finding $\frac{dz}{dt}$ of $z(x,y)= x^2y^3$ , $x(t)= 2t^3$ , $y(t) = 3t^2$

Question

finding $\frac{dz}{dt}$ of $z(x,y)= x^2y^3$ , $x(t)= 2t^3$ , $y(t) = 3t^2$

First I found

$\frac{\partial z}{\partial x} = 2xy^3$

$\frac{\partial z}{\partial y} = 3 x^2 y^2$

$ x’(t) = 6t^2 $

$y’(t)= 6t $

$\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} $

And thus I got $\frac{dz}{dt} = 12 xy^3t^2 + 18y^2x^2t$

But the answer is $1296t^{11}$

Where have I went wrong ?

Answer 1

You should substitute back the $x(t)$ and $y(t)$: $$12xy^3t^2+18y^2x^2t=$$ $$12*2t^3*(3t^2)^3*t^2+18*(3t^2)^2*(2t^3)^2*t=$$ $$(12*2*3^3+18*3^2*2^2)t^{11}=$$ $$1296t^{11}$$