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please help see que… - QuestionCove
OpenStudy (anonymous):

please help see question and equation below - calculus one

3 years ago
OpenStudy (anonymous):

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 \[\frac{ 5\pi }{ 3 }\]

3 years ago
OpenStudy (freckles):

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?

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

@e.mccormick @jdoe0001

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

@amistre64

3 years ago
OpenStudy (freckles):

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

3 years ago
OpenStudy (amistre64):

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

3 years ago
OpenStudy (amistre64):

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

3 years ago
OpenStudy (amistre64):

at least thats how im reading it

3 years ago
OpenStudy (amistre64):

freckles and i have the same outcome

3 years ago
OpenStudy (freckles):

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.

3 years ago
OpenStudy (amistre64):

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

3 years ago
OpenStudy (freckles):

or on the sphere (whatever)

3 years ago
OpenStudy (anonymous):

hmm ok sorry my computer is being slow!!

3 years ago
OpenStudy (amistre64):

it may be the site

3 years ago
OpenStudy (anonymous):

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

3 years ago
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