Part 1: Relationship between $r$ and $t$

Given that sand is being poured at a constant rate onto a flat surface and forms a pile in the shape of a cone, we are to find how the radius $r$ of the base of the cone varies with time $t$, given that $r(0) = 0$.

Let:

• $r$ be the radius of the cone's base at time $t$.
• $h$ be the height of the cone.
• $V$ be the volume of the cone.

The volume $V$ of a cone is given by the formula: $V = \frac{1}{3} \pi r^2 h$ We are told that the slope of the cone is constant, which implies that the ratio of the height $h$ to the radius $r$ remains constant. Therefore, we can express $h$ in terms of $r$: $h = kr$ where $k$ is a constant representing the fixed ratio of height to radius (since the slope is constant).

Substituting $h = kr$ into the volume formula, we get: $V = \frac{1}{3} \pi r^2 (kr) = \frac{1}{3} \pi k r^3$ Thus, the volume of the cone is proportional to $r^3$: $V = C r^3$ where $C = \frac{1}{3} \pi k$ is a constant.

Since sand is being poured at a constant rate, the rate of change of the volume $\frac{dV}{dt}$ is constant. Let $\frac{dV}{dt} = Q$, where $Q$ is the constant rate of change of the volume: $\frac{dV}{dt} = Q$ But from the volume expression $V = C r^3$, we also have: $\frac{dV}{dt} = 3 C r^2 \frac{dr}{dt}$ Equating the two expressions for $\frac{dV}{dt}$: $3 C r^2 \frac{dr}{dt} = Q$ Solving for $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{Q}{3 C r^2}$

To find $r(t)$, we can separate the variables and integrate: $r^2 dr = \frac{Q}{3C} dt$ Integrating both sides: $\int r^2 dr = \int \frac{Q}{3C} dt$ $\frac{r^3}{3} = \frac{Qt}{3C} + \text{constant}$ Since $r(0) = 0$, the constant of integration is zero. Thus, we have: $\frac{r^3}{3} = \frac{Qt}{3C}$ Simplifying: $r^3 = \frac{Qt}{C}$ Therefore, the radius $r$ varies with time $t$ as: $r(t) = \left(\frac{Qt}{C}\right)^{\frac{1}{3}}$

Part 2: Time to Reach Radius $2r_1$

Let $t_1$ be the time it takes for the radius to reach $r_1$. From the expression for $r(t)$, we have: $r_1 = \left(\frac{Qt_1}{C}\right)^{\frac{1}{3}}$ Now, we want to find the time $t_2$ it takes for the radius to reach $2r_1$. Using the same formula for $r(t)$, we have: $2r_1 = \left(\frac{Qt_2}{C}\right)^{\frac{1}{3}}$ Taking the ratio of these two equations: $\frac{2r_1}{r_1} = \left(\frac{t_2}{t_1}\right)^{\frac{1}{3}}$ Simplifying: $2 = \left(\frac{t_2}{t_1}\right)^{\frac{1}{3}}$ Cubing both sides: $8 = \frac{t_2}{t_1}$ Thus, the time $t_2$ is: $t_2 = 8 t_1$

Final Answer:

• The radius $r$ varies with time as $r(t) = \left(\frac{Qt}{C}\right)^{\frac{1}{3}}$.
• If it takes a time $t_1$ for the radius to reach $r_1$, it will take a time $8 t_1$ for the radius to reach $2r_1$.

Do you want any details or have further questions?

Here are 5 relative questions to expand on the problem:

1. How does the height of the cone change with time if the slope is constant?
2. What would happen if the slope of the cone were allowed to vary with time?
3. Can you derive the volume flow rate $Q$ based on experimental data?
4. What if the sand were poured at a non-constant rate? How would the equations change?
5. How would you modify this model for different cone shapes or materials?

Tip: When integrating, always check the initial conditions to solve for any constants of integration.