1. **Problem statement:** Given the drug entry rate function $$R(t) = \frac{80t}{t^2 + 4}$$ for $$t \geq 0$$, answer the following: (a) Find the rate of drug entering the bloodstream after a long time. (b) Find when the drug entry rate is maximum and determine that maximum rate. (c) Find the total amount of drug that enters the bloodstream in the first 6 hours. 2. **Formula and rules:** - To find the long-term rate, evaluate the limit $$\lim_{t \to \infty} R(t)$$. - To find the maximum rate, find critical points by setting the derivative $$R'(t)$$ to zero and check for maxima. - To find total amount entered over time interval $$[0,6]$$, integrate $$R(t)$$ over that interval. 3. **Step (a): Long-term rate** $$ \lim_{t \to \infty} \frac{80t}{t^2 + 4} = \lim_{t \to \infty} \frac{80t}{t^2(1 + \frac{4}{t^2})} = \lim_{t \to \infty} \frac{80}{t(1 + \frac{4}{t^2})} = 0 $$ So, the rate approaches zero as time goes to infinity. 4. **Step (b): Maximum rate** Find $$R'(t)$$: $$ R(t) = \frac{80t}{t^2 + 4} = 80t (t^2 + 4)^{-1} $$ Using quotient rule or product rule: $$ R'(t) = \frac{(80)(t^2 + 4) - 80t(2t)}{(t^2 + 4)^2} = \frac{80(t^2 + 4) - 160t^2}{(t^2 + 4)^2} = \frac{80t^2 + 320 - 160t^2}{(t^2 + 4)^2} = \frac{-80t^2 + 320}{(t^2 + 4)^2} $$ Set numerator to zero for critical points: $$ -80t^2 + 320 = 0 \implies t^2 = 4 \implies t = 2 \text{ (since } t \geq 0\text{)} $$ Check if maximum by second derivative or sign test: For $$t < 2$$, numerator positive; for $$t > 2$$, numerator negative, so $$t=2$$ is a maximum. Calculate maximum rate: $$ R(2) = \frac{80 \times 2}{2^2 + 4} = \frac{160}{4 + 4} = \frac{160}{8} = 20 $$ 5. **Step (c): Total amount in first 6 hours** Calculate: $$ \int_0^6 R(t) dt = \int_0^6 \frac{80t}{t^2 + 4} dt $$ Use substitution: let $$u = t^2 + 4$$, then $$du = 2t dt$$, so $$t dt = \frac{du}{2}$$. Rewrite integral: $$ \int_0^6 \frac{80t}{t^2 + 4} dt = 80 \int_0^6 \frac{t}{t^2 + 4} dt = 80 \int_{u=4}^{u=6^2+4=40} \frac{1}{u} \cdot \frac{du}{2} = 40 \int_4^{40} \frac{1}{u} du = 40 [\ln|u|]_4^{40} = 40 (\ln 40 - \ln 4) = 40 \ln \frac{40}{4} = 40 \ln 10 $$ Numerical approximation: $$ 40 \ln 10 \approx 40 \times 2.302585 = 92.1034 $$ **Final answers:** (a) Rate after a long time: $$0$$ mg/hr (b) Maximum rate is $$20$$ mg/hr at $$t = 2$$ hours (c) Total amount in first 6 hours: $$40 \ln 10 \approx 92.1$$ mg