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 Question from Steven, a student: Water is running out of a conical container 12 feet in diameter and 8 feet deep (vertex down) and filling a spherical balloon. At the instant the depth of the water in the cone is 4 feet, the radius of the sphere is approximately 4 feet. The rate of change of the depth of the water in the cone at the instant is approximately ______________ times the rate of change of the radius of the balloon. i think that the rates are equal, since the radius and the depth are equivalent at that point. But I do not know if this is a trick question or not since I did not do any calculations to reach my conclusion. Any help will be appreciated. Thank you.

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?
Penny

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