Molten plastic is injected into the centre of a circular mold of constant height, H, through a small hole. The rate of injection is such that the radius of the plastic inside the mould increases so that
r(t)= 3t^2 -2t^3 where the mold has maximum radius of one unit. Compute the rate of change of volume of plastic in the mold, V (t), given that V = H(pie)r^2. At what time is the mold filling fastest and what is the value of dV/dt at that time?
To compute the rate of change of volume of the plastic in the mold, we need to find dV/dt. We are given that V = H(pi)e(r^2), where r is the radius of the plastic inside the mold.
We can start by finding an expression for V in terms of t, using the given expression for r: r(t) = 3t^2 - 2t^3.
Substituting this expression for r into the formula for V, we get:
V(t) = H(pi)e(3t^2 - 2t^3)^2
Simplifying this expression:
V(t) = H(pi)e(9t^4 - 12t^5 + 4t^6)
Now, to find dV/dt, we can use the power rule of differentiation. Differentiating V(t) with respect to t, we get:
dV/dt = H(pi)e(36t^3 - 60t^4 + 24t^5)
To find the time at which the mold is filled fastest, we need to find the maximum value of dV/dt. For this, we can set dV/dt equal to zero and solve for t:
36t^3 - 60t^4 + 24t^5 = 0
Factoring out t^3, we have:
t^3(36 - 60t + 24t^2) = 0
This equation is satisfied when either t^3 = 0 or 36 - 60t + 24t^2 = 0.
The first case, t^3 = 0, implies that t = 0. However, since we are considering the time t when the mold is filling, we discard this solution as it corresponds to the initial time when no plastic has been injected into the mold yet.
To solve the second equation, 36 - 60t + 24t^2 = 0, we can factor it as follows:
12(3 - 5t + 2t^2) = 0
This equation is satisfied when either 3 - 5t + 2t^2 = 0 or t = 3/2.
To solve 3 - 5t + 2t^2 = 0, we can use factoring:
(3 - 2t)(1 - t) = 0
This equation is satisfied when either 3 - 2t = 0 or 1 - t = 0.
Solving these equations, we get two potential values for t: t = 3/2 or t = 3/2.
Therefore, the mold is filling fastest at t = 3/2 units of time.
To find the value of dV/dt at t = 3/2, we substitute this value into the expression for dV/dt:
dV/dt = H(pi)e(36(3/2)^3 - 60(3/2)^4 + 24(3/2)^5) = H(pi)e(243 - 405/16 + 81/16)
Simplifying this expression, we get:
dV/dt = H(pi)e(339/16)
So, at t = 3/2 units of time, the value of dV/dt is H(pi)e(339/16).