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(A) Given that $\frac{dx}{dt} = c$ and $\frac{dy}{dt} = c$. Since $c$ is a constant,the second derivatives are $\frac{d^2x}{dt^2} = 0$ and $\frac{d^2y

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

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(A) Given that $\frac{dx}{dt} = c$ and $\frac{dy}{dt} = c$. Since $c$ is a constant,the second derivatives are $\frac{d^2x}{dt^2} = 0$ and $\frac{d^2y}{dt^2} = 0$.The path is given by $z = ax^3 + by^2$.To find the velocity component in the $z$-direction,differentiate with respect to time $t$:$\frac{dz}{dt} = 3ax^2 \frac{dx}{dt} + 2by \frac{dy}{dt} = 3ax^2(c) + 2by(c) = 3acx^2 + 2bcy$.To find the acceleration component in the $z$-direction,differentiate $\frac{dz}{dt}$ with respect to time $t$:$\frac{d^2z}{dt^2} = \frac{d}{dt}(3acx^2 + 2bcy) = 3ac(2x \frac{dx}{dt}) + 2bc(\frac{dy}{dt})$.Substituting $\frac{dx}{dt} = c$ and $\frac{dy}{dt} = c$:$\frac{d^2z}{dt^2} = 6acx(c) + 2bc(c) = 6ac^2x + 2bc^2$.The acceleration vector is $\vec{a} = \frac{d^2x}{dt^2} \hat{i} + \frac{d^2y}{dt^2} \hat{j} + \frac{d^2z}{dt^2} \hat{k}$.Substituting the values,we get $\vec{a} = 0 \hat{i} + 0 \hat{j} + (6ac^2x + 2bc^2) \hat{k} = (6ac^2x + 2bc^2) \hat{k}$.

$A$ 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 constants. The acceleration of the particle is:

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