A spherical party balloon is being inflated with helium pumped in at a rate of 12 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 1 ft?
V = 4/3*PI*r^3
dV/dt = 12 = 4*PI*r^2*dr/dt = 4*PI*1^2*dr/dt
Solve for dr/dt, the rate at which the radius is growing
To find how fast the radius is growing at the instant when it reaches 1 ft, we first need to determine the relationship between the volume and the radius of a sphere.
The formula for the volume of a sphere is given by V = (4/3) * pi * r^3, where V is the volume and r is the radius.
Now, let's differentiate both sides of the equation with respect to time (t), using the chain rule:
dV/dt = d/dt [(4/3) * pi * r^3]
On the right side, we have a constant (4/3) * pi, so it can be taken out of the differentiation:
dV/dt = (4/3) * pi * d/dt [r^3]
Next, we apply the power rule for differentiation:
dV/dt = (4/3) * pi * 3 * r^2 * dr/dt
Since we are given the rate of change of volume (dV/dt) as 12 cubic feet per minute, and we need to find the rate of change of radius (dr/dt) when the radius (r) is 1 ft, we can substitute these values into the equation:
12 = (4/3) * pi * 3 * (1^2) * dr/dt
Simplifying further:
12 = (4/3) * pi * 3 * dr/dt
Dividing both sides by (4/3) * pi * 3:
dr/dt = 12 / [(4/3) * pi * 3]
dr/dt = 12 / [4pi]
Therefore, the rate at which the radius is growing at the instant when it reaches 1 ft is 12 / (4pi) cubic feet per minute.