Water is slowing at the rate of 50m^3/min from a concrete conical reservoir of the base radio us 45m and height 6m How fast is the water level falling when the water is 5 m deep? Give answer in cm/min
I'll show work on what I did on there
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\[V = \frac{ 1 }{ 3 }*\pi*r^2*h\]
I have a question
I called the height 6 but wouldn't that h have to be 6-h than?
Find: dh/dt Write the volume in terms of just h by using similar triangles to relate r and h. \[\frac{ 45 }{ 6 }=\frac{ r }{ h }\] \[r=\frac{ 45 }{ 6 }*h\]
Why wouldn't it be 6-h?
Oh wait
the entire height is 6m, the arbitrary height is h for the smaller triangle
Now you have a relation between r and h by similar triangles... put that r into the Area \[A = \frac{ 1 }{ 3 }*\pi*(\frac{ 45 }{ 6 }h)^2*h\] Simplify that and take the time derivative of the whole thing
But if It was for the large one and the small one was 6 it would be h-6 than? For that part of the height
I'm sorry if I'm confusing you
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