



 
Hi Steven, Suppose the radius of the water in the cone at time $t$ seconds is $r = r(t),$ the depth is $h = h(t),$ and he volume is $V_C = V_C(t),$ then \[V_C = \frac13 \pi \; r^2 h.\] Use similar triangles to find a relationship between $r$ and $h$ and use it to write $V_C$ as a function of $h$ alone. Differentiate both sides of this expression with respect to $t$ to find $\large \frac{dV_C}{dt}$ in terms of $h$ and $\large \frac{dh}{dt}.$ Suppose the spherical balloon has radius $R = R(t)$ at time $t$ and the volume is $V_S = V_S(t)$ then \[V_S = \frac43 \pi R^3.\] Differentiate both sides of this expression with respect to $t$ to find $\large \frac{dV_S}{dt}$ in terms of $R$ and $\large \frac{dR}{dt}.$ Since the water is running out of the cone at the same rate as it is running into the sphere \[\frac{dV_C}{dt} = \frac{dV_S}{dt}.\] Can you complete the problem from here?  


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