1.) How fast (centimeters per minute) is the water level falling when the water is 5m deep? 2.) How fast is the radius of the water's surface changing then? Answer in centimeters per minute. Related Rates Q: Water is flowing at the rate of 50 m^3/min from a shallow concrete conical reservoir (vertex down) of base radius 45m and height 6m. THANKS SO MUCH!!!

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The volume of the cone at h metres is

`V = int_0^u r^2 = 1/3r^3|_0^u = 1/3u^3`  cubic metres

where `u = R h/H = 45h/6`

So `V = 1/3(45/6)^3h^3 = 140.6h^3`

`(dV)/(dt) = 50 m^3`  per minute

Now, using the chain rule

`(dt)/(dh) = ((dt)/(dV))((dV)/(dh))`

` `Differentiating V(h) we...

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The volume of the cone at h metres is

`V = int_0^u r^2 = 1/3r^3|_0^u = 1/3u^3`  cubic metres

where `u = R h/H = 45h/6`

So `V = 1/3(45/6)^3h^3 = 140.6h^3`

`(dV)/(dt) = 50 m^3`  per minute

Now, using the chain rule

`(dt)/(dh) = ((dt)/(dV))((dV)/(dh))`

` `Differentiating V(h) we have `(dV)/(dh) = 3(140.6)h^2 = 421.9h^2`

and we also have `(dt)/(dV) = 1/50` by simply inverting `(dV)/(dt)`

So `(dt)/(dh) = (1/50)(421.9h^2)`

`implies (dh)/(dt) = 50/(421.9h^2)`

So, when h = 5m,

`(dh)/(dt) = 50/((421.9)(25)) = 0.00474` metres per minute

`= 0.474` centimetres per minute

The equation for the radius at h metres is

`u = R h/H = 45h/6 = 7.5h`

Again using the chain rule

`(du)/(dt) = ((du)/(dh))((dh)/(dt))`

` `Now, `(du)/(dh) = 7.5`  

So when h = 5

` ` `(du)/(dt) = 7.5/0.474 = 15.82` centimetres per minute

1) 0.474 cm/min 2) 15.82 cm/min

 

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