Mathematics

Mathematics

Problem

If x=sin⁡tcos⁡2tx = \sin t \cos 2t and y=cos⁡tsin⁡2ty = \cos t \sin 2t, then at t=π4t = \frac{\pi}{4}, the value of dydx\frac{dy}{dx} is equal to (a) -2 (b) 2 (c) 12\frac{1}{2} (d) −12-\frac{1}{2}
Intuitive explanation

The key is to find dydx\frac{dy}{dx} using the chain rule: dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}. You'll need to differentiate both xx and yy with respect to tt using the product rule, then evaluate both derivatives at t=π4t = \frac{\pi}{4}. The trigonometric values at this angle are simple (sin⁡π4=cos⁡π4=22\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}), so the arithmetic becomes manageable once you've set up the derivatives correctly.

Intuitive explanation

The key is to find dydx\frac{dy}{dx} using the chain rule: dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}. You'll need to differentiate both xx and yy with respect to tt using the product rule, then evaluate both derivatives at t=π4t = \frac{\pi}{4}. The trigonometric values at this angle are simple (sin⁡π4=cos⁡π4=22\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}), so the arithmetic becomes manageable once you've set up the derivatives correctly.

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