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A particle moves along a curve 3y2=6x3+83y^2 = 6x^3 + 8 such that its y-coordinate is changing at the rate of 9 cm/sec at point (1,2). Then the rate of a change of x-coordinate at the same point is :

Solution

✅ Correct Option: 4

Differentiate with respect to tt: 6ydydt=18x2dxdt6y\dfrac{dy}{dt}=18x^2\dfrac{dx}{dt}.

At (1,2)(1,2) with dydt=9\dfrac{dy}{dt}=9: 6(2)(9)=18(1)dxdt6(2)(9)=18(1)\dfrac{dx}{dt}.

So dxdt=10818=6\dfrac{dx}{dt}=\dfrac{108}{18}=6 cm/sec.

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