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 ( π /2 - t/10 ). 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-t²-75). 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). 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 W at t=3 is calculated by using the formula L(x) = W(a) + W'(a)⋅(x - a), where W(3) = 2.5 and W'(3) needs to be found. By evaluating L(3.5), we can approximate the volume of water at t=3.5 minutes.

Step-by-step explanation:

The task is to find the locally linear approximation of the volume of water function W at t=3 and use it to approximate the volume of water in the tub at t=3.5. Since we are given W(3) = 2.5 cubic feet and F(t) is the rate of water flowing into the tub, the derivative W'(t) will be the flow rate, F(t), at time t. We calculate the derivative W'(3) = F(3) = arctan( π /2 - 3/10 ).

The linear approximation of W at a point a can be given by the equation L(x) = W(a) + W'(a)⋅(x - a), where W(a) is the function value and W'(a) is the derivative at point a. Here, W(3) = 2.5, and let's calculate W'(3) to find the slope of this approximation.

Using L(x) with W'(3) and W(3), we can now approximate the volume at t = 3.5 by evaluating L(3.5). This local linear approximation gives us a quick method to estimate the volume at a point close to t = 3 without integrating the flow rate function over time.