22

our answers should come out the same, after some substitution perhaps

Answer 23

@perl yeah I think the question is poorly worded, it's already a function of t, because r is a function of t.

I think one of the trickiest things is that people aren't able to correctly write out what their function is, like perhaps they write z(t) or z(x,y) but really they have something like z(x(t),y(t))

Answer 24

did you use partial diferentiation notation intentionaly

would it be wrong to write
dV/dt = -k* dV/dt , with just normal script d

Answer 25

since you are not treating the other independent variable as a constant

Answer 26

To show that it works for this\[\LARGE \frac{d V}{d t}=-k \frac{d V}{d r}\] \[\LARGE V(t)=\frac{4}{3} \pi (-kt+c)^3 \\ \LARGE V'(t)=4\pi (-kt+c)^2 (-k)\] \[\LARGE V(r)=\frac{4}{3} \pi r^3 \\ \LARGE V'(r)=4 \pi r^2 \]
\[\LARGE 4\pi (r)^2 (-k)=-k*(4\pi r^2)\]

Yeah I guess I sorta just wrote them with partials for no reason, but it doesn't really matter since partial derivative of a single variable function is really the same thing. I don't know, I just felt like it and didn't think too much about that lol.

Answer 27

ok just checking.

Answer 28

i have seen people use partials with one variable , but this is like a composite function

V(r(t))

Answer 29

Also there is not an "other independent variable" here, only t is independent, radius, surface area, and volume all depend on this variable only.

Answer 30

ok

Answer 31

i meant, V is a function of r, and r is a function of time. so you could say V( r, t) , but thats gets messy

Answer 32

No, that would be wrong like saying V(S,r,t)

Answer 33

ok then its just V(r(t))

Answer 34

Maybe you could get fancy and write something like V(S(r(t))) that could be perfectly fine although super weird lol

Answer 35

V(r(t)) = g(t)

Answer 36

right, lets not complicate it further

Answer 37

i just needed to clarify it in my head :)

Answer 38

did you finally get the same answer
dV/dt = -C * V^(2/3) ?

Answer 39

You can derive some interesting things in tensor calculus by saying things like

y(x(y))=y

where the outside y is a function and x(y) is the inverse but y by itself is an independent variable.

Take the derivative to get:

\[\LARGE y'(x(y))*x'(y)=1\] Then you can turn x(y) into just x and get: \[\LARGE y'(x)=\frac{1}{x'(y)}\] which is the same as saying inverse functions have reciprocal slopes, which makes sense since they are just reflected across the line y=x

Answer 40

right

Answer 41

Oh I just showed that my answer satisfies the differential equation above so I'm done checking lol. Just sorta playing around idk