Use the chain rule to find the value Use the chain rule to find the value of   $\frac{dz}{dt}$   when  $t = 3$.
$z  =  5x^2 − 2y,     x  =  t^3 + 5,     y  =  5t − 6$
I could really use some help answering this question. Please assist if you can.
This is my way of solving the question. I sub t=3 in equation x and y.
$x  =  (3)^3 + 5 = 14$
$y  =  5(3) − 6 = 9$
And then I sub z and y into the z equation.
$z  =  5(14)^2 − 2(9)=962$
I was wondering where to go from here or whether to use a different approach.
 A: Since $x$ and $y$ are both functions of $t$, it follows that $z$ is a function of $t$. Using the Multivariable Chain Rule (see this) we get
$$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}.$$
We have $\frac{\partial z}{\partial x}=10x$ and  $\frac{\partial z}{\partial y}=-2$. Also, $\frac{dx}{dt}=3t^2$ and $\frac{dy}{dt}=5$. At $t=3$, we get 
$$x=3^3+5=32\quad\text{and}\quad \frac{dx}{dt}=3(3^2)=27.$$
Thus, 
$$\frac{dz}{dt}=10(32)\cdot 27+(-2)5=8630.$$  
A: Hint: Note the difference between $\frac{dz}{dt}$ and $z$. What you have calculated is the explicit value of $z$ when $t=3$; however, the question is asking for the rate in change of z with respect to the change in t, or $\frac{dz}{dt}$.
Given a function of two variables $x$ and $y$, we use the chain rule:
\begin{equation}\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \end{equation}
The idea that motivates this chain rule is we want to use an expression that will enable us to account for the changes in each of the orthogonal directions $x$ and $y$, and then each of the changes $\Delta x$ and $\Delta y$ with respect to the change $\Delta t$ (since both $x$ and $y$ are themselves functions of $t$). Only once we've done all the calculus can we say that we've obtained $\frac{dz}{dt}$.
