Climate change. The most abundant greenhouse gas is carbon dioxide. According to the UN's "worst-case scenario" prediction the amount of carbon dioxide in the atmosphere (in parts of volume per million) can be approximated by \( C(t) \approx 277 \mathrm{e}^{0.00353 t} \) where \( t \) is the time in years since 1750 . a. What does the model predict the amount of carbon dioxide in the atmosphere is now? How accurate is this? b. What does the model predict the amount of carbon dioxide in the atmosphere will be in 2025 ? 2050 ? 2100 ? c. According to the model when will the level surpass 700 parts per million?

**a. Prediction for “now”** Assume “now” refers to the current year 2023. Then the time since 1750 is \[ t=2023-1750=273. \] The model gives \[ C(273)=277e^{0.00353\cdot 273}. \] First, compute the exponent: \[ 0.00353\times273\approx0.96369. \] Then, \[ e^{0.96369}\approx2.621. \] Thus, \[ C(273)\approx277\times2.621\approx726. \] So the model predicts about 726 parts per million (ppm) of carbon dioxide in 2023. In reality, current measurements are around 415–420 ppm. This shows that the model (which is based on a “worst‑case scenario”) overestimates the present-day concentration. --- **b. Predictions for future years** 1. **For 2025:** The time since 1750 is \[ t=2025-1750=275. \] Then, \[ C(275)=277e^{0.00353\cdot 275}. \] Compute the exponent: \[ 0.00353\times275\approx0.97075 \quad \text{and} \quad e^{0.97075}\approx2.639. \] Thus, \[ C(275)\approx277\times2.639\approx732 \text{ ppm}. \] 2. **For 2050:** Here, \[ t=2050-1750=300. \] Then, \[ C(300)=277e^{0.00353\cdot 300}. \] Compute the exponent: \[ 0.00353\times300=1.059 \quad \text{and} \quad e^{1.059}\approx2.884. \] Thus, \[ C(300)\approx277\times2.884\approx800 \text{ ppm}. \] 3. **For 2100:** Here, \[ t=2100-1750=350. \] Then, \[ C(350)=277e^{0.00353\cdot 350}. \] Compute the exponent: \[ 0.00353\times350\approx1.2355 \quad \text{and} \quad e^{1.2355}\approx3.439. \] Thus, \[ C(350)\approx277\times3.439\approx953 \text{ ppm}. \] --- **c. When will the level surpass 700 ppm?** We need to solve the equation \[ 277e^{0.00353t}=700. \] Dividing both sides by 277 gives \[ e^{0.00353t}=\frac{700}{277}\approx2.529. \] Taking the natural logarithm of both sides: \[ 0.00353t=\ln(2.529). \] Calculate the logarithm: \[ \ln(2.529)\approx0.928. \] Then, solving for \(t\): \[ t\approx\frac{0.928}{0.00353}\approx263. \] Since \(t\) is the number of years after 1750, the level reaches 700 ppm in the year \[ 1750+263\approx2013. \]