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The volume. V cm^3, of a cone of height h is (pi)h^3/12. If h increases at a constant rate of 0.2cm/s and the initial height is 2cm, express V in interms of t and find the rate of change of V at time t.
 10 months ago
 10 months ago
The volume. V cm^3, of a cone of height h is (pi)h^3/12. If h increases at a constant rate of 0.2cm/s and the initial height is 2cm, express V in interms of t and find the rate of change of V at time t.
 10 months ago
 10 months ago

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myininayaBest ResponseYou've already chosen the best response.1
So we have \[V(x)=\frac{\pi }{12} (h(x))^3\] We will let x represent the variable time for now.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Find derivative of both sides.
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
Both V and h are functions of time this implies that both V and h are functions of x
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
So derivative of V is V' and derivative of h is h'
 10 months ago

myininayaBest ResponseYou've already chosen the best response.1
So tell me what does V' equal?
 10 months ago
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