Respuesta :
Answer:
5 cm/min
Step-by-step explanation:
A spherical balloon is being inflated with helium at a rate of 20π cm³/min.
The volume of a sphere is given by :
[tex]V=\dfrac{4}{3}\pi r^3[/tex] ...(1)
Differentiate equation (1) wrt t as follows :
[tex]\dfrac{dV}{dt}=3\times \dfrac{4}{3}\pi r^2\times \dfrac{dr}{dt}\\\\\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}\ ....(1)[/tex]
Put dV/dt = 20π cm³/min and r = 1 cm, we can fin dr/dt
[tex]\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}\\\\20\pi =4\pi (1)\times \dfrac{dr}{dt}\\\\\dfrac{dr}{dt}=5\ cm/min[/tex]
So, the radius of changing at the rate of 5 cm/min.
Using implicit differentiation, it is found that the radius is changing at a rate of 5 cm/min.
The volume of a sphere of radius r is given by:
[tex]V = \frac{4\pi r^3}{3}[/tex]
Applying implicit differentiation, the rate of change is of:
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
- A spherical balloon is being inflated with helium at a rate of 20π cm3/min, hence [tex]\frac{dV}{dt} = 20\pi[/tex]
- The radius is of 1 cm, hence [tex]r = 1[/tex].
Then:
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
[tex]20\pi = 4\pi (1)^2\frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt} = \frac{20}{4}[/tex]
[tex]\frac{dr}{dt} = 5[/tex]
The radius is changing at a rate of 5 cm/min.
A similar problem is given at https://brainly.com/question/14004755
