The volume, V cm3 , of a cone height h is
(pi)(h^3)/ 12.
If h increases at a constant rate
of 0.2 cm/sec and the initial height is 2 cm, express V in terms of t and find the rate of
change of V at time t.
To express V in terms of t, we can use the formula for the volume of a cone and substitute the given information.
The formula for the volume of a cone is:
V = (π)(h^3)/12
Given:
h increases at a constant rate of 0.2 cm/sec
Initial height, h₀ = 2 cm
We can express h in terms of t, the time elapsed since the initial height:
h = h₀ + rt
where r is the rate of change of h, which is 0.2 cm/sec.
Substituting the values into the formula, we get:
V = (π)((h₀ + rt)^3)/12
Now, to find the rate of change of V at time t, we need to differentiate V with respect to t and evaluate it at time t.
dV/dt = (π/12) * 3(h₀ + rt)^2 * r
Therefore, the rate of change of V at time t is (π/4) * r * (h₀ + rt)^2.
To express the volume, V, in terms of time, t, we need to find a relationship between V and t. Since the height, h, is changing at a constant rate of 0.2 cm/sec, we can express h as a function of t.
Let's assume that initially, at t = 0, the height is 2 cm. As time passes, the height increases by 0.2 cm each second. Thus, we can express h as a function of t using the equation:
h(t) = 2 + 0.2t.
Next, let's substitute this expression for h into the formula for the volume of a cone:
V = (πh^3) / 12.
Replacing h with 2 + 0.2t, we have:
V = (π(2 + 0.2t)^3) / 12.
Now, to find the rate of change of V at time t, we need to differentiate the volume equation with respect to t. Differentiating both sides of the equation, we get:
dV/dt = (π/12) * 3(2 + 0.2t)^2 * 0.2.
Simplifying this equation further, we have:
dV/dt = (0.6π) * (2 + 0.2t)^2.
Therefore, the rate of change of V at time t is (0.6π) * (2 + 0.2t)^2 cm^3/sec.