A spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2cm/min. At what rates are the volume and surface area of the balloon increasing when the radius is 5cm?
The radius of a circle is increase at a 3 c.m/min. Find the rate of change of the area when
A)r=2c.m B)r=3c.m
To find the rates at which the volume and surface area of the balloon are increasing, we can use the formulas for the volume and surface area of a sphere:
1. Volume of a sphere: V = (4/3) * π * r^3
2. Surface area of a sphere: A = 4 * π * r^2
Given that the radius of the balloon is increasing at a rate of 2 cm/min, we can differentiate both formulas with respect to time (t) using implicit differentiation:
1. Differentiating the volume formula with respect to time:
dV/dt = d/dt [(4/3) * π * r^3]
= (4/3) * π * 3r^2 * (dr/dt)
= 4πr^2 * (dr/dt)
2. Differentiating the surface area formula with respect to time:
dA/dt = d/dt [4 * π * r^2]
= 4π * 2r * (dr/dt)
= 8πr * (dr/dt)
Now we can substitute r = 5 cm and dr/dt = 2 cm/min into the above equations to find the rates at which the volume and surface area of the balloon are increasing:
1. Rate of increase in volume:
dV/dt = 4π(5^2)(2)
= 4π(25)(2)
= 200π cm^3/min
2. Rate of increase in surface area:
dA/dt = 8π(5)(2)
= 8π(10)
= 80π cm^2/min
Therefore, when the radius of the balloon is 5 cm, the volume is increasing at a rate of 200π cm^3/min and the surface area is increasing at a rate of 80π cm^2/min.
To find the rates at which the volume and surface area of the balloon are increasing, we can use the formulas for the volume and surface area of a sphere in terms of its radius.
The volume of a sphere is given by the formula: V = (4/3)πr³
The surface area of a sphere is given by the formula: A = 4πr²
We are given that the radius of the balloon is increasing at a rate of 2 cm/min. We can differentiate the volume and surface area formulas with respect to time (t) to find the rates at which they are changing.
Differentiating the volume formula, we get:
dV/dt = d/dt[(4/3)πr³]
dV/dt = 4πr²(dr/dt)
Differentiating the surface area formula, we get:
dA/dt = d/dt[4πr²]
dA/dt = 8πr(dr/dt)
Now, we need to find the values of dV/dt and dA/dt when the radius (r) is 5 cm. To do this, we substitute r = 5 cm and dr/dt = 2 cm/min into the respective formulas.
For the volume:
dV/dt = 4π(5)²(2) = 200π cm³/min
For the surface area:
dA/dt = 8π(5)(2) = 80π cm²/min
So, when the radius of the balloon is 5 cm, the volume is increasing at a rate of 200π cm³/min and the surface area is increasing at a rate of 80π cm²/min.
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
when r = 5 and dr/dt = 2
dV/dt = 4π(25)(2) = 200π cm^3/min
SA = 4πr^2
d(SA)/dt = 8πr dr/dt
= 8π(5)(2) = 80π cm^2/min