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Exercise 2.1 · Q13

Q.A particle moves along the curve 6y=x3+26y = x^3 + 2. Find the points on the curve at which the yy-coordinate is changing 88 times as fast as the xx-coordinate.

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Given 6y=x3+26y=x^3+2. Since dydt=dydx⋅dxdt\dfrac{dy}{dt}=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}, the condition "yy-coordinate changing 88 times as fast as the xx-coordinate" means dydt=8dxdt\dfrac{dy}{dt}=8\dfrac{dx}{dt}, i.e. dydx=8\dfrac{dy}{dx}=8.

Differentiating the curve: 6dydx=3x2⇒dydx=x226\dfrac{dy}{dx}=3x^2 \Rightarrow \dfrac{dy}{dx}=\dfrac{x^2}{2}.

Setting x22=8⇒x2=16⇒x=±4\dfrac{x^2}{2}=8 \Rightarrow x^2=16 \Rightarrow x=\pm4. …

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