Suppose a spherical snowball is melting at a constant rate of 0.20in per hour. How fast is the volume changing when the radius is 10in?
v = 4/3 pi r^3
dv/dt = 4 pi r^2 dr/dt
Now just use your numbers
-251.327
To find out how fast the volume is changing, we need to use the concept of related rates. This involves using calculus and the chain rule to differentiate the equation that relates the volume and the radius of the snowball.
The formula for the volume of a sphere is V = (4/3)πr³, where V is the volume and r is the radius.
We know that the radius is changing with time, but we want to find out how fast the volume is changing at a specific point when the radius is 10in.
To find this, we need to differentiate the volume equation with respect to time. Let's call the rate of change of the radius (∆r/∆t) as dr/dt and the rate of change of the volume (∆V/∆t) as dV/dt.
So, to differentiate the volume equation, we use the chain rule:
dV/dt = dV/dr * dr/dt
Now we need to find dV/dt when r = 10in. To do this, we need to find dV/dr, which is the derivative of the volume equation with respect to the radius:
dV/dr = 4πr²
Next, substitute r = 10in into dV/dr:
dV/dr = 4π(10)² = 400π in²
Now, we need to find dr/dt. We are given that the radius is changing at a constant rate of 0.20in per hour, so:
dr/dt = 0.20 in/hr
Finally, we can find dV/dt when r = 10in by substituting the values we calculated into the related rates equation:
dV/dt = dV/dr * dr/dt
dV/dt = (400π in²) * (0.20 in/hr)
dV/dt = 80π in³/hr
Therefore, when the radius is 10in, the volume of the snowball is changing at a rate of 80π in³/hr.