Question

Linda is still bored and is now pouring sugar onto the floor. The poured-out sugar forms a conical pile whose height is three-quarters of its radius and whose height is growing at a rate of \(1.5\) inches per second. How fast is Linda pouring the sugar at the instant that the pile of

Short answer

The rate of change is$24{\text{inches}}^3/\sec$

Step-by-step solution

Step 1. Given information

Given height is three-quarters of its radius and whose height is growing at a rate of $1.5$inches per second

Substitute the value of height and differentiate to calculate the rate of change

Calculating, we get

$$\begin{gathered} V=\frac{1}{3}\pi \times {\left(\frac{4}{3}h\right)}^2\times h \\ V=\frac{1}{3}\pi \times \frac{16}{9}{h}^3 \end{gathered}$$

Differentiating, we get

$$\begin{gathered} \frac{dV}{dt}=\frac{1}{3}\pi \times \frac{16}{9}\times 3h^2\frac{dh}{dt} \\ \frac{dV}{dt}=\frac{16}{9}\times 9\times 1.5 \\ \frac{dV}{dt}=24{\text{inches}}^3/\sec \end{gathered}$$