Matching Levels and Slopes in a Piecewise Tracer Model
An environmental engineer models the concentration f(t) of a tracer in a flow chamber for 0 ≤ t ≤ 2 minutes. Measurements suggest f(t)=t^2+t+b for 0≤t<1 and f(t)=c√t+d for 1≤t≤2, with d=0 fixed by calibration. The engineer needs the model to be differentiable at t=1 so that a downstream sensitivity calculation is valid. A colleague argues that matching the two one-sided limits is enough; another insists that matching derivatives is also necessary. The engineer also plans to compute the total accumulated tracer by integrating f from 0 to 2, and later wants to know whether the model has a well-defined second derivative at t=1. The parameters b and c are still unknown, and no numerical smoothing is allowed. Determine what values of b and c make the model differentiable at t=1, then assess the integral and second-derivative questions under those values.
1. Find b and c so that f is differentiable at t=1. Show the continuity and derivative conditions.
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Differentiability at t=1 requires continuity first. Left limit: f(1^-)=1^2+1+b=2+b. Right value: f(1)=c√1+0=c. Continuity gives c=2+b. Left derivative: d/dt(t^2+t+b)=2t+1, so at 1 it is 3. Right derivative: d/dt(c√t)=c/(2√t), so at 1 it is c/2. Match: c/2=3, so c=6. Then b=c-2=4. Check: f(1^-)=2+4=6=f(1^+). Thus b=4, c=6.
2. Compute ∫_0^2 f(t) dt using b=4, c=6.
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With b=4 and c=6, split at t=1. ∫_0^1 (t^2+t+4) dt = [t^3/3+t^2/2+4t]_0^1 = 1/3+1/2+4 = 29/6. ∫_1^2 6√t dt = 6·(2/3)t^{3/2}|_1^2 = 4(2√2-1)=8√2-4. Total = 29/6+8√2-4 = 5/6+8√2. Numerically this is about 12.15.
3. Does f''(1) exist? Justify.
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For t<1, f''(t)=2. For t>1, f'(t)=3/√t and f''(t)=-3/(2t^{3/2}), so the right-hand second derivative at 1 is -3/2. The left-hand second derivative at 1 is 2. Since 2 ≠ -3/2, the two-sided second derivative f''(1) does not exist. The first derivative f'(1) does exist and equals 3. This shows differentiability does not imply twice differentiability.