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Related rate for growing conical sand pile

Sand is being poured onto a pile, forming a cone whose height is always equal to its radius, at the point in time considered.

Sand pours from a chute onto a conical pile at a constant rate of 10 ft3/min10 \text{ ft}^3/\text{min}. As the pile grows, its shape stays such that the height hh always equals the radius rr of the base. At the moment when the pile's height is 55 ft, how fast is the height increasing?

Hint

Start with the volume of a cone, V=13πr2hV = \frac{1}{3}\pi r^2 h, substitute r=hr = h to write VV as a function of hh alone, then differentiate both sides with respect to tt using the chain rule before plugging in numbers.

Step-by-step solution

The volume of a cone is V=13πr2hV = \frac{1}{3}\pi r^2 h. Since the pile's shape keeps r=hr = h at all times, substitute to get V=13πh2⋅h=13πh3V = \frac{1}{3}\pi h^2 \cdot h = \frac{1}{3}\pi h^3, a single-variable relation between the volume and height.

Differentiate both sides with respect to tt: dVdt=13π⋅3h2dhdt=πh2dhdt\frac{dV}{dt} = \frac{1}{3}\pi \cdot 3h^2 \frac{dh}{dt} = \pi h^2 \frac{dh}{dt}. We know dVdt=10\frac{dV}{dt} = 10 ft3^3/min and h=5h = 5 ft, so 10=π(5)2dhdt=25πdhdt10 = \pi (5)^2 \frac{dh}{dt} = 25\pi \frac{dh}{dt}.

Solving gives dhdt=1025π=25π\frac{dh}{dt} = \frac{10}{25\pi} = \frac{2}{5\pi} ft/min, which is approximately 0.1270.127 ft/min.

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