Question

For time 0 < t < 10, water is flowing into a small tub at a rate given by the function F defined by F(t) = arctan(t). For time 5 < t < 10, water is leaking from the tub at a rate given by the function L defined by L(t) = 0.03(20t - 75). Both F(t) and L(t) are measured in cubic feet per minute, and t is measured in minutes. The volume of water in the tub, in cubic feet, at time t minutes is given by W(t).

(a) At time t = 3, there are 2.5 cubic feet of water in the tub. Write an equation for the locally linear approximation of W at t = 3 and use it to approximate the volume of water in the tub at time t = 3.5.

Answer 1

The locally linear approximation of the function W at t = 3 is 2.5 + 0.1(t - 3), giving an approximate volume of 2.55 cubic feet of water in the tub at t = 3.5.

Step-by-step explanation:

The student is tasked with finding the locally linear approximation of the function W(t), which represents the volume of water in the tub at time t. This can be done by computing the derivative of W(t) and using it to approximate the change in volume around t = 3 minutes. With W(t) being the integral of water inflow minus outflow, and since outflow L(t) starts at t = 5, we only consider the inflow F(t) = arctan(t) for t = 3. The derivative, F'(t) = 1/(1+t^2), at t = 3 is F'(3) = 1/10. Knowing that there are 2.5 cubic feet at t = 3, the linear approximation for W near t=3 is W(t) ≈ W(3) + W'(3)(t - 3), which simplifies to W(t) ≈ 2.5 + 0.1(t - 3).

Applying the approximation for t = 3.5 gives us W(3.5) ≈ 2.5 + 0.1(3.5 - 3) = 2.55 cubic feet. Thus, at time t = 3.5, the volume of water in the tub is approximately 2.55 cubic feet.