Differentiating Parametric Equations: dy/dx when x a cos t - t sin t and y a sin t - t cos t

In this article, we will explore the process of finding the derivative (frac{dy}{dx}) when (x acos t - tsin t) and (y asin t - tcos t). This involves using the chain rule and implicit differentiation, showcasing a detailed step-by-step approach.

Introduction to Parametric Equations

Parametric equations are a set of equations that express the coordinates of a curve as functions of a parameter. In this case, the parameter is (t). The equations are given by:

[x acos t - tsin t] [y asin t - tcos t]

Step-by-Step Solution

Step 1: Differentiating x and y with Respect to t

To find (frac{dy}{dx}), we first need to differentiate both (x) and (y) with respect to (t). We will use standard differentiation rules and the product rule as necessary.

Differentiating x with respect to t

[x acos t - tsin t]

Using the product rule and incorporating the constant factor (a): [frac{dx}{dt} afrac{d}{dt}(cos t) - frac{d}{dt}(tsin t)] Applying the derivatives: [frac{dx}{dt} a(-sin t) - (sin t - tcos t)] [frac{dx}{dt} asin t - sin t tcos t] [frac{dx}{dt} asin t - sin t tcos t atcos t]

Differentiating y with respect to t

[y asin t - tcos t]

Again, using the product rule and the constant factor (a): [frac{dy}{dt} afrac{d}{dt}(sin t) - frac{d}{dt}(tcos t)] Applying the derivatives: [frac{dy}{dt} acos t - (cos t - tsin t)] [frac{dy}{dt} acos t - cos t tsin t] [frac{dy}{dt} acos t - cos t tsin t atsin t]

Step 2: Finding (frac{dy}{dx}) Using the Chain Rule

The chain rule states that: [frac{dy}{dx} frac{frac{dy}{dt}}{frac{dx}{dt}}] Substituting the derivatives obtained from the previous step: [frac{dy}{dx} frac{atsin t}{atcos t} frac{sin t}{cos t} tan t]

Conclusion

Thus, the derivative (frac{dy}{dx}) when (x acos t - tsin t) and (y asin t - tcos t) is: (frac{dy}{dx} tan t)

This result shows that the relationship between (y) and (x) in terms of the parameter (t) can be expressed using the tangent function.