Answer:
a) [tex]\frac{dv}{dr} = 16\pi cubic ft /ft[/tex]
b) Approximate error of the volume increase when the radius changes from 2 to 2.4 ft = 0.6 that is
[tex]\frac{dv}{v} =0.6 cubic ft /ft[/tex]
Step-by-step explanation:
Explanation:-
a) Given Volume of spherical balloon
[tex]V = \frac{4\pi r^{3} }{3}[/tex] …(i)
Differentiating with respective to the radius 'r'
[tex]\frac{dv}{dt} = \frac{4\pi 3r^{2} }{3}[/tex]
[tex]\frac{dv}{dt} = 4 \pi r^{2}[/tex]
At r=2
[tex](\frac{dv}{dr})r_{=2} =4\pi (4)=16\pi cubeft/ft[/tex]
b) Step(1):-
Given Volume of spherical balloon
[tex]V = \frac{4\pi r^{3} }{3}[/tex]
[tex]logV = log(\frac{4\pi r^{3} }{3})[/tex]
we property of logarithmic log(ab) = log a +log b
taking logarithmic on both sides , we get
[tex]logv = log(\frac{4\pi }{3} )+log(r^{3})[/tex] …(ii)
Step2:-
we will use [tex]\frac{d(logx)}{dx} = \frac{1}{x}[/tex]
[tex]\frac{1}{v}[/tex] δ v = 0 + [tex]3 \frac{1}{r}[/tex] δ r
Given data radius changes from 2 to 2.4 ft
r+ δ r = 2.4 =2+0.4
here r =2 and δ r =0.4
Now [tex]\frac{1}{v}[/tex] δ v = [tex]3 X \frac{1}{2} X 0.4[/tex]
[tex]\frac{1}{v}[/tex] δ v = 0.6
Approximate error in Volume = 0.6 cubic/ft
Approximate error of the volume increase when the radius changes from 2 to 2.4 ft = 0.6 that is
[tex]\frac{dv}{v} =0.6 cubic ft /ft[/tex]
