AP Calculus: Applications to Physics and Economics (Work, Mass, Present/Future Value)

Applications to Physics and Economics (Work, Mass, Present/Future Value)

Concept Summary

• Work (Physics): Force applied over a distance, calculated as the integral of force with respect to displacement; measures energy transferred.
• Mass (Variable Density): Total mass of an object with non-uniform density, found by integrating density over its length/area/volume.
• Present Value (Economics): Current worth of future cash flows, discounted by an interest rate; used to compare financial options.
• Future Value (Economics): Value of an investment at a future date, compounded by an interest rate; models growth over time.
• Continuous Income Stream: A flow of money over time, modeled by integrating the rate of income with respect to time, often discounted.

Core Questions

WHAT (definitional)

Q: What is work in calculus-based physics? A: Work is the integral of a variable force ( F(x) ) over the displacement interval ([a, b]), ( W = \int_a^b F(x) \, dx ). Trap/Clarification: Work is not ( F \cdot d ) unless force is constant; always integrate for variable forces.

Q: What is present value of a continuous income stream? A: The integral of the income rate ( R(t) ) discounted by ( e^{-rt} ), ( PV = \int_0^T R(t)e^{-rt} \, dt ), where ( r ) is the interest rate. Trap/Clarification: Present value decreases as the interest rate ( r ) increases (opposite of future value).

WHY (causal/explanatory)

Q: Why is integration used for work/mass calculations? A: Integration sums infinitesimal contributions (e.g., force over tiny distances, density over tiny slices) to account for variability. Trap/Clarification: Using average values (e.g., average force) fails for non-linear distributions.

Q: Why is discounting necessary for present value? A: Money today is worth more than the same amount in the future due to potential interest earnings; discounting adjusts for this time value. Trap/Clarification: Discounting ignores inflation unless explicitly modeled (AP problems assume real rates).

HOW (process/application)

Q: How do you calculate work to pump liquid out of a tank? A: Integrate the weight of horizontal slices of liquid (( \rho g \cdot \text{volume} )) times the distance each slice is lifted: ( W = \int_a^b \rho g \cdot A(y) \cdot D(y) \, dy ). Trap/Clarification: The distance ( D(y) ) is not just ( y ); it’s the height from the slice to the outlet (often ( h - y )).

Q: How is future value of a continuous income stream calculated? A: Integrate the income rate ( R(t) ) compounded forward: ( FV = \int_0^T R(t)e^{r(T-t)} \, dt ). Trap/Clarification: The exponent is ( r(T-t) ), not ( rt ); time remaining matters.

CAN (conditions/possibilities)

Q: Can work be negative? A: Yes, if the force opposes the direction of displacement (e.g., friction). Trap/Clarification: Negative work reduces total energy but is still valid in calculations.

Q: Under what conditions is the present value of a perpetuity finite? A: When the discount rate ( r > 0 ); the integral ( \int_0^\infty R e^{-rt} \, dt ) converges to ( R/r ). Trap/Clarification: A perpetuity with ( r = 0 ) has infinite present value (no discounting).

Quick Facts & Traps

• Fact: Hooke’s Law for springs: ( F(x) = kx ); work to stretch/compress is ( \int_0^x kx \, dx = \frac{1}{2}kx^2 ).
• Trap: Units mismatch-Reality: Force must be in Newtons (kg·m/s²), distance in meters for work in Joules.
• Fact: Variable density mass: For a rod with density ( \rho(x) ), mass is ( \int_a^b \rho(x) \, dx ).
• Trap: Density as mass-Reality: Density is mass per unit length/area/volume; integrate to get total mass.
• Fact: Continuous compounding uses ( e^{rt} ); discrete compounding uses ( (1 + r/n)^{nt} ).
• Trap: Exponent sign error-Reality: Present value uses ( e^{-rt} ); future value uses ( e^{r(T-t)} ).

Rapid-Fire True/False

• Statement: The work to lift a bucket of water from a well is the same as the work to lower it. Answer: FALSE Why the common mistake happens: Ignores that force (weight) and displacement directions are opposite when lowering.

• Statement: If a rod’s density doubles, its mass doubles. Answer: TRUE (if length is unchanged) Why the common mistake happens: Assumes density is constant or forgets to integrate.

• Statement: The present value of a future sum increases as the interest rate increases. Answer: FALSE Why the common mistake happens: Confuses present value (discounted) with future value (compounded).