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Worked Examples · Example 31

Q.Find dydx\frac{dy}{dx}, if x=acos⁡θx = a\cos\theta, y=asin⁡θy = a\sin\theta.

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✓ Free question

For parametric equations x=acos⁡θx = a\cos\theta, y=asin⁡θy = a\sin\theta, the derivative dydx\frac{dy}{dx} is found by dividing dydθ\frac{dy}{d\theta} by dxdθ\frac{dx}{d\theta}, giving dydx=−cot⁡θ\frac{dy}{dx} = -\cot\theta.

When a curve is given in parametric form — where both xx and yy are expressed in terms of a third variable (here θ\theta) — you cannot directly differentiate yy with respect to xx as a single function. Instead, you use the chain rule in reverse:

dydx=dy/dθdx/dθ\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}

provided dx/dθ≠0dx/d\theta \neq 0. This works because dy/dθ=(dy/dx)⋅(dx/dθ)dy/d\theta = (dy/dx) \cdot (dx/d\theta), so solving for dy/dxdy/dx gives the ratio.

Let’s apply this to the given equations.

  1. Differentiate xx with respect to θ\theta x=acos⁡θx = a\cos\theta The derivative of cos⁡θ\cos\theta is −sin⁡θ-\sin\theta, so

dxdθ=−asin⁡θ\frac{dx}{d\theta} = -a\sin\theta

  1. Differentiate yy with respect to θ\theta y=asin⁡θy = a\sin\theta The derivative of sin⁡θ\sin\theta is cos⁡θ\cos\theta, so

dydθ=acos⁡θ\frac{dy}{d\theta} = a\cos\theta

  1. Form the ratio

dydx=dy/dθdx/dθ=acos⁡θ−asin⁡θ\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\cos\theta}{-a\sin\theta}

The aa cancels (provided a≠0a \neq 0, which is true for a non-degenerate circle), leaving

dydx=−cos⁡θsin⁡θ=−cot⁡θ\frac{dy}{dx} = -\frac{\cos\theta}{\sin\theta} = -\cot\theta

Watch out

A common mistake is to forget the negative sign from the derivative of cos⁡θ\cos\theta, or to accidentally invert the ratio. Always check: dydx\frac{dy}{dx} should be the slope of the tangent — for a circle x2+y2=a2x^2 + y^2 = a^2, the slope at angle θ\theta is −cot⁡θ-\cot\theta, which matches.

Tip

You can verify this result by eliminating θ\theta: x2+y2=a2x^2 + y^2 = a^2 (a circle). Implicit differentiation gives 2x+2ydydx=02x + 2y \frac{dy}{dx} = 0, so dydx=−x/y\frac{dy}{dx} = -x/y. Substituting x=acos⁡θx = a\cos\theta, y=asin⁡θy = a\sin\theta yields −cot⁡θ-\cot\theta — consistent.

✓Final answer

The derivative is −cot⁡θ\boxed{-\cot\theta}.

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