Question

Your parents bought you a brand new car at graduation! Well, we can dream, right? Assume that the value of the car is given by the function, V(t) = 25,000e⁻⁰.¹ᵗ, where v is the value, in dollars, of the car and t is the time, in years, from the date of purchase. A. How much was the car worth at the dealership (t = 0)?

B. How much did the car depreciate during its fourth year (from t = 3 to t = 4)?
C. When you graduate from college, assuming you are on the 4-year plan, what is the instantaneous rate of change in dollars per year) of the value at t = 4 ? Why is this different than your answer in part B?
D. When will the value of the car reach $12,500?

Answer 1

A. The car was worth $25,000 at the dealership (t = 0). B. The depreciation during the fourth year can be calculated using the difference in value between t = 3 and t = 4. C. The instantaneous rate of change in dollars per year at t = 4 can be found using the derivative of the given function. D. The time when the value of the car reaches $12,500 can be found by setting V(t) = 12,500 in the given function and solving for t.

Step-by-step explanation:

A. To find the value of the car at the dealership (t = 0), we can substitute t = 0 into the given function: V(0) = 25,000e^(-0.1*0) = 25,000e^0 = 25,000 * 1 = 25,000 dollars. Therefore, the car was worth $25,000 at the dealership (t = 0).

B. To find how much the car depreciated during its fourth year (from t = 3 to t = 4), we can calculate the difference in value between these two-time points: V(3) - V(4) = 25,000e^(-0.1*3) - 25,000e^(-0.1*4). Using a calculator, we can find this difference.

C. The instantaneous rate of change in dollars per year of the value of the car at t = 4 can be found by calculating the derivative of the given function concerning t and evaluating it at t = 4. This gives us the slope of the tangent line to the curve at that point. The reason this is different from the depreciation during the fourth year is that depreciation is calculated over a specific time interval, while the instantaneous rate of change represents the rate of change at a specific point in time.

D. To find when the value of the car will reach $12,500, we can set V(t) = 12,500 in the given function and solve for t: 12,500 = 25,000e^(-0.1*t). We can rearrange this equation and use logarithms to solve for t.

Answer 2

The initial value of the car is $25,000, and the amount of depreciation during the fourth year, along with the instantaneous rate of change at t = 4, can be determined through calculations involving the given function V(t). The time at which the car's value reaches $12,500 is computed using logarithms.

Step-by-step explanation:

Calculating the Value of a Car Over Time

Let's address each part of the student's question step by step:

Initial value of the car (t = 0): To determine the value of the car at the dealership, we plug t = 0 into the provided function V(t) = 25,000e⁻⁰ⁱ¹. Since e⁻⁰ⁱ¹ is 1 when t=0, the value V(0) is simply $25,000.

Depreciation during the fourth year (from t = 3 to t = 4): We calculate V(3) and V(4) using the function and find the difference. V(4) will be lower than V(3), indicating a depreciation over the year.

Instantaneous rate of change at t = 4: This requires differentiating the function V(t) to find V'(t) and then evaluating V'(4) to find the rate of depreciation at exactly t = 4 years.

Time until the car's value is $12,500: We need to solve the equation V(t) = 12,500 for t. This will involve natural logarithms due to the exponential function.

Note that the instantaneous rate of change is different from the depreciation value between t = 3 and t = 4 because the former is a rate at an exact moment, while the latter is a measure of change over an interval of time.

The value of the car is continuously decreasing, so the rate of change at any instant is slightly different from average changes over an interval.