Question

11. A conical water tank with vertex down has a radius of 10 ft at the
top and is 24ft high. If water flows into the tank at a rate of
20ft/min, how fast is the depth of the water increasing when the
water is 16ft deep?​

Answer 1

Answer:

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Step-by-step explanation:

Given:

r=10

h=24

dV/dt=20

Find: dh/dt, h=16

r/h=10/24

r=5/12h

V=1/3πr^2h

V= 1/3π(5/12h)^2h

V=25/432πh^3

dV/dt=25/144πh^2 dh/dt

20=25/144π(16)^2 dh/dt

dh/dt=0.14

Answer 2

The rate that the depth of the water increases is 0.127 ft/min.

Given:

A conical water tank with vertex down has a radius of 10 ft at the top and is 24ft high. The water flows into the tank at a rate of 20ft/min.

To find:

How fast is the depth of the water increasing when the water is 16ft deep?​  

Solution:

The radius of the cone, r  = 10 ft  

The height of the cone, h = 24 ft

Do a cross-section on the cone as shown in the figure,

from the two similar triangles, 10/24 = r/h

=> r = 10h/24 = 5h/12      

The formula for the volume of a cone,  V = 1/3πr²h    

=> V = 1/3π(5h/12)²h     [ ∵  r = 5h/12 ]      

=> V = 1/3π(25/144)h³  

Differentiate V with respect to time 't'  

=> dV/dt = 1/3π(25/144) 3(h²) dh/dt

=> dV/dt = π(25/144)(h²) dh/dt        

Here dV/dt will equal the rate of flow of water

=> 20 = π(25/144)(24²) dh/dt  

=> 4/π = 5/144 (576) dh/dt

=> 4/π = 5/1 (2) dh/dt      

=> dh/dt = 4/5π  

=> dh/dt = 0.127127...

=> dh/dt = 0.127 (approx)

Therefore,

The rate that the depth of the water increases is 0.127 ft/min.  

 

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