A financial analyst is modeling the value of an investment. The value V(t) of the investment, in thousands of dollars, after t years, is given by V(t) = 10 * (1.05)^t. What is the instantaneous rate of change of the investment's value at t = 2 years?
- A11.025 thousand dollars/year
- B0.525 thousand dollars/year
- C1.05 thousand dollars/year
- D0.531 thousand dollars/year
Show answer & explanationAnswer & explanation
Correct answer: D. 0.531 thousand dollars/year
The instantaneous rate of change is given by the derivative of the function. For V(t) = 10 * (1.05)^t, the derivative V'(t) is 10 * (1.05)^t * ln(1.05). Now, evaluate V'(2): V'(2) = 10 * (1.05)^2 * ln(1.05) = 10 * 1.1025 * 0.04879 ≈ 0.531. (Note: On the Digital SAT, questions requiring derivatives will often provide a formula or context that allows approximation, or the numbers will be simpler. Given this is a Digital SAT item, the expectation is that a calculator is available and the concept of derivative (even if not explicitly named) is being tested via rate of change.)
Why the other options are wrong
- A. This is the value of V(2), the investment value, not its rate of change.
- B. This might be 10 * ln(1.05) * 1.05, a slight miscalculation of the exponent.
- C. This is likely 10 * ln(1.05) or a similar misinterpretation.
Instantaneous Rate of Change of Exponential Functions
The instantaneous rate of change of an exponential function f(x) = a * b^x is given by its derivative, f'(x) = a * b^x * ln(b). This measures how fast the function's value is changing at a specific point.
- Requires calculus (derivatives).
- The natural logarithm (ln) of the base is a key component.
- Evaluated at a specific x-value.
Memory trick: Derivative of A B to the X is A B to the X Times Natural Log B!