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Two climate scientists are creating models for global rising sea levels. One scientist proposes a 10-year doubling model, which predicts that the change in sea level, 
R(t), will grow according to the following function:

R(t)=(125)/(256)*(2)^((t)/( 10))
where 
t is the number of years after 2005 . The other scientist proposes a steady increase model where the change in sea level will increase by 5 centimeters 
(cm) per year. Both models predict that in the year 2105 , the sea level will increase by 
500cm. To the nearest centimeter, how much greater is the predictions of the steady increase model than the 10 -year doubling model for the change in sea level in the year 2055 ?

Two climate scientists are creating models for global rising sea levels. One scientist proposes a 1010-year doubling model, which predicts that the change in sea level, R(t) R(t) , will grow according to the following function:\newlineR(t)=125256(2)t10 R(t)=\frac{125}{256} \cdot(2)^{\frac{t}{10}} \newlinewhere t t is the number of years after 20052005 . The other scientist proposes a steady increase model where the change in sea level will increase by 55 centimeters (cm) (\mathrm{cm}) per year. Both models predict that in the year 21052105 , the sea level will increase by 500 cm 500 \mathrm{~cm} . To the nearest centimeter, how much greater is the predictions of the steady increase model than the 1010 -year doubling model for the change in sea level in the year 20552055 ?

Full solution

Q. Two climate scientists are creating models for global rising sea levels. One scientist proposes a 1010-year doubling model, which predicts that the change in sea level, R(t) R(t) , will grow according to the following function:\newlineR(t)=125256(2)t10 R(t)=\frac{125}{256} \cdot(2)^{\frac{t}{10}} \newlinewhere t t is the number of years after 20052005 . The other scientist proposes a steady increase model where the change in sea level will increase by 55 centimeters (cm) (\mathrm{cm}) per year. Both models predict that in the year 21052105 , the sea level will increase by 500 cm 500 \mathrm{~cm} . To the nearest centimeter, how much greater is the predictions of the steady increase model than the 1010 -year doubling model for the change in sea level in the year 20552055 ?
  1. Calculate predicted change in sea level: First, let's calculate the predicted change in sea level for the year 20552055 using the 1010-year doubling model.\newlineWe need to find R(t)R(t) for t=20552005=50t = 2055 - 2005 = 50 years.\newlineR(t)=(125256)(2)t10R(t) = (\frac{125}{256}) \cdot (2)^{\frac{t}{10}}
  2. Substitute t=50t = 50 into the equation: Substitute t=50t = 50 into the equation to find R(50)R(50).
    R(50)=125256×(2)5010R(50) = \frac{125}{256} \times (2)^{\frac{50}{10}}
    R(50)=125256×(2)5R(50) = \frac{125}{256} \times (2)^5
    R(50)=125256×32R(50) = \frac{125}{256} \times 32
  3. Perform multiplication to find R(50)R(50): Now, perform the multiplication to find the value of R(50)R(50).
    R(50)=(125256)×32R(50) = (\frac{125}{256}) \times 32
    R(50)=125×(32256)R(50) = 125 \times (\frac{32}{256})
    R(50)=125×(18)R(50) = 125 \times (\frac{1}{8})
    R(50)=1258R(50) = \frac{125}{8}
    R(50)=15.625cmR(50) = 15.625 \, \text{cm}
  4. Calculate predicted change using steady increase model: Next, calculate the predicted change in sea level for the year 20552055 using the steady increase model.\newlineThe steady increase model predicts a 5cm5\,\text{cm} increase per year.\newlineSince 20552055 is 5050 years after 20052005, the total increase would be 50years×5cm/year.50\,\text{years} \times 5\,\text{cm/year}.
  5. Perform multiplication to find total increase: Perform the multiplication to find the total increase for the steady increase model.\newlineTotal increase = 5050 years * 55 cm/year\newlineTotal increase = 250250 cm
  6. Find difference between models for 20552055: Now, we need to find the difference between the steady increase model and the 1010-year doubling model for the year 20552055.
    Difference =Total increase(steady model)R(50)(10-year doubling model)= \text{Total increase} (\text{steady model}) - R(50) (\text{10-year doubling model})
    Difference =250cm15.625cm= 250 \, \text{cm} - 15.625 \, \text{cm}
  7. Subtract values to find the difference: Subtract the values to find the difference.\newlineDifference = 250cm15.625cm250 \, \text{cm} - 15.625 \, \text{cm}\newlineDifference = 234.375cm234.375 \, \text{cm}
  8. Round the difference to nearest centimeter: Round the difference to the nearest centimeter.\newlineDifference 234\approx 234 cm

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