Using differentials, estimate the amount of paint need to apply a coat of paint 2mm thick to a sphere with a...
$begingroup$
So first I use the surface area of a sphere which is $$ SA=4pi r^2$$
but after that I have no idea what to do next because I shouldn't apply $frac{d}{dx}$ until I have substituted in something for $r$ right?
calculus
asked Dec 9 '18 at 1:00
Eric BrownEric Brown
737
$endgroup$
add a comment |
$begingroup$
So first I use the surface area of a sphere which is $$ SA=4pi r^2$$
but after that I have no idea what to do next because I shouldn't apply $frac{d}{dx}$ until I have substituted in something for $r$ right?
calculus
asked Dec 9 '18 at 1:00
Eric BrownEric Brown
737
$endgroup$
1
$begingroup$
Looks to me as if you're supposed to simplify $frac 43 pi (r+dr)^3 - frac 43 pi r^3$. And it simplifies to $4pi r^2 dr$.
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– I like Serena
Dec 9 '18 at 1:09
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@IlikeSerena. It does not simplify to that. It is approximated by that.
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– William Elliot
Dec 9 '18 at 2:32
add a comment |
$begingroup$
So first I use the surface area of a sphere which is $$ SA=4pi r^2$$
but after that I have no idea what to do next because I shouldn't apply $frac{d}{dx}$ until I have substituted in something for $r$ right?
calculus
asked Dec 9 '18 at 1:00
Eric BrownEric Brown
737
$endgroup$
So first I use the surface area of a sphere which is $$ SA=4pi r^2$$
but after that I have no idea what to do next because I shouldn't apply $frac{d}{dx}$ until I have substituted in something for $r$ right?
calculus
calculus
asked Dec 9 '18 at 1:00
Eric BrownEric Brown
737
asked Dec 9 '18 at 1:00
Eric BrownEric Brown
737
asked Dec 9 '18 at 1:00
Eric BrownEric Brown
737
asked Dec 9 '18 at 1:00
Eric BrownEric Brown
737
asked Dec 9 '18 at 1:00
Eric BrownEric Brown
737
737
1
$begingroup$
Looks to me as if you're supposed to simplify $frac 43 pi (r+dr)^3 - frac 43 pi r^3$. And it simplifies to $4pi r^2 dr$.
$endgroup$
– I like Serena
Dec 9 '18 at 1:09
$begingroup$
@IlikeSerena. It does not simplify to that. It is approximated by that.
$endgroup$
– William Elliot
Dec 9 '18 at 2:32
add a comment |
1
$begingroup$
Looks to me as if you're supposed to simplify $frac 43 pi (r+dr)^3 - frac 43 pi r^3$. And it simplifies to $4pi r^2 dr$.
$endgroup$
– I like Serena
Dec 9 '18 at 1:09
$begingroup$
@IlikeSerena. It does not simplify to that. It is approximated by that.
$endgroup$
– William Elliot
Dec 9 '18 at 2:32
1
1
$begingroup$
Looks to me as if you're supposed to simplify $frac 43 pi (r+dr)^3 - frac 43 pi r^3$. And it simplifies to $4pi r^2 dr$.
$endgroup$
– I like Serena
Dec 9 '18 at 1:09
$begingroup$
Looks to me as if you're supposed to simplify $frac 43 pi (r+dr)^3 - frac 43 pi r^3$. And it simplifies to $4pi r^2 dr$.
$endgroup$
– I like Serena
Dec 9 '18 at 1:09
$begingroup$
@IlikeSerena. It does not simplify to that. It is approximated by that.
$endgroup$
– William Elliot
Dec 9 '18 at 2:32
$begingroup$
@IlikeSerena. It does not simplify to that. It is approximated by that.
$endgroup$
– William Elliot
Dec 9 '18 at 2:32
add a comment |
1 Answer
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$begingroup$
$V = 4/3×pi r^3$
$dV/dr = 4pi r^2$
Lower estimate for paint volume is
$dV = 4pi r^2 dr$
dr = 2mm, r = 30cm. Calculate dV in cm$^3.$
answered Dec 9 '18 at 2:52
William ElliotWilliam Elliot
7,6072720
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$V = 4/3×pi r^3$
$dV/dr = 4pi r^2$
Lower estimate for paint volume is
$dV = 4pi r^2 dr$
dr = 2mm, r = 30cm. Calculate dV in cm$^3.$
answered Dec 9 '18 at 2:52
William ElliotWilliam Elliot
7,6072720
$endgroup$
add a comment |
$begingroup$
$V = 4/3×pi r^3$
$dV/dr = 4pi r^2$
Lower estimate for paint volume is
$dV = 4pi r^2 dr$
dr = 2mm, r = 30cm. Calculate dV in cm$^3.$
answered Dec 9 '18 at 2:52
William ElliotWilliam Elliot
7,6072720
$endgroup$
add a comment |
$begingroup$
$V = 4/3×pi r^3$
$dV/dr = 4pi r^2$
Lower estimate for paint volume is
$dV = 4pi r^2 dr$
dr = 2mm, r = 30cm. Calculate dV in cm$^3.$
answered Dec 9 '18 at 2:52
William ElliotWilliam Elliot
7,6072720
$endgroup$
$V = 4/3×pi r^3$
$dV/dr = 4pi r^2$
Lower estimate for paint volume is
$dV = 4pi r^2 dr$
dr = 2mm, r = 30cm. Calculate dV in cm$^3.$
answered Dec 9 '18 at 2:52
William ElliotWilliam Elliot
7,6072720
answered Dec 9 '18 at 2:52
William ElliotWilliam Elliot
7,6072720
answered Dec 9 '18 at 2:52
William ElliotWilliam Elliot
7,6072720
answered Dec 9 '18 at 2:52
William ElliotWilliam Elliot
7,6072720
7,6072720
add a comment |
add a comment |
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1
$begingroup$
Looks to me as if you're supposed to simplify $frac 43 pi (r+dr)^3 - frac 43 pi r^3$. And it simplifies to $4pi r^2 dr$.
$endgroup$
– I like Serena
Dec 9 '18 at 1:09
$begingroup$
@IlikeSerena. It does not simplify to that. It is approximated by that.
$endgroup$
– William Elliot
Dec 9 '18 at 2:32