19.3 Applications of antiderivatives

Antiderivatives are powerful tools for understanding motion and solving differential equations. They let us find position from velocity, velocity from acceleration, and calculate distances traveled. This process is like working backwards from the rate of change to the original function.

Separable differential equations take this idea further, using antiderivatives to solve complex problems. By separating variables and integrating both sides, we can find solutions to equations that describe real-world phenomena. It's like unscrambling a puzzle to reveal the underlying relationships.

Antiderivatives and Their Applications

Position from velocity function

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• Antiderivatives determine position functions from given velocity functions
  • Velocity $v(t)$ measures rate of change of position with respect to time
  • Position $s(t)$ is the antiderivative of velocity
• Integrate velocity function with respect to time to find position function
  • $s(t) = \int v(t) dt$
  • Constant of integration $[C](https://www.fiveableKeyTerm:c)$ represents initial position at time $t=0$
• Example: Given $v(t) = 3t^2 + 2t$, position function is $s(t) = \int (3t^2 + 2t) dt = t^3 + t^2 + C$
  • If initial position is known (e.g., $s(0) = 1$), solve for $C$ to determine specific position function

Velocity from acceleration function

• Antiderivatives find velocity functions from given acceleration functions
  • Acceleration $a(t)$ measures rate of change of velocity with respect to time
  • Velocity $v(t)$ is the antiderivative of acceleration
• Integrate acceleration function with respect to time to find velocity function
  • $v(t) = \int a(t) dt$
  • Constant of integration $C$ represents initial velocity at time $t=0$
• Example: Given $a(t) = 6t + 2$, velocity function is $v(t) = \int (6t + 2) dt = 3t^2 + 2t + C$
  • If initial velocity is known (e.g., $v(0) = 5$), solve for $C$ to determine specific velocity function

Applications of antiderivatives in motion

• Antiderivatives calculate distance, displacement, and total distance traveled
  • Distance is total path length traveled regardless of direction
  • Displacement is shortest distance between start and end points
  • Total distance traveled sums absolute values of distances in each direction
• Find distance by integrating absolute value of velocity function over time interval
  • $\text{Distance} = \int_{t_1}^{t_2} |v(t)| dt$
  • Example: $\int_0^2 |3t| dt = 6$ units of distance traveled from $t=0$ to $t=2$
• Find displacement by evaluating position function at start and end times and subtracting
  • $\text{Displacement} = s(t_2) - s(t_1)$
  • Example: If $s(t) = t^2 + 1$, displacement from $t=1$ to $t=4$ is $s(4) - s(1) = 15$
• Find total distance traveled by splitting time interval into subintervals where velocity doesn't change sign
  • Calculate distances in each subinterval and add together
  • Example: If $v(t) = t-2$ from $t=0$ to $t=4$, split into $[0,2]$ and $[2,4]$, find distances, and add

Solving separable differential equations

• Separable differential equations are solved using antiderivatives
  • Separable differential equation has form $\frac{dy}{dx} = f(x)g(y)$
  • Variables can be separated with $x$ terms on one side and $y$ terms on other
• Steps to solve separable differential equation:
  1. Separate variables by moving $x$ terms to one side and $y$ terms to other side
  2. Integrate both sides of equation with respect to their variables
  3. Solve for $y$ as function of $x$ using constant of integration $C$
• Example: Solve differential equation $\frac{dy}{dx} = xy$
  1. Separate variables: $\frac{dy}{y} = x dx$
  2. Integrate both sides: $\int \frac{dy}{y} = \int x dx$ gives $\ln |y| = \frac{1}{2}x^2 + C$
  3. Solve for $y$: $y = \pm e^{\frac{1}{2}x^2 + C}$ (note the $\pm$ from absolute value)