Question

The Temperature of water in tub at time tt is modeled by a strictly increasing, twice-differentiable function WW, where W(t)W(t) is measured in degrees Fahrenheit and tt is measured in minutes. At time t=0t=0, the temperature of the water is 55∘∘F. The water is heated for 30 minutes, beginning at time t=0t=0. Values of W(t)W(t) at selected times tt for the first 20 minutes are given in the table above.(d) For 20≤t≤2520≤t≤25, the function WW that models the water tempature has first derivative given by W′(t)=0.4t√cos(0.06t)W′(t)=0.4tcos⁡(0.06t). Based on the model, what is the temperature of the water at time t=25t=25?

Answer 1

Answer:

  w(25) = 140 °F

Step-by-step explanation:

You want the value of w(25) given w(0) = 55 and w'(t) = 0.4t√cos(0.06t).

Integral

The derivative function does not have a closed-form integral, but we can get a nice solution using a graphing calculator. The attachment shows the function w(t) and its value at t=25.

  w(25) = 140 . . . . . degrees F

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Additional comment

When this calculator's integration result can be compared to a closed-form expression or apps that provide similar solutions, it is generally found accurate to 12 significant figures. "Confidence is high."