Question

please help see question and equation below - calculus one

Answer 1

imagine slicing through a sphere with a plane (sheet of paper) the smaller piece is produced is called a spherical cap. its volume is
\[V= \pi h^2(3r-h)/3\]
where r is the radius and h is the the thickness
find dr/dh for a sphere with a volume of \[\frac{ 5\pi }{ 3 }\]

Answer 2

well we will have to find r and h that correspond to the V=5pi/3 thing.
We are also going to differentiate r with respect to h.
Do we know a relationship between just r and h for a sphere?

Answer 3

well second part of problem says evaluate this derivative when r=2 and h=1

Answer 4

@e.mccormick @jdoe0001

Answer 5

i have the answer from back of the book if it helps answer is saying
\[\frac{ h-2r }{ h }\]

Answer 6

@amistre64

Answer 7

so \[\frac{5\pi}{3}=\frac{\pi h^2(3r-h)}{3} => 5=h^2(3r-h)\]
see what you get when you try to find dr/dh

Answer 8

\[V= \pi h^2(3r-h)/3\]
\[V'= \pi/3~[ (h^2)'(3r-h)+h^2(3r-h)']\]
\[V'= \pi/3~[ 2h~h'(3r-h)+h^2(3r'-h')]\]
now, h' = dh/dh = 1, sooo
\[V'= \pi/3~[ 2h~(3r-h)+h^2(3r'-1)]\]

Answer 9

since we are given V as a constant, V' = 0 and we can solve for r' in terms of h

Answer 10

at least thats how im reading it

Answer 11

freckles and i have the same outcome

Answer 12

i deleted that at first thinking V was changing for a second. then i realized V wasn't changing (like this had nothing to do with volume changing over time). Just the radius and height is changing as we move from one point to another point inside the sphere.

Answer 13

\[0= \pi/3~[ 2h~(3r-h)+h^2(3r'-1)]\]
\[0= 2h~(3r-h)+h^2(3r'-1)\]
\[-2h~(3r-h)=h^2(3r'-1)\]
\[-\frac2h~(3r-h)=3r'-1\]
\[-\frac2h~(3r-h)+1=3r'\]
\[-\frac2{3h}~(3r-h)+\frac13=r'\]

Answer 14

or on the sphere (whatever)

Answer 15

hmm ok sorry my computer is being slow!!

Answer 16

it may be the site

Answer 17

That's true!! Wel Thanks for the help wil write this dwn and ask my teacher tomorrow