An elastic spherical ballon is being blown up so that the radius is increasing at a rate of 1cm/sec. Calculate the rate at which the volume of the ballon is increasing when the radius is 5cm
v = 4π/3 r^3
dv/dt = 4π r^2 dr/dt
when r=5,
dv/dt = 4π * 25 * 1 = 100π cm^3/s
To calculate the rate at which the volume of the balloon is increasing when the radius is 5 cm, we will use the formula for the volume of a sphere:
V = (4/3)πr^3
where V is the volume and r is the radius.
To find the rate at which the volume is changing with respect to time, we can differentiate this equation with respect to time (t):
dV/dt = d/dt((4/3)πr^3)
To do this, we apply the chain rule. The derivative of r^3 with respect to t can be calculated using the power rule:
d(r^3)/dt = 3r^2(dr/dt)
Since we know that dr/dt (the rate at which the radius is changing) is given as 1 cm/sec, we can substitute this value:
dV/dt = (4/3)π(3r^2)(1)
Now we can plug in the radius value of 5 cm:
dV/dt = (4/3)π(3(5^2))(1)
Simplifying:
dV/dt = 20π cm^3/sec
Therefore, when the radius is 5 cm, the rate at which the volume of the balloon is increasing is 20π cm^3/sec.