Related Rates
Consider the volume of a sphere, `V=4/3 pi r^3`.
Suppose that you fill the ballon with air
at a constant rate, `100" cm"^3/"s"`. How does the radius increase?
The intersting aspect of this situation is that both the volume and the radius are changing with time. Hence, implicit derivatives!
Let's differentiate the volume equation with repect to time:
`d/dt[V] = d/dt[4/3 pi r^3]`
`(dV)/dt = 4/3 pi 3 r^2 (dr)/dt`
`(dV)/dt = 4 pi r^2 (dr)/dt`
So, if the change in volume is constant, `100" cm"^3/"s"`, then the change in the radius is a function of the radius.
`(dr)/dt = 100/(4 pi r^2)`
Think about why this makes sense? The increase of the radius is inversely proportional to the square of the raduis.
Consider a conical tank:
Next
- `d/(dt)[V(t)] = d/dt[1/3 pi r^2 h]`