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Difficulty: Very hardLinear and Exponential Growth

The value of a financial asset, V(t)V(t), in dollars, is modeled as a function of time tt, in years, for 0≤t≤40 \le t \le 4. The growth of the asset's value is described by two different models over this period:

- For 0≤t≤20 \le t \le 2, the value of the asset increases by a constant percentage of r%r\% per year.
- For 2≤t≤42 \le t \le 4, the value of the asset increases by a constant amount of dd dollars per year.

At t=0t = 0, the value of the asset is 100100 dollars, and at t=1t = 1, the value of the asset is 120120 dollars. If the average rate of change of the asset's value from t=0t = 0 to t=4t = 4 is 2525 dollars per year, what is the value of the asset, in dollars, at t=3t = 3?

  1. A
    169
  2. B
    170
  3. 172Answer
  4. D
    175

Answer

172
To find the value of the asset at t=3t = 3, we first determine its value at t=2t = 2 using the exponential growth model. Since the value increases by a constant percentage from t=0t=0 to t=2t=2, the value is modeled by V(t)=V(0)×(1+r100)tV(t) = V(0) \times (1 + \frac{r}{100})^t. Given V(0)=100V(0) = 100 and V(1)=120V(1) = 120, the annual growth factor is 1.21.2. Thus, V(2)=100×(1.2)2=144V(2) = 100 \times (1.2)^2 = 144 dollars.

Next, the linear growth from t=2t = 2 to t=4t = 4 is modeled by V(t)=V(2)+d(t−2)=144+d(t−2)V(t) = V(2) + d(t - 2) = 144 + d(t - 2), which gives V(4)=144+2dV(4) = 144 + 2d dollars. The average rate of change of the asset's value from t=0t = 0 to t=4t = 4 is 2525 dollars per year, so we set up the equation: V(4)−V(0)4=25  ⟹  (144+2d)−1004=25\frac{V(4) - V(0)}{4} = 25 \implies \frac{(144 + 2d) - 100}{4} = 25. Solving for dd gives d=28d = 28 dollars per year. Finally, we evaluate the value at t=3t = 3: V(3)=144+28(3−2)=172V(3) = 144 + 28(3 - 2) = 172 dollars.

Step-by-Step Solution

1
Identify the growth model for 0≤t≤20 \le t \le 2 and write the corresponding equation using the initial values.
Since the value increases by a constant percentage each year, the growth is exponential: V(t)=V(0)×(1+r100)tV(t) = V(0) \times (1 + \frac{r}{100})^t. Given V(0)=100V(0) = 100 and V(1)=120V(1) = 120, we have 120=100(1+r100)1  ⟹  1+r100=1.2120 = 100(1 + \frac{r}{100})^1 \implies 1 + \frac{r}{100} = 1.2.
Establishing the exponential model allows us to find the growth factor and calculate the value of the asset at the boundary of the two models (t=2t = 2).
2
Calculate the value of the asset at t=2t = 2 using the exponential growth model.
V(2)=100×(1.2)2=144V(2) = 100 \times (1.2)^2 = 144 dollars.
The value at t=2t = 2 serves as the initial value for the linear growth model in the second interval.
3
Write the growth model for the interval 2≤t≤42 \le t \le 4.
The growth is linear, so V(t)=V(2)+d(t−2)=144+d(t−2)V(t) = V(2) + d(t - 2) = 144 + d(t - 2) for t≥2t \ge 2. At t=4t = 4, the value is V(4)=144+2dV(4) = 144 + 2d dollars.
Expressing the value at t=4t = 4 in terms of the constant rate dd allows us to use the average rate of change information to solve for dd.
4
Set up the equation for the average rate of change over the entire interval from t=0t = 0 to t=4t = 4 and solve for dd.
The average rate of change is given by V(4)−V(0)4−0=25\frac{V(4) - V(0)}{4 - 0} = 25. Substituting the values, we get (144+2d)−1004=25  ⟹  44+2d4=25  ⟹  44+2d=100  ⟹  2d=56  ⟹  d=28\frac{(144 + 2d) - 100}{4} = 25 \implies \frac{44 + 2d}{4} = 25 \implies 44 + 2d = 100 \implies 2d = 56 \implies d = 28 dollars per year.
The average rate of change relates the final value at t=4t=4 to the initial value at t=0t=0, enabling us to determine the linear rate of change dd.
5
Calculate the value of the asset at t=3t = 3 using the linear model.
V(3)=V(2)+d(3−2)=144+28(1)=172V(3) = V(2) + d(3 - 2) = 144 + 28(1) = 172 dollars.
Substituting t=3t = 3 into the linear growth model yields the final requested value.

Key Concept

Linear and Exponential Growth

Alternative Method

Instead of setting up the linear equation from t=2t = 2 to t=4t = 4, we can write V(4)V(4) directly as V(0)+4×25=200V(0) + 4 \times 25 = 200 dollars using the definition of average rate of change. Since V(2)=144V(2) = 144 dollars, the total change over the linear interval [2,4][2, 4] is 200−144=56200 - 144 = 56 dollars. Because the growth is linear during this interval, the change over one year (from t=2t = 2 to t=3t = 3) must be exactly half of the change over two years (from t=2t = 2 to t=4t = 4). Thus, the increase from t=2t = 2 to t=3t = 3 is 56/2=2856 / 2 = 28 dollars, yielding V(3)=144+28=172V(3) = 144 + 28 = 172 dollars.
Estimated Time:3m 0s
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Topics

Linear and Exponential Growth

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