Respuesta :
Answer:
The volume of cone is increasing at a rate 2512 cubic inches per second.
Step-by-step explanation:
We are given the following in the question:
[tex]\dfrac{dr}{dt} = 3\text{ inches per second}\\\\\dfrac{dh}{dt} = -3\text{ inches per second}[/tex]
Radius = 40 inches
Height = 30 inches
The volume of cone is given by:
[tex]V = \dfrac{1}{3}\pi r^2 h[/tex]
Rate of change of volume is given by:
[tex]\dfrac{dV}{dt} = \dfrac{1}{3}\pi (2r\dfrac{dr}{dt}h+\dfrac{dh}{dt}r^2)[/tex]
Putting the values, we get,
[tex]\dfrac{dV}{dt} = \dfrac{1}{3}(3.14) \big(2(40)(3)(30)+(-3)(40)^2\big)\\\\\dfrac{dV}{dt} = \dfrac{1}{3}(3.14)(2400)\\\\\dfrac{dV}{dt} = 2512[/tex]
Thus, the volume of cone is increasing at a rate 2512 cubic inches per second.
