1. **Problem Statement:** We have a battery level function $$B(t) = 100\left(1 - \frac{t}{10} e^{-0.2t}\right)$$ for $$t \in [0,10]$$ hours. We need to find: a. The maximum and minimum battery levels during the first 10 hours. b. The time when the battery is discharging most rapidly (greatest rate of decrease). c. Sketch the graph labeling maximum, minimum, and steepest descent points. Also, explain the product rule used in this problem. 2. **Understanding the function:** The function is a product inside the parentheses: $$\frac{t}{10} e^{-0.2t}$$. 3. **Product Rule Explanation:** The product rule states that for two functions $$u(t)$$ and $$v(t)$$, the derivative is: $$\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)$$ Here, $$u(t) = \frac{t}{10}$$ and $$v(t) = e^{-0.2t}$$. 4. **Find the derivative $$B'(t)$$ to locate extrema:** $$B(t) = 100\left(1 - u(t)v(t)\right) = 100 - 100u(t)v(t)$$ So, $$B'(t) = -100 \frac{d}{dt}[u(t)v(t)] = -100 \left(u'(t)v(t) + u(t)v'(t)\right)$$ Calculate derivatives: - $$u'(t) = \frac{1}{10}$$ - $$v'(t) = \frac{d}{dt} e^{-0.2t} = -0.2 e^{-0.2t}$$ Therefore, $$B'(t) = -100 \left( \frac{1}{10} e^{-0.2t} + \frac{t}{10} (-0.2 e^{-0.2t}) \right) = -100 \left( \frac{1}{10} e^{-0.2t} - \frac{0.2t}{10} e^{-0.2t} \right)$$ Simplify: $$B'(t) = -100 \cdot \frac{e^{-0.2t}}{10} (1 - 0.2t) = -10 e^{-0.2t} (1 - 0.2t)$$ 5. **Find critical points by setting $$B'(t) = 0$$:** Since $$e^{-0.2t} > 0$$ for all $$t$$, $$-10 e^{-0.2t} (1 - 0.2t) = 0 \implies 1 - 0.2t = 0 \implies t = 5$$ 6. **Evaluate $$B(t)$$ at endpoints and critical point:** - At $$t=0$$: $$B(0) = 100\left(1 - 0 \cdot e^{0}\right) = 100$$ - At $$t=5$$: $$B(5) = 100\left(1 - \frac{5}{10} e^{-0.2 \times 5}\right) = 100\left(1 - 0.5 e^{-1}\right)$$ Calculate $$e^{-1} \approx 0.3679$$: $$B(5) \approx 100(1 - 0.5 \times 0.3679) = 100(1 - 0.18395) = 81.605$$ - At $$t=10$$: $$B(10) = 100\left(1 - 1 \times e^{-2}\right) = 100(1 - e^{-2})$$ Calculate $$e^{-2} \approx 0.1353$$: $$B(10) \approx 100(1 - 0.1353) = 86.47$$ 7. **Determine max and min:** - $$B(0) = 100$$ (maximum) - $$B(5) \approx 81.605$$ (minimum) - $$B(10) \approx 86.47$$ So, max battery level is 100% at $$t=0$$, min is about 81.6% at $$t=5$$. 8. **Find when battery discharges most rapidly:** This is when the rate of decrease $$B'(t)$$ is most negative, i.e., minimum of $$B'(t)$$. Recall: $$B'(t) = -10 e^{-0.2t} (1 - 0.2t)$$ Find $$B''(t)$$ to locate extrema of $$B'(t)$$: Calculate: $$B''(t) = \frac{d}{dt}[-10 e^{-0.2t} (1 - 0.2t)]$$ Use product rule again with: - $$f(t) = e^{-0.2t}$$ - $$g(t) = (1 - 0.2t)$$ $$B''(t) = -10 \left(f'(t) g(t) + f(t) g'(t)\right)$$ Calculate derivatives: - $$f'(t) = -0.2 e^{-0.2t}$$ - $$g'(t) = -0.2$$ So, $$B''(t) = -10 \left(-0.2 e^{-0.2t} (1 - 0.2t) + e^{-0.2t} (-0.2)\right) = -10 e^{-0.2t} \left(-0.2 (1 - 0.2t) - 0.2\right)$$ Simplify inside parentheses: $$-0.2 (1 - 0.2t) - 0.2 = -0.2 + 0.04t - 0.2 = 0.04t - 0.4$$ Therefore, $$B''(t) = -10 e^{-0.2t} (0.04t - 0.4)$$ Set $$B''(t) = 0$$ to find extrema of $$B'(t)$$: Since $$e^{-0.2t} > 0$$, $$0.04t - 0.4 = 0 \implies t = 10$$ 9. **Check sign of $$B''(t)$$ around $$t=10$$:** - For $$t < 10$$, say $$t=9$$: $$0.04(9) - 0.4 = 0.36 - 0.4 = -0.04 < 0$$ - For $$t > 10$$, say $$t=11$$: $$0.44 - 0.4 = 0.04 > 0$$ Since $$B''(t)$$ changes from negative to positive at $$t=10$$, $$B'(t)$$ has a minimum at $$t=10$$. 10. **Evaluate $$B'(10)$$:** $$B'(10) = -10 e^{-2} (1 - 2) = -10 e^{-2} (-1) = 10 e^{-2} \approx 10 \times 0.1353 = 1.353$$ This is positive, meaning battery is increasing at $$t=10$$, which contradicts intuition. Re-examine step 8: The battery level function is decreasing initially but the derivative sign needs careful interpretation. Actually, since $$B'(t) = -10 e^{-0.2t} (1 - 0.2t)$$: - For $$t < 5$$, $$1 - 0.2t > 0$$, so $$B'(t) < 0$$ (battery decreasing). - For $$t > 5$$, $$1 - 0.2t < 0$$, so $$B'(t) > 0$$ (battery increasing). So battery discharges until $$t=5$$, then starts increasing. The steepest descent is at the minimum of $$B'(t)$$ for $$t < 5$$. Find critical points of $$B'(t)$$ in $$[0,5]$$ by setting $$B''(t) = 0$$: $$B''(t) = -10 e^{-0.2t} (0.04t - 0.4) = 0$$ $$0.04t - 0.4 = 0 \implies t=10$$ (outside $$[0,5]$$). Since $$B''(t) < 0$$ for $$t < 10$$, $$B'(t)$$ is decreasing on $$[0,5]$$. Evaluate $$B'(0)$$ and $$B'(5)$$: - $$B'(0) = -10 e^{0} (1 - 0) = -10$$ - $$B'(5) = -10 e^{-1} (1 - 1) = 0$$ Since $$B'(t)$$ increases from $$-10$$ to $$0$$ on $$[0,5]$$, the most rapid decrease is at $$t=0$$. 11. **Summary:** - Maximum battery level: 100% at $$t=0$$ - Minimum battery level: approx 81.6% at $$t=5$$ - Battery discharges most rapidly at $$t=0$$ (initial moment) 12. **Graph sketch:** - Start at 100% at $$t=0$$ - Decreases to minimum at $$t=5$$ - Then increases slightly to about 86.5% at $$t=10$$ 13. **Final answers:** - a. Max battery level = 100% at $$t=0$$, Min battery level $$\approx 81.6\%$$ at $$t=5$$ - b. Battery discharges most rapidly at $$t=0$$ - c. Graph shows a decrease from 100% to 81.6% then slight increase to 86.5% **Product rule** was used to differentiate $$u(t)v(t)$$ where $$u(t) = \frac{t}{10}$$ and $$v(t) = e^{-0.2t}$$, applying: $$\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)$$ This allowed us to find $$B'(t)$$ and analyze the battery behavior.