- 3 m/s
We are given the rate at which the volume of the fluid is increasing, = 108π, and are
looking for the rate at which the height of the fluid is increasing, . Thus, we need to find a
way to relate the volume of a cylinder to its height. We know that the volume of a cylinder is V
= πr^2 h. We will want to take the derivative of the equation with respect to t. Note that the
radius of the tank doesn’t change as the volume changes, therefore r is a constant in this
problem, not a variable. We get = πr^2 . Because r is not changing, = 0, and we
would have ended up with the same derivative.
Now we can plug in = 108π and r = 6, and solve for . We get 108π = π(6)^2 . Thus,
= 3 m/s.
- 380 volts/s
We are given the rate at which the current is decreasing, = −4, and the rate at which the
resistance is increasing, = 20, and are looking for the rate at which the voltage is changing.
We are also given an equation that relates the three variables: V = IR. So, we simply take the
derivative of the equation with respect to t (Product Rule!): . Now we can
use our equation to find R, which we will need to plug into the derivative equation: 100 = 20R,
so R = 5. Next, we plug = −4, = 20, I = 20, and R = 5 into the derivative equation:
= (20)(20) + (5)(−4) = 380 volts/s.
- t = 3
In order to find where the particle is changing direction, we need to find where the velocity of
the particle changes sign. The velocity function of the particle is the derivative of the position
