Respuesta :
Answer:
[tex]\frac{dr}{dt}= \frac{7}{4\pi (\frac{1101}{4\pi })^{\frac{2}{3}}}[/tex]
Step-by-step explanation:
we know that volume of the spherical balloon = [tex]\frac{4}{3}\times \pi \times r^{3}[/tex]
given [tex]\frac{dv}{dt}[/tex] = 7
differentiating v with respect to time
[tex]\frac{dv}{dt}=4\pi r^{2}\frac{dr}{dt}[/tex]
given [tex]367 = \frac{4}{3}\pi r^{3}[/tex]
[tex]r = (\frac{1101}{4\pi })^{\frac{1}{3}}[/tex]
[tex]\frac{dr}{dt}= \frac{7}{4\pi (\frac{1101}{4\pi })^{\frac{2}{3}}}[/tex]
Answer:
The rate of change of radius of the spherical balloon is 0.0282 cm per sec .
Step-by-step explanation:
Given as :
The rate at which air is filled inside balloon = 7 cm³ per sec
Let The radius = r cm
And The volume of balloon = v = 367 cm³
let The change of radius = [tex]\frac{\mathrm{d} r}{\mathrm{d} t}[/tex]
∵ volume = 367 cm³
∵ volume of sphere = [tex]\frac{4}{3} \pi[/tex] r³
i.e 367 = [tex]\frac{4}{3}[/tex] × [tex]\frac{22}{7}[/tex] × r³
or, r³ = [tex]\frac{7707}{88}[/tex]
So, r³ = 87.57
∴ r = [tex]\sqrt[3]{87.57}[/tex] = 4.44 cm
I.e r = 4.44 cm
Now, [tex]\frac{\mathrm{d} v}{\mathrm{d} t}[/tex] = [tex]\frac{\mathrm{d} \frac{4}{3}\Pi r^{3}}{\mathrm{d} t}[/tex]
Or, [tex]\frac{\mathrm{d} v}{\mathrm{d} t}[/tex] = 4[tex]\pi[/tex]r²×[tex]\frac{\mathrm{d} r}{\mathrm{d} t}[/tex]
Or, 7 = 4[tex]\pi[/tex](4.44)²×[tex]\frac{\mathrm{d} r}{\mathrm{d} t}[/tex]
Or, 7 = 4 × 3.14 × 19.71 × [tex]\frac{\mathrm{d} r}{\mathrm{d} t}[/tex]
Or, 7 = 247.55 × [tex]\frac{\mathrm{d} r}{\mathrm{d} t}[/tex]
∴ [tex]\frac{\mathrm{d} r}{\mathrm{d} t}[/tex] = [tex]\frac{7}{247.55}[/tex]
Or, [tex]\frac{\mathrm{d} r}{\mathrm{d} t}[/tex] = 0.0282 cm per sec
Hence, The rate of change of radius of the spherical balloon is 0.0282 cm per sec . Answer
