Understanding Applications of Calculus in Physics

Calculus is a fundamental tool in physics, enabling us to understand motion, change, and the behavior of physical systems. In this article, we will explore practical examples that demonstrate how calculus is used to solve real-world physics problems.
By Jamie

1. Motion and Velocity

In physics, understanding motion is crucial. Calculus allows us to analyze how objects move over time. Let’s take the example of a car accelerating.

Example: Acceleration of a Car

  • Problem: A car’s position is given by the function \(s(t) = 4t^2 + 2t\), where \(s\) is in meters and \(t\) is in seconds.

  • Solution: To find the velocity, we differentiate the position function with respect to time:

    [
    v(t) = \frac{ds}{dt} = \frac{d}{dt}(4t^2 + 2t) = 8t + 2
    ]

  • Interpretation: The velocity function \(v(t) = 8t + 2\) tells us how fast the car is moving at any time \(t\). For example, at \(t = 3\) seconds, the velocity is \(v(3) = 8 \cdot 3 + 2 = 26\) m/s.

2. Area Under the Curve

Calculus helps us find areas under curves, which has numerous applications in physics, such as calculating work done.

Example: Work Done by a Variable Force

  • Problem: A force \(F(x) = 3x^2\) (in Newtons) is applied to move an object along a straight line from \(x = 0\) to \(x = 4\) meters.

  • Solution: The work done is the integral of the force over the displacement:

    [
    W = \int_{0}^{4} F(x) \, dx = \int_{0}^{4} 3x^2 \, dx
    ]

    Calculating the integral:

    [
    W = 3 \left[ \frac{x^3}{3} \right]_{0}^{4} = 3 \left( \frac{4^3}{3} - \frac{0^3}{3} \right) = 3 \cdot \frac{64}{3} = 64 \, \text{Joules}
    ]

  • Interpretation: The work done by the force to move the object is 64 Joules.

3. Projectile Motion

Calculus is essential in analyzing projectile motion, which involves both horizontal and vertical components.

Example: Maximum Height of a Projectile

  • Problem: A projectile is launched with an initial velocity of \(v_0 = 20\) m/s at an angle of \(45^\circ\). The height function is given by:

    [
    h(t) = v_0 \sin(\theta) t - \frac{1}{2}gt^2
    ]
    where \(g = 9.81 \, \text{m/s}^2\).

  • Solution: First, calculate \(h(t)\):

    [
    h(t) = 20 \sin(45^\circ) t - \frac{1}{2} \cdot 9.81 t^2
    ]

    Using \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\):

    [
    h(t) = 20 \cdot \frac{\sqrt{2}}{2} t - 4.905 t^2
    ]

    To find the maximum height, set the first derivative equal to zero:

    [
    \frac{dh}{dt} = 10\sqrt{2} - 9.81t = 0 \Rightarrow t = \frac{10\sqrt{2}}{9.81} \approx 1.44 \, \text{seconds}
    ]

    Now, substitute \(t\) back into \(h(t)\):

    [
    h(1.44) = 20 \cdot \frac{\sqrt{2}}{2} \cdot 1.44 - 4.905 \cdot (1.44)^2 \approx 14.4 \, \text{meters}
    ]

  • Interpretation: The maximum height reached by the projectile is approximately 14.4 meters.

Conclusion

These examples illustrate how calculus is applied in various physics problems, from analyzing motion to calculating work and understanding projectile dynamics. Mastering these concepts enables a deeper comprehension of the physical world.