A particle moves in space alone the path $z=a{x}^3+b{y}^2$ 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

Given that $\frac{dx}{dt}=\frac{dy}{dt}=c\therefore \frac{d^{2}x}{d{t}^2}=\frac{d^{2}y}{d{t}^2}=0$

$\begin{array}{l}Furtherz=a{x}^3+b{y}^2\\ \therefore \frac{dz}{dt}=3a{x}^2\frac{dx}{dt}+2by\frac{dy}{dt}\\ =3ac{x}^2+2bcy\left(\frac{dx}{dt}=c=\frac{dy}{dt}\right)\\ \therefore \frac{d^{2}z}{d{t}^2}=6acx\left(\frac{dx}{dt}\right)+2bc\left(\frac{dy}{dt}\right)=6a{c}^{2}x+2b{c}^2\\ Now,accelerationofparticleis\\ \overrightarrow{a}=\frac{d^{2}x}{d{t}^2}i+\frac{d^{2}y}{d{t}^2}\widehat{j}+\frac{d^{2}z}{d{t}^2}\widehat{k}\\ =(6a{c}^{2}x+2b{c}^2)\widehat{k}\end{array}$