# A Deep Dive Into the World Derivative of Polar Coordinates

In polar coordinates, the derivative involves both the radial \(r\) and angular \(\theta\) components. The rate of change of the Cartesian coordinates \(x\) and \(y\) is calculated using the product rule, accounting for changes in both \(r\) and \(\theta\) with respect to time.

### Derivative of Polar Coordinates

In polar coordinates, a point in the plane is represented by \((r, \theta)\), where:

- \(r\) is the radial distance from the origin.
- \(\theta\) is the angle from the positive \(x-axis\).

When differentiating in polar coordinates, we typically seek to find how the position of a point changes as a function of \(r\) and \(\theta\). This often requires converting between polar and Cartesian coordinates, which are related by the following equations:

\[

x = r \cos(\theta)

\]

\[y = r \sin(\theta)

\]

### Derivative of Polar Coordinates with Respect to Time

If a point’s polar coordinates \((r, \theta)\) change with time, then we are interested in finding the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). To compute these, we differentiate \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\) with respect to time using the **product rule** and **chain rule**.

**Derivative of \(x = r \cos(\theta)\):**

\[

\frac{dx}{dt} = \frac{d}{dt} \left(r \cos(\theta)\right)

\]

Using the product rule:

\[

\frac{dx}{dt} = \frac{dr}{dt} \cos(\theta) – r \sin(\theta) \frac{d\theta}{dt}

\]

Thus, the rate of change of \(x\) depends on both \(\frac{dr}{dt}\) (how fast \(r\) is changing) and \(\frac{d\theta}{dt}\) (how fast \(\theta\) is changing).

**Derivative of \(y = r \sin(\theta)\):**

\[

\frac{dy}{dt} = \frac{d}{dt} \left(r \sin(\theta)\right)

\]

Using the product rule again:

\[

\frac{dy}{dt} = \frac{dr}{dt} \sin(\theta) + r \cos(\theta) \frac{d\theta}{dt}

\]

These equations show how the Cartesian components \(x\) and \(y\) change as \(r\) and \(\theta\) change with time.

### Example: A Particle Moving in Polar Coordinates

Suppose a particle moves such that its radial distance changes as \(r(t) = 2t\) (i.e., \(r\) increases linearly with time), and its angular position changes as \(\theta(t) = \frac{\pi}{4}t\) (i.e., it rotates counterclockwise at a constant angular velocity). Let’s find the particle’s velocity \((\frac{dx}{dt}, \frac{dy}{dt})\) at any time \(t\).

#### Step 1: Compute \(\frac{dr}{dt}\) and \(\frac{d\theta}{dt}\)

Given \(r(t) = 2t\), we differentiate with respect to time:

\[

\frac{dr}{dt} = 2

\]

Given \(\theta(t) = \frac{\pi}{4}t\), we differentiate with respect to time:

\[

\frac{d\theta}{dt} = \frac{\pi}{4}

\]

#### Step 2: Compute \(\frac{dx}{dt}\)

Using the formula:

\[

\frac{dx}{dt} = \frac{dr}{dt} \cos(\theta) – r \sin(\theta) \frac{d\theta}{dt}

\]

Substitute \(\frac{dr}{dt} = 2\), \(\frac{d\theta}{dt} = \frac{\pi}{4}\), and \(r(t) = 2t\):

\[

\frac{dx}{dt} = 2 \cos\left(\frac{\pi}{4}t\right) – 2t \sin\left(\frac{\pi}{4}t\right) \cdot \frac{\pi}{4}

\]

This expression represents the rate of change of the (x)-coordinate at time (t).

#### Step 3: Compute \(\frac{dy}{dt}\)

Using the formula:

\[

\frac{dy}{dt} = \frac{dr}{dt} \sin(\theta) + r \cos(\theta) \frac{d\theta}{dt}

\]

Substitute the known values:

\[

\frac{dy}{dt} = 2 \sin\left(\frac{\pi}{4}t\right) + 2t \cos\left(\frac{\pi}{4}t\right) \cdot \frac{\pi}{4}

\]

This expression represents the rate of change of the \(y-coordinate\) at time \(t\).

### Geometrical Interpretation of the Velocity:

**Radial velocity**\(\frac{dr}{dt}\) represents how fast the particle is moving toward or away from the origin (in the direction of \(r\)).**Angular velocity**\(r\frac{d\theta}{dt}\) represents how fast the particle is moving around the origin in a circular path (in the direction of \(\theta)\).

The particle’s total velocity in Cartesian coordinates can be found by combining these two components, and it shows how both the radial and angular motion contribute to the overall motion.

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