Chapter 6: Applications of Integration – Velocity, Net Change, and Physical Applications

6.1 Velocity and Net Change

Displacement and Distance Traveled

In calculus, integration is used to analyze motion along a straight line by relating position, velocity, and acceleration. The displacement of an object is the net change in its position over a time interval, while the distance traveled is the total length of the path, regardless of direction.

• Displacement: If s(t) is the position function, the displacement over [a, b] is s(b) - s(a).

• Distance Traveled: The total distance is the integral of the speed (the absolute value of velocity): $\int_a^b |v(t)|\,dt$

• Velocity and Position: Velocity is the derivative of position: v(t) = s'(t).

• Fundamental Theorem of Calculus (FTC): Relates the integral of velocity to displacement: $\int_a^b v(t)\,dt = s(b) - s(a)$

Example: Jogger's Motion

Consider a jogger with velocity function v(t) = 2t^2 - 8t + 6 for 0 \leq t \leq 3 (in mi/hr). We analyze the motion as follows:

• Direction of Motion: The jogger moves in the positive direction when v(t) > 0 and in the negative direction when v(t) < 0.

• Displacement on Intervals:

  • On [0, 1]: $\int_0^1 v(t)\,dt = \frac{8}{3}$

  • On [1, 3]: $\int_1^3 v(t)\,dt = -\frac{8}{3}$

  • On [0, 3]: $\int_0^3 v(t)\,dt = 0$

• Distance Traveled: $\int_0^3 |v(t)|\,dt = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}$

Position from Velocity

Given the velocity function and initial position, the position at any time t is found by integrating velocity:

• Theorem 6.1: $s(t) = s(0) + \int_0^t v(x)\,dx$

Example: If v(t) = \frac{1}{4} \sin t, s(0) = -\frac{1}{4}, then $s(t) = -\frac{1}{4} + \int_0^t \frac{1}{4} \sin x\,dx = -\frac{1}{4} \cos t$

Piecewise Velocity Example

For a skydiver with a piecewise velocity function, the total distance is the sum of integrals over each interval:

• $D = \int_0^{19} 80\,dt + \int_{19}^{21} (783 - 37t)\,dt + \int_{21}^{40} 6\,dt$

Velocity from Acceleration

Given acceleration a(t) and initial velocity v(0), the velocity is:

• $v(t) = v(0) + \int_0^t a(x)\,dx$

Example: Projectile Motion

An artillery shell is fired upward with v(0) = 300 m/s from s(0) = 30 m. Gravity acts with a(t) = -9.8 m/s2.

• Velocity: $v(t) = 300 - 9.8t$

• Position: $s(t) = 30 + 300t - 4.9t^2$

• Highest Point: Occurs when $v(t) = 0 \Rightarrow t \approx 30.6$ s, $s(30.6) = 4621.8$ m

• Time to Ground: Solve $30 + 300t - 4.9t^2 = 0 \Rightarrow t \approx 61.3$ s

6.1 Net Change and Future Value

Net Change Theorem

The Net Change Theorem states that the net change in a quantity Q(t) over [a, b] is the integral of its rate of change:

• $Q(b) - Q(a) = \int_a^b Q'(t)\,dt$

• Future value from initial value: $Q(b) = Q(a) + \int_a^b Q'(t)\,dt$

Example: Population Growth

A town has initial population 250 and grows at rate $P'(t) = 30(1 + \sqrt{t})$.

• Net Change (10 to 20 years): $P(20) - P(10) = \int_{10}^{20} 30(1 + \sqrt{t})\,dt = 30(t + \frac{2}{3} t^{3/2})\Big|_{10}^{20} \approx 1456$

• Population at time t: $P(t) = 250 + 30(t + \frac{2}{3} t^{3/2})$

Summary Tables

Velocity–Displacement Problems
Position: $s(t)$
Velocity: $s'(t) = v(t)$
Displacement: $s(b) - s(a) = \int_a^b v(t)\,dt$
Future position: $s(t) = s(0) + \int_0^t v(x)\,dx$

General Problems
Quantity: $Q(t)$ (e.g., volume, population)
Rate of change: $Q'(t)$
Net change: $Q(b) - Q(a) = \int_a^b Q'(t)\,dt$
Future value: $Q(t) = Q(0) + \int_0^t Q'(x)\,dx$