While building a snowman, a large snowball is created at the rate of 5 inches per hour. How fast is the volume of the snowball changing at the instant the snowball has a radius of 10 inches?
... at a rate of 5 inches/hr sounds like a linear rate of change.
I will assume that the radius is changing at that rate
V = (4/3)pi r^3
dV/dt = 4pi r^2 dr/dt
dV/dt = 4pi(100)(5)
= 2000pi cubic inches/hr
To find the rate at which the volume of the snowball is changing, we can use the formula for the volume of a sphere:
V = (4/3) * π * r^3
Where V is the volume and r is the radius of the snowball.
Now, let's take the derivative of both sides of the equation with respect to time t:
dV/dt = (d/dt) [(4/3) * π * r^3]
The derivative of V with respect to t represents the rate at which the volume is changing, and the derivative of (4/3) * π * r^3 is:
dV/dt = (4/3) * π * (3r^2) * (dr/dt)
Since the radius is given as 10 inches and the snowball is growing, we know the radius is increasing, so dr/dt is positive.
We are given that the snowball is growing at a rate of 5 inches per hour, so dr/dt = 5 inches/hour.
Plugging in the values into the equation, we get:
dV/dt = (4/3) * π * (3(10^2)) * 5
Simplifying further:
dV/dt = (4/3) * π * (300) * 5
dV/dt = 2000π
Therefore, at the instant the snowball has a radius of 10 inches, the volume of the snowball is changing at a rate of 2000π cubic inches per hour.
To find the rate at which the volume of the snowball is changing, we need to find the derivative of the volume with respect to time.
Let's start by finding an equation that relates the volume of the snowball and its radius. The volume of a sphere is given by the formula: V = (4/3) * π * r^3, where V is the volume and r is the radius.
Taking the derivative of both sides with respect to time (t), we get:
dV/dt = (4/3) * π * (3r^2 * dr/dt),
where dV/dt represents the rate at which the volume is changing over time, and dr/dt represents the rate at which the radius is changing over time.
Now, we have the equation to find the rate of change of the volume with respect to time. We are given that the radius is 10 inches, so we can substitute r = 10 into the equation:
dV/dt = (4/3) * π * (3 * 10^2 * dr/dt).
To find the rate at which the volume is changing, we need to know the rate at which the radius is changing, dr/dt. Unfortunately, this information is not given in the question. We would need to be provided with the rate at which the radius is changing over time.
If you have the information about dr/dt, you can substitute it into the equation above to find the rate at which the volume is changing at the instant the snowball has a radius of 10 inches.