ACT (Enhanced)MathematicsMedium
A financial analyst is modeling the value of a stock. The value V(t) in dollars after t months is given by the function V(t) = 50 * e^(0.02t). What is the instantaneous rate of change of the stock value at t = 10 months?
- A1.00 * e^(0.2)
- B50 * e^(0.02)
- C1.00 * e^(0.02)
- D50 * e^(0.2)
Show answer & explanationAnswer & explanation
Correct answer: A. 1.00 * e^(0.2)
To find the instantaneous rate of change, we need to find the derivative of V(t). The derivative of a * e^(kx) is a * k * e^(kx). Here, a=50 and k=0.02. So, V'(t) = 50 * 0.02 * e^(0.02t) = 1 * e^(0.02t). At t=10, V'(10) = 1 * e^(0.02 * 10) = 1 * e^(0.2).
Why the other options are wrong
- B. This is the original function evaluated at t=1, not the derivative at t=10.
- C. This would be V'(1) = 1 * e^(0.02), not V'(10).
- D. This incorrectly retains the original coefficient of 50 and incorrectly applies the derivative rule.
Derivative of e^(kx)
The derivative of an exponential function of the form f(x) = a * e^(kx) is f'(x) = a * k * e^(kx).
- Used to find instantaneous rates of change.
- 'k' is the constant in the exponent.
- 'a' is the coefficient.
Memory trick: The 'K' in 'K'x 'K'omes 'K'icking out front!