A particle moves along the curve y = x^{3/2} | vedclass

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(A) Let the distance of the particle from the origin be $r$. Then $r^2 = x^2 + y^2$.Given $y = x^{3/2}$,so $y^2 = x^3$. Thus,$r^2 = x^2 + x^3$.Differe

Vedclass · Accepted Answer

$4$

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(A) Let the distance of the particle from the origin be $r$. Then $r^2 = x^2 + y^2$.Given $y = x^{3/2}$,so $y^2 = x^3$. Thus,$r^2 = x^2 + x^3$.Differentiating with respect to $t$,we get $2r \frac{dr}{dt} = (2x + 3x^2) \frac{dx}{dt}$.Given $\frac{dr}{dt} = 11$ and $x = 3$. Then $y = 3^{3/2} = 3\sqrt{3}$.The distance $r = \sqrt{x^2 + y^2} = \sqrt{3^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6$.Substituting the values into the derivative equation:$2(6)(11) = (2(3) + 3(3^2)) \frac{dx}{dt}$$132 = (6 + 27) \frac{dx}{dt}$$132 = 33 \frac{dx}{dt}$$\frac{dx}{dt} = \frac{132}{33} = 4$.

$A$ particle moves along the curve $y = x^{3/2}$ in the first quadrant in such a way that its distance from the origin increases at the rate of $11$ units per second. The value of $\frac{dx}{dt}$ when $x = 3$ is

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