



 
MJ, Let's try a somewhat simpler problem first.
If $V(t)$ is the volume of the balloon in $m^3$ at time $t$ hours and $r(t)$ is the radius on the balloon in $m$ at time $t$ hours then \[V(t) = \frac43 \pi \; r(t)^3.\] You can differentiate this expression with respect to $t$ to obtain \[V^{\;\prime}(t) = 4 \pi \; r(t)^2 r^\prime(t)\] You are told that at some specific time $t^*$ the radius is 2 m and the radius is increasing at 1 m/hr. That is $r(t^*) = 2$ m and $ r^\prime(t^*) = 1$ m/hr. Thus at this time \[V^{\;\prime}(t^*) = 4 \pi \; r(t^*)^2 r^\prime(t^*) = 4 \pi \times 2^2 \times 1 = 16 \pi\; m^3 \mbox{ per hour}\] Since the water is being pumped into the balloon at a constant rate, $V(t)$ is a constant and hence $V(t) = 16 \pi\; m^3 \mbox{ per hour}$ for all $t.$ Your problem is quite similar. You have a conical shape and the volume $V(t)$ of a cone is given by \[V(t) = \frac13 \pi \; r(t)^2 h(t)\] where $r(t)$ is the radius and $h(t)$ is the height. The complication you have is that $V(t)$ is a function of two variables $r(t)$ and $h(t)$ and you only know values of $r(t)$ and $r^\prime(t)$ so somehow you need to eliminate $h(t)$ from the expression for the volume. Read the problem again and you see that "the height is always 2/3 the radius" and hence $h(t) = \frac23 r(t).$ Substitute this into the expression for the volume and you will then have the volume as a function of one variable $r(t).$ Now you can proceed as in my problem with the balloon. I hope this helps,  


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