Question

Sand is pouring from a pipe at the rate of \[12{\text{ c}}{{\text{m}}^3}{\text{/s}}\]. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 3 cm?

Answer 1

- Hint: First of all, consider the radius and height of the forming sand cone as variables and find their relation between them by the given condition. The rate of height of the sand cone increasing is given by the differentiation of height of the cone with respect to time.

Complete step-by-step solution -

Given that sand is pouring from a pipe and falling sand from a cone.
Let \[r\] and \[h\] be the radius of the sand cone respectively and \[v\] be the volume of the cone.
Sand from cone of the ground in such a way that the height of the cone is always one-sixth of the radius i.e.,
\[
  h = \dfrac{1}{6}r \\
  r = 6h..............................................................\left( 1 \right) \\
\]
And given that the sand is pouring from a pipe at the rate of \[12{\text{ c}}{{\text{m}}^3}{\text{/s}}\] i.e., rate of volume of a cone w.r.t is \[12{\text{ c}}{{\text{m}}^3}{\text{/s}}\] i.e., \[\dfrac{{dv}}{{dt}} = 12{\text{ c}}{{\text{m}}^3}{\text{/s}}\]
And we need to find how fast height of the cone is increasing when height is 4 cm i.e., to find \[\dfrac{{dh}}{{dt}}\] when \[h = 4{\text{ cm}}\]
We have
\[\dfrac{{dv}}{{dt}} = 12{\text{ c}}{{\text{m}}^3}{\text{/s}}\]
We know that the volume of the cone with radius \[r\] and height \[h\] is given by \[v = \dfrac{1}{3}\pi {r^2}h\].
i.e., \[v = \dfrac{1}{3}\pi {r^2}h\]
\[
  v = \dfrac{1}{3}\pi {\left( {6h} \right)^2}h \\
  v = \dfrac{1}{3}\pi \times 36{h^2} \times h \\
  v = 12\pi {h^3} \\
\]
Differentiating w.r.t \[x\], we have
\[
  \dfrac{{dv}}{{dt}} = \dfrac{{d\left( {12\pi {h^3}} \right)}}{{dt}} \\
  \dfrac{{dv}}{{dt}} = 12\pi \dfrac{{d\left( {{h^3}} \right)}}{{dt}} \\
  \dfrac{{dv}}{{dt}} = 12\pi \times 3{h^2} \times \dfrac{{dh}}{{dt}} \\
  12 = 12\pi \times 3{h^2} \times \dfrac{{dh}}{{dt}} \\
  \dfrac{{dh}}{{dt}} = \dfrac{{12}}{{12 \times 3{h^2}}} \\
\]
We need \[\dfrac{{dh}}{{dt}}\] when \[h = 4{\text{ cm}}\]
\[
  {\left( {\dfrac{{dh}}{{dt}}} \right)_{h = 4}} = \dfrac{{12}}{{12 \times 3 \times 4 \times 4}} \\
  {\left( {\dfrac{{dh}}{{dt}}} \right)_{h = 4}} = \dfrac{1}{{48\pi }} \\
\]
Since height is in cm and time in sec
So, \[{\left( {\dfrac{{dh}}{{dt}}} \right)_{h = 4}} = \dfrac{1}{{48\pi }}{\text{ cm/sec}}\]
Thus, the rate of the sand cone is increasing at the rate of \[\dfrac{1}{{48\pi }}{\text{ cm/sec}}\].

Note: The volume of the cone increases as the height of the cone. Given the height of the cone is always one-sixth of the radius, this means the radius of the cone is also increasing as the height and volume of the cone increases. The volume of the cone with radius \[r\] and height \[h\] is given by \[v = \dfrac{1}{3}\pi {r^2}h\].