A particle moves in space along the path z = | vedclass

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Question

Given that $:$

$x-axis$ is perpendicular to plane inwards

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{c} \quad 	he

Vedclass · Accepted Answer

$(6ac^2x + 2bc^2 ) \, \widehat k$

Vedclass · Answer

Given that $:$

$x-axis$ is perpendicular to plane inwards

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{c} \quad 	herefore \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=0$

Further $z=a x^{3}+b y^{2}$

$	herefore \quad \frac{d z}{d t}=3 a x^{2} \frac{d x}{d t}+2 b y \frac{d y}{d t}$

$=3 a c x^{2}+2 b c y\left[\frac{d x}{d t}=c=\frac{d y}{d t}ight]$

$	herefore \frac{\mathrm{d}^{2} z}{\mathrm{dt}^{2}}=6 \mathrm{acx}\left[\frac{\mathrm{dx}}{\mathrm{dt}}ight]+2 \mathrm{bc}\left[\frac{\mathrm{dy}}{\mathrm{dt}}ight]=6 \mathrm{ac}^{2} \mathrm{x}+2 \mathrm{bc}^{2}$

Now acceleration of particle is

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}} \hat{\mathrm{i}}+\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}} \hat{\mathrm{j}}+\frac{\mathrm{d}^{2} \mathrm{z}}{\mathrm{dt}^{2}} \hat{\mathrm{k}}=\left(6 \mathrm{ac}^{2} \mathrm{x}+2 \mathrm{bc}^{2}ight) \hat{\mathrm{k}}$

A particle moves in space along the path $z = ax^3 + by^2$ in such a way that $\frac{dx}{dt} = c = \frac{dy}{dt}.$ Where $a, b$ and $c$ are contants. The acceleration of the particle is

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