Respuesta :
Answer:
The rate the radius of the balloon shrinking at the moment the radius is 5 centimeters is 0.1783 cm/s
Explanation:
Here we have
dV/dt = 56 cm³/s
[tex]\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi\cdot3r^2 \frac{dr}{dt}[/tex]
When the radius is 5 cm we have
[tex]56 \hspace {0.09cm}cm^3/s= \frac{4}{3}\pi\cdot3\cdot 5^2 \cdot \frac{dr}{dt} = 314.16 \times \frac{dr}{dt}[/tex]
Therefore,
[tex]56 \hspace {0.09cm}cm^3/s= 314.16 \hspace {0.09cm}cm^2\times \frac{dr}{dt}[/tex]
From which,
[tex]\frac{dr}{dt} = 56 \hspace {0.09cm}cm^3/s \div314.16 \hspace {0.09cm}cm^2[/tex]
[tex]\frac{dr}{dt} = 0.1783 \hspace {0.09cm}cm/s[/tex]
The rate the radius of the balloon shrinking at the moment the radius is 5 centimeters = 0.1783 cm/s.
Answer:
Radius is increasing at the rate of 0.1783 cm/s
Explanation:
Let r be the radius of the balloon and V be its volume at any time t.
Thus, V= (4/3)πr³
Now, let's differentiate both sides with respect to t and we obtain;
dV/dt = 4πr²•(dr/dt)
From the question, we are given that; dV/dt = 56 cm³/s
Thus,
56 = 4πr²•(dr/dt)
dr/dt = 56/(4πr²)
So,we want to find dr/dt at r=5,
Thus,
dr/dt = 56/(4π•5²)
dr/dt = 0.56/π = 0.1783 cm/s
