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EXERCISE 1.4 · Q167

Q.If x=acos⁡3tx=a\cos^3t, y=asin⁡3ty=a\sin^3t, show that dydx=−(yx)1/3\dfrac{dy}{dx}=-\left(\dfrac{y}{x}\right)^{1/3}.

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We have x=acos⁡3tx=a\cos^3t, y=asin⁡3ty=a\sin^3t.

Step 1. dxdt=−3acos⁡2tsin⁡t\dfrac{dx}{dt}=-3a\cos^2t\sin t

Step 2. dydt=3asin⁡2tcos⁡t\dfrac{dy}{dt}=3a\sin^2t\cos t

Step 3.

dydx=3asin⁡2tcos⁡t−3acos⁡2tsin⁡t=−sin⁡tcos⁡t=−tan⁡t\frac{dy}{dx}=\frac{3a\sin^2t\cos t}{-3a\cos^2t\sin t}=-\frac{\sin t}{\cos t}=-\tan t

Step 4. Also, yx=asin⁡3tacos⁡3t=tan⁡3t\dfrac{y}{x}=\dfrac{a\sin^3t}{a\cos^3t}=\tan^3t, so …

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