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coffee maker has the shape of a cone 10cm high with a radius of 6 cm at the top. The coffee maker is filled with water, and the water is allowed to drain out the bottom of the cone. Water is flowing out of the filter at a rate of 5 cubic centimerters per second. At what rate is the water level in the cone falling when it is 5cm from the bottom of the cone.

2 Answers

6 votes

Final answer:

The rate at which the water level in the cone is falling when it is 5cm from the bottom is 0 cm/s.

Step-by-step explanation:

To find the rate at which the water level in the cone is falling when it is 5cm from the bottom, we can use related rates. We know that the volume of a cone is given by V = (1/3)πr^2h, where r is the radius and h is the height. Given that the height is decreasing at a rate of 5cm/s, we need to find the rate at which the radius is changing when the height is 5cm.

We can start by finding the relationship between the radius and the height. Since the cone has a fixed shape, similar triangles can be used to show that the ratio of the radius to the height is constant. Therefore, we can write r/h = R/H, where R and H are the given radius and height (6cm and 10cm respectively).

Now we can use this relationship to find the rate at which the radius is changing. Differentiating both sides of the equation with respect to time gives (dr/dt)/h = (dR/dt)/H. We are given that (dR/dt) = 0 (since the radius of the coffee maker is not changing), and we need to find (dr/dt) when h = 5cm.

Substituting the known values into the equation, we have (dr/dt)/5 = 0/10. Solving for (dr/dt), we find that (dr/dt) = 0 cm/s. Therefore, the water level in the cone is not falling at all when it is 5cm from the bottom.

User Anagha
by
5.0k points
7 votes

Answer: the rate of the water level in the cone falling when it is 5cm from the bottom of the cone is -0.1768 cm/sec

Step-by-step explanation:

Given the data in the question;

H/h = R/r

10/5 = 6/r

r 1 = 30/10 = 3

Also

10/h = 6/r

r2 = 6h/10 = 3/5 × h

Volume = 1/3πr²h

V = 1/3 × π × (3/5 × h)² × h

V = 1/3 × π × 9/25 × h³

V = 3π/25 × h³

Now

dv/dt = 3π/25 × 3 × h² × dh/dt

-5 = 3π/25 × 3 × 25 × dh/dt

-5 = 3π × 3 × dh/dt

-5 = 9π × dh/dt

dh/dt = -5/9π

dh/dt = -0.1768 cm/sec

therefore, the rate of the water level in the cone falling when it is 5cm from the bottom of the cone is -0.1768 cm/sec

User Amoroso
by
4.7k points