At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

p=m/v
now i know that the change in density is going to be due to the change i mass and volume

so would it be
\[dp=\frac{dm}{dv}\]

p is a linear function also

|dw:1358678795614:dw|

What subject is this?

it's calculus-based physics

|dw:1358679271874:dw|

\[dp \neq \frac{ dm }{ dv }\]

\[p=\frac{dm}{dv}\]

\[dp = \frac{ vdm - mdv }{ v^2 }\]

correct so where does \[pdv=dm\] come from

by any chance, is the density function given? p(x)?

ehh i think it was something like (-.0000007+10^4)r

it is (-.000005r+10^4)

So p(x) is (-.000005r+10^4)?

pdv = dm is wrong

p(x) given?

p=-.0000004r+10^4

i need somehow to relate the change of mass nad volume to the change in radius

or change in density

|dw:1358680137198:dw|

what my teacher wrote for a very similar problem was
\[pdv=dm\]
\[p(4\pi r^2)=dm\]

|dw:1358680208001:dw|

|dw:1358680330123:dw|

now is it adding 10^4 or multiplying 10^4?

adding

this is what he hsa i just need help understanding where the dm is coming from

|dw:1358680376740:dw|

only difference is he has R-d at top

|dw:1358680521783:dw|

What do you mean?

on the integral?

|dw:1358680673477:dw|

forogot int sign

|dw:1358680956774:dw|

|dw:1358681105717:dw|

\[ p=\frac{m}{v}\]
is it because
p is essentially the change in density

so \[p=\Delta p=\frac{\Delta m}{\Delta v}\]

\[p=\frac{dm}{dv}\]

\[pdv=dm\]

|dw:1358681240467:dw|

or would it actually be
\[pdr=\frac{dm}{dv}\]
\[pdvdr=dm\]

Alright, I hope some of my work was of some use.