Partial derivative with chain rule

Question

$$z (x,y) = xy^2 + x^2y ,\quad x(t)= at^2 ,\quad y(t)= 2at $$ Find $\dfrac{dz}{dt}$

First , this is the chain rule formula I am using

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

I found the following

$$\frac{\partial z}{\partial x} = y^2 + 2xy$$ therefore, $z_x = 4a^2t^2 + 4a^2t^3. $

$$ \frac{\partial z}{\partial y} = 2xy + x^2 $$ therefore, $z_y =4a^2t^3 + a^2t^4. $

$$ \frac{dx}{dt} = t^2 + 2at $$

$$ \frac{dy}{dt} = 2t + 2a $$

I substituted all the values into the above chain rule formula and got $$ 8a^2t^4 + 12a^3t^3 + 6a^2t^5 + 10a^3t^4 $$

This is not the required answer which is $16a^3t^3 + 10a^3t^4 $

Where have I gone wrong ?

Answer 1

Well,$$x=at^2\implies \frac{dx}{dt}=2at\\y=2at\implies \frac{dy}{dt}=2a$$assuming $a$ is a constant. If it is not, then you need $\frac{da}{dt}$ terms which are not included (but I believe it probably is a constant).

Answer 2

$\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ are correct, but

$\frac{dx}{dt} = \frac{d(at^2)}{dt}=2at$

$\frac{dy}{dt} = \frac{d(2at)}{dt}=2a$

Now you'll find the correct answer.

Answer 3

i already know Mathjax & Latex code here's the answer,this is easy you only have to substitute the x's and y's

$$z(x,y)=at^2{\left(2at\right)}^2+{\left(at^2\right)}^2(2at)$$ $$z(t)=4a^3t^4+2a^3t^5$$ $$\frac{d}{dt}\left[z(t)=4a^3t^4+2a^3t^5\right]$$ $$\mathbf{\frac{dz}{dt}=16a^3t^3+10a^3t^4}$$

That's the answer