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Exercise 5.6 · Q10

Q.Find dydx\frac{dy}{dx} in the following: x=a(cos⁡θ+θsin⁡θ),y=a(sin⁡θ−θcos⁡θ)x = a (\cos \theta + \theta \sin \theta), y = a (\sin \theta - \theta \cos \theta)

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For parametric equations, dydx=dy/dθdx/dθ\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}. Here, differentiating and simplifying gives dydx=tan⁡θ\frac{dy}{dx} = \tan \theta.

We are given xx and yy as functions of a parameter θ\theta, not directly as y=f(x)y = f(x). This is a classic parametric differentiation problem. The chain rule tells us that if both xx and yy are differentiable functions of θ\theta, then

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

provided dx/dθ≠0dx/d\theta \neq 0. The intuition is simple: a small change in θ\theta causes a small change in both xx and yy; the ratio of those changes (as the change shrinks to zero) gives the slope of the curve at that point.

Let’s compute each derivative carefully.

  1. Differentiate xx with respect to θ\theta. x=a(cos⁡θ+θsin⁡θ)x = a(\cos \theta + \theta \sin \theta). The constant aa factors out. Differentiate term by term:
    • Derivative of cos⁡θ\cos \theta is −sin⁡θ-\sin \theta.
    • For θsin⁡θ\theta \sin \theta, use the product rule: derivative is (1)(sin⁡θ)+(θ)(cos⁡θ)=sin⁡θ+θcos⁡θ(1)(\sin \theta) + (\theta)(\cos \theta) = \sin \theta + \theta \cos \theta. So

dxdθ=a(−sin⁡θ+sin⁡θ+θcos⁡θ)=aθcos⁡θ.\frac{dx}{d\theta} = a\bigl(-\sin \theta + \sin \theta + \theta \cos \theta\bigr) = a \theta \cos \theta.

  1. Differentiate yy with respect to θ\theta. y=a(sin⁡θ−θcos⁡θ)y = a(\sin \theta - \theta \cos \theta). Again aa factors out. Differentiate:
    • Derivative of sin⁡θ\sin \theta is cos⁡θ\cos \theta.
    • For −θcos⁡θ-\theta \cos \theta, product rule: derivative is −(1)(cos⁡θ)−(θ)(−sin⁡θ)=−cos⁡θ+θsin⁡θ-(1)(\cos \theta) - (\theta)(-\sin \theta) = -\cos \theta + \theta \sin \theta. So

dydθ=a(cos⁡θ−cos⁡θ+θsin⁡θ)=aθsin⁡θ.\frac{dy}{d\theta} = a\bigl(\cos \theta - \cos \theta + \theta \sin \theta\bigr) = a \theta \sin \theta.

  1. Form the ratio. dydx=aθsin⁡θaθcos⁡θ=sin⁡θcos⁡θ=tan⁡θ,\frac{dy}{dx} = \frac{a \theta \sin \theta}{a \theta \cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta, …

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