ACT (Enhanced)MathematicsHard
A financial analyst is modeling the value of a stock. The value V(t) in dollars after t months is given by V(t) = 50e^(0.04t). What is the instantaneous rate of change of the stock value after 6 months?
- A50.00 dollars/month
- B2.00 dollars/month
- C0.04 dollars/month
- D2.54 dollars/month
Show answer & explanationAnswer & explanation
Correct answer: D. 2.54 dollars/month
The instantaneous rate of change is given by the derivative of the function. For V(t) = Ae^(kt), the derivative V'(t) = Ake^(kt). Calculate V'(6) using this formula.
Why the other options are wrong
- A. Incorrect. This is the initial value of the stock, not its rate of change.
- B. Incorrect. This is 50 * 0.04 = 2, which is the initial rate of change (at t=0) but not at t=6 months.
- C. Incorrect. This is the growth rate constant (k), not the rate of change of the stock value.
Derivative of e^(kx)
The derivative of an exponential function f(x) = Ae^(kx) with respect to x is f'(x) = Ake^(kx). This represents the instantaneous rate of change.
- The derivative of e^(kx) is k * e^(kx).
- The constant A multiplies the derivative.
- It's used to find instantaneous growth or decay rates.
Memory trick: The 'Exponent Power' becomes the 'Multiplier' in the 'Rate Change'.