Sand is pouring from a pipe at the rate of 12 cm3/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 t heradius of the base. How fast is the sand cone increasing when the height is 4 cm?

Solution

The volume of a cone with radius r and height h is given by the formula

$V = \frac{1}{3} π r^{2} h$

According to the question,

$h = \frac{1}{6} r \Rightarrow r = 6 h$

Substituting in the formula,

$∴ V = \frac{1}{3} π {(6 h)}^{2} h = 12 π h^{3}$

The rate of change of the volume with respect to time is

$\frac{dv}{dt} = 12 π \frac{d}{dh} {(h)}^{3} x \frac{dh}{dt} . . . . . . . [By chain rule] = 12 π {(3 h)}^{2} x \frac{dh}{dt} = 36 π h^{2} x \frac{dh}{dt} . . . . . . . . . (i) Given that \frac{dv}{dt} = 12 {cm}^{3} / s Substituting the values \frac{dv}{dt} = 12 and h = 4 in equation (i), we have, 12 = 36 π {(4)}^{2} x \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{12}{36 π (16)} \Rightarrow \frac{dh}{dt} = \frac{1}{48 π} Hence, the height of the sand cone is increasing at the rate of \frac{1}{48 π} cm / s .$

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Application Of Derivatives

Sand is pouring from a pipe at the rate of 12 cm3/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 t heradius of the base. How fast is the sand cone increasing when the height is 4 cm?

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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