Hello, Danielle428!
Sand falls from a conveyor belt at the rate of 10 m<sup>3</sup>/min onto the top of a conical pile.
The height of the pile is always three eighths of the base diameter.
How fast are the a) height and b) radius changing when the pile is 4 m high?
Answer in centimeters per minute.
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The volume of a cone is: \(\displaystyle \:V\;=\;\frac{1}{3}\pi r^2h\)
We are told that the height is \(\displaystyle \frac{3}{8}\) of the base diameter: \(\displaystyle \,h\:=\;\frac{3}{8}(2r)\)
\(\displaystyle \;\;\)Hence: \(\displaystyle \,\)
[1] \(\displaystyle h\,=\,\frac{3}{4}r\) . . . or: \(\displaystyle \,\)
[2] \(\displaystyle r\,=\,\frac{4}{3}h\)
(a) We want \(\displaystyle \frac{dh}{dt}\) when \(\displaystyle h\,=\,4\).
Substitute
[2] into the volume formula: \(\displaystyle \:V\;=\;\frac{1}{3}\pi\left(\frac{4}{3}h\right)^2h\:=\:\frac{16\pi}{27}h^3\)
Differentiate with respect to time: \(\displaystyle \:\frac{dV}{dt}\:=\:\frac{16\pi}{9}h^2\left(\frac{dh}{dt}\right)\;\;\Rightarrow\;\;\frac{dh}{dt}\:=\:\frac{9}{16\pi h^2}\left(\frac{dV}{dt}\right)\)
Since \(\displaystyle h\,=\,4\) and \(\displaystyle \frac{dV}{dt}\,=\,10\), we have: \(\displaystyle \L\:\frac{dh}{dt}\:=\:\frac{9}{16\pi(4^2)}(10)\:=\:\frac{45}{128\pi}\:\approx\:0.11\) m/min.
(b) We want \(\displaystyle \frac{dr}{dt}\) when \(\displaystyle h\,=\,4\).
Substitute
[1] into the volume formula: \(\displaystyle \:V\:=\:\frac{1}{3}\pi r^2\left(\frac{3}{4}r\right) \:=\:\frac{\pi}{4}r^3\)
Differentiate with respect to time: \(\displaystyle \:\frac{dV}{dt}\:=\:\frac{3\pi}{4}r^2\left(\frac{dr}{dt}\right)\;\;\Rightarrow\;\;\frac{dr}{dt}\:=\:\frac{4}{3\pi r^2}\left(\frac{dV}{dt}\right)\)
When \(\displaystyle h\,=\,4,\:r\,=\,\frac{16}{3}\) . . . so we have: \(\displaystyle \,\frac{dr}{dt}\:=\:\frac{4}{3\pi\left(\frac{16}{3}\right)^2}(10)\)
Therefore: \(\displaystyle \L\,\frac{dr}{dt}\:=\:\frac{15}{32\pi}\:\approx\:0.15\) m/min.
