Respuesta :
Answer:
0.45 ft/min
Step-by-step explanation:
Given:-
- The flow rate of the gravel, [tex]\frac{dV}{dt} = 35 \frac{ft^3}{min}[/tex]
- The base diameter ( d ) of cone = x
- The height ( h ) of cone = x
Find:-
How fast is the height of the pile increasing when the pile is 10 ft high?
Solution:-
- The constant flow rate of gravel dumped onto the conveyor belt is given to be 35 ft^3 / min.
- The gravel pile up into a heap of a conical shape such that base diameter ( d ) and the height ( h ) always remain the same. That is these parameter increase at the same rate.
- We develop a function of volume ( V ) of the heap piled up on conveyor belt in a conical shape as follows:
[tex]V = \frac{\pi }{12}*d^2*h\\\\V = \frac{\pi }{12}*x^3[/tex]
- Now we know that the volume ( V ) is a function of its base diameter and height ( x ). Where x is an implicit function of time ( t ). We will develop a rate of change expression of the volume of gravel piled as follows Use the chain rule of ordinary derivatives:
[tex]\frac{dV}{dt} = \frac{dV}{dx} * \frac{dx}{dt}\\\\\frac{dV}{dt} = \frac{\pi }{4} x^2 * \frac{dx}{dt}\\\\\frac{dx}{dt} = \frac{\frac{dV}{dt}}{\frac{\pi }{4} x^2}[/tex]
- Determine the rate of change of height ( h ) using the relation developed above when height is 10 ft:
[tex]h = x\\\\\frac{dh}{dt} = \frac{dx}{dt} = \frac{35 \frac{ft^3}{min} }{\frac{\pi }{4}*10^2 ft^2 } \\\\\frac{dh}{dt} = \frac{dx}{dt} = 0.45 \frac{ft}{min}[/tex]
