We are tasked with finding the rate of change of the volume of a cone with respect to time, given that the radius and height of the cone are changing with time.

The volume $V$ of a cone is given by the formula: $V = \frac{1}{3} \pi r^2 h$ Where:

• $V$ is the volume of the cone,
• $r$ is the radius of the cone's base,
• $h$ is the height of the cone.

Step 1: Differentiate the volume formula with respect to time

We want to find the rate of change of the volume with respect to time $\frac{dV}{dt}$. To do so, we use implicit differentiation on the volume formula: $V = \frac{1}{3} \pi r^2 h$ Using the product rule and chain rule: $\frac{dV}{dt} = \frac{1}{3} \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right)$ Where:

• $\frac{dr}{dt}$ is the rate of change of the radius,
• $\frac{dh}{dt}$ is the rate of change of the height.

Step 2: Plug in the given values

We know that at time $t = 3$ days:

• $r = 100$ cm,
• $h = 50$ cm,
• $\frac{dh}{dt} = -2$ cm/day (since the height is decreasing at 2 cm/day),
• $\frac{dr}{dt}$ is unknown.

We also need to know if there is any relationship between the radius $r$ and height $h$ as they change with time. Since it's not given directly, let's assume the ratio between the radius and height is constant. At $t = 3$, we have: $\frac{r}{h} = \frac{100}{50} = 2$ Thus, we have the relation: $r = 2h$ Differentiating both sides of this equation with respect to time gives: $\frac{dr}{dt} = 2 \frac{dh}{dt}$ Since $\frac{dh}{dt} = -2$ cm/day, we find: $\frac{dr}{dt} = 2 \times (-2) = -4 \, \text{cm/day}$

Step 3: Substitute values into the rate of change formula

Now, substitute all the known values into the equation for $\frac{dV}{dt}$: $\frac{dV}{dt} = \frac{1}{3} \pi \left( 2(100) (-4) (50) + (100)^2 (-2) \right)$ Simplifying: $\frac{dV}{dt} = \frac{1}{3} \pi \left( -80000 + (-20000) \right)$ $\frac{dV}{dt} = \frac{1}{3} \pi (-100000)$ $\frac{dV}{dt} = -\frac{100000}{3} \pi \, \text{cubic centimeters per day}$

Thus, the rate of change of the volume of the cone at $t = 3$ days is: $\frac{dV}{dt} = -\frac{100000}{3} \pi \, \text{cubic centimeters per day}$

Final Answer:

The rate of change of the volume is approximately: $\frac{dV}{dt} \approx -104719.755 \, \text{cubic centimeters per day}$

Would you like any further details or clarification? Here are some related questions:

1. How does the volume change with respect to the radius alone, assuming the height is constant?
2. What would happen to the rate of change of volume if the radius was increasing instead of decreasing?
3. How could you model a situation where both the height and radius are changing independently of each other?
4. How can the formula for the volume of a cone be adapted if the height is kept constant and only the radius is changing?
5. What other real-life applications involve cones where this type of rate of change might be used?

Tip: In problems like this, always look for relationships between variables (like $r = 2h$) to simplify the differentiation process.