The point on the curve 6y = x^3 + 2 at which | vedclass

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(A) Given the curve equation $6y = x^3 + 2$. Differentiating both sides with respect to $t$,we get $6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$. This simpli

Vedclass · Accepted Answer

$(4, 11)$

Vedclass · Answer

(A) Given the curve equation $6y = x^3 + 2$. Differentiating both sides with respect to $t$,we get $6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$. This simplifies to $2 \frac{dy}{dt} = x^2 \frac{dx}{dt}$. We are given that the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate,so $\frac{dy}{dt} = 8 \frac{dx}{dt}$. Substituting this into the derivative equation: $2(8 \frac{dx}{dt}) = x^2 \frac{dx}{dt}$. Assuming $\frac{dx}{dt} 
eq 0$,we have $16 = x^2$,which gives $x = 4$ or $x = -4$. If $x = 4$,then $6y = (4)^3 + 2 = 64 + 2 = 66$,so $y = 11$. The point is $(4, 11)$. If $x = -4$,then $6y = (-4)^3 + 2 = -64 + 2 = -62$,so $y = -31/3$. Comparing with the given options,the correct point is $(4, 11)$.

The point on the curve $6y = x^3 + 2$ at which the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate is:

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