The height of a cylinder is increasing at

Answered question

2022-04-12

The height of a cylinder is increasing at a constant rate of 2 meters per second, and the volume is decreasing at a rate of 800 cubic meters per second. At the instant when the radius of the cylinder is 5 meters and the volume is 648 cubic meters, what is the rate of change of the radius? The volume of a cylinder can be found with the equationV=πr2h. Round your answer to three decimal places.


 

 

Answer & Explanation

user_27qwe

user_27qwe

Skilled2023-04-27Added 375 answers

Given information:
- Height of the cylinder is increasing at a constant rate of dhdt=2 m/s
- Volume of the cylinder is decreasing at a rate of dVdt=800 m3/s
- Radius of the cylinder at the instant is r=5 m
- Volume of the cylinder at the instant is V=648 m3
We need to find the rate of change of the radius drdt at the instant.
We are given that the volume of a cylinder can be found with the formula V=πr2h. To find the rate of change of the radius drdt, we need to differentiate this formula with respect to time t:
dVdt=ddt(πr2h)
Using product rule of differentiation, we get:
dVdt=π(2rdrdth+r2dhdt)
Substituting the given values, we get:
800=π(2(5)drdt(h)+(5)2(2))
800=π(10hdrdt+50)
Simplifying and solving for drdt, we get:
drdt=800/π5010h=80/π5h
To find the value of h at the instant, we use the formula for volume of the cylinder:
V=πr2h
648=π(5)2h
h=64825π
Substituting this value in the expression for drdt, we get:
drdt=80/π5648/(25π)
drdt=50081π m/s (rounded to three decimal places)
Therefore, the rate of change of the radius at the instant is drdt=50081π m/s.

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