An NRCC student attempts alleviate calculus stress by purchasing an ice cream co

Question

An NRCC student attempts alleviate calculus stress by purchasing an ice cream cone Ben & Jerry's ( DQ RIP). After the purchase and eating the top portion down to the rim of the cone, the student becomes distracted while talking to another student standing in line to buy their stress-relief cone. Since multi-talking is 30 percent less productive, the student focuses on the conversation and forgets about the ice cream and the ice cream cone ( a sugar cone- a true mathematical conical shape) begins to melt out the bottom at a rate of 1 cm -3 / min. If the cone is 6 cm tall and has a diameter that is one-third the height, how fast is the height of the ice cream in the cone changing when the ice cream is 2 cm below the top of the cone?

Explanation / Answer

dv/dt =1cm^3/min

h = 6 cm and d = (1/3) *6 =2 cm ;r =1cm ; d = 2r ;

dh/dt = ? h = cm

v=(1/3) * pi * r ^2 * h = volume of cone

differentiating on both sides wrt " t ";

dv/dt = (1/3) * pi *(2*r ) dr/dt

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