A volcanic cone spews fine particles of rock which land in the surrounding area and cause the size of the cone to increase. (a) Suppose that a volcanologist takes measurements that show that when the cone is 100 meters wide and 50 meters tall, its height is increasing at 0.5 meters/day. If the assump- tion is made that the cone will retain its height-width proportion, what is the rate at which the volume of rock is being released by the volcano? (b) Suppose that the volcanologist takes more measurements and realizes the original as- sumption was poorly made, because the radius is

the width is the diameter I guess. So the radius of the cone is 50m and the height is 50m too the rate of change is 0.5 for both the radius and the height The volume of the cone is \[V=1/3 \pi r^2h\]

So the volume is 1/3*pi*50^2*50=130899.69 after 1 step the radius is 50.5 the height is 50.5 so the volume 1/3*pi*50.5^2*50.5=134866.09 the difference is 3966.39 And that is the answer.every day the volume of the cone will be bigger by 3966

Hope it makes sense

the answer is 1250 pi....which isnt your answer

That is 3927

Pretty close :)

yup thanks

could you help with (b) Suppose that the volcanologist takes more measurements and realizes the original as- sumption was poorly made, because the radius is increasing at 0.4 meters/day. What would be the new finding for the rate at which the volume of rock is being released by the volcano?

It just changes the numbers V=1/3pi* 50^3 V(new)=1/3pi 50.4^3

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