Question

A container has the shape of an open right circular cone, as shown. The height of the container is 10 cm and the diameter of the opening is 10 cm. water in the container is evaporating so that its depth h is changing at a constant rate of -3/10cm/hr. (v=1/3pir^2h). A) find the volume when h=5cm and B) fine the rate of change of the volume of water when h=5

Answer 1

a) Volume = (1/12)pi*h^3, with height = 5cm. 

b) You should be able to differentiate V = (1/12)pi*h^3 with respect to h, and you were given dh/dt = -0.3 cm/hr. 

does that make sense?

Answer 2

The volume when h = 5cm is 8.18 $\rm cm^3$ and the rate of change of the volume of water when h=5 is -1.47 cm/hr and this can be determined by using the given data.

Given :

• A container has the shape of an open right circular cone.
• The height of the container is 10 cm and the diameter of the opening is 10 cm.
• Water in the container is evaporating so that its depth h is changing at a constant rate of -3/10 cm/hr.

The volume of the right circular cone is given by:

$\rm V = \dfrac{1}{3}\pi r^2 h$

where h is the height and r is the radius of the right circular cone.

At h = 5 cm the volume of the right circular cone is given by:

$\rm V = \dfrac{1}{3}\times \pi \times \left(\dfrac{h}{4}\right)^2\times h$

$\rm V=\dfrac{\pi}{48}\times h^3$   --- (1)

A) Now, at h = 5cm the value of the volume of a right circular cone is:

$\rm V=\dfrac{\pi}{48}\times (5)^3$

V = 8.18 cm cube

B) The rate of change of the volume of water when h=5 is determined by differentiating the equation (1) with respect to time.

$\rm \dfrac{dV}{dt}=\dfrac{\pi}{48}\times 3h^2\times \dfrac{dh}{dt}$

Substitute the value of h and dh/dt in the above expression.

$\rm \dfrac{dV}{dt}=\dfrac{\pi}{48}\times 3(5)^2\times -\dfrac{3}{10}$

$\rm \dfrac{dV}{dt}=\dfrac{\pi}{48}\times 3(5)^2\times -\dfrac{3}{10}$

$\rm \dfrac{dV}{dt} =-1.47\;cm/hr$

For more information, refer to the link given below:

https://brainly.com/question/11897796