Answer:
![0.2122\text{ ft per min}](https://tex.z-dn.net/?f=0.2122%5Ctext%7B%20ft%20per%20min%7D)
Step-by-step explanation:
Let r be the radius ( in feet ) of the cone,
So, the diameter of the cone = 2 × radius = 2r feet,
∵ The diameter of the base of the cone is three times the height,
If h represents the height of the cone,
⇒ 2r = 3 × h,
⇒
feet,
∵ Volume of a cone,
![V=\frac{1}{3}\pi R^2 H](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20R%5E2%20H)
Where,
R = radius, H = height,
Thus, the volume of the given cone is,
![V=\frac{1}{3}\pi r^2 h](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2%20h)
![=\frac{1}{3}\pi (\frac{3h}{2})^2 h](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%28%5Cfrac%7B3h%7D%7B2%7D%29%5E2%20h)
![=\frac{3}{4}\pi h^3](https://tex.z-dn.net/?f=%3D%5Cfrac%7B3%7D%7B4%7D%5Cpi%20h%5E3)
Differentiating with respect to t ( time )
![\frac{dV}{dt}=\frac{9}{4}\pi h^2\frac{dh}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D%5Cfrac%7B9%7D%7B4%7D%5Cpi%20h%5E2%5Cfrac%7Bdh%7D%7Bdt%7D)
We have,
![\frac{dV}{dt}=6\text{ cubic ft per min},h=2\text{ feet}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D6%5Ctext%7B%20cubic%20ft%20per%20min%7D%2Ch%3D2%5Ctext%7B%20feet%7D)
![6=\frac{9}{4}\pi (2)^2 \frac{dh}{dt}](https://tex.z-dn.net/?f=6%3D%5Cfrac%7B9%7D%7B4%7D%5Cpi%20%282%29%5E2%20%5Cfrac%7Bdh%7D%7Bdt%7D)
![6=9\pi \frac{dh}{dt}](https://tex.z-dn.net/?f=6%3D9%5Cpi%20%5Cfrac%7Bdh%7D%7Bdt%7D)
![\implies \frac{dh}{dt}=\frac{6}{9\pi }\approx 0.2122\text{ ft per min}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7Bdh%7D%7Bdt%7D%3D%5Cfrac%7B6%7D%7B9%5Cpi%20%7D%5Capprox%200.2122%5Ctext%7B%20ft%20per%20min%7D)
Answer:
B. The Parameter are different from Zero
Step-by-step explanation:
Here is the solution for the eqns.