Consider a box with dimensions x, y, and z. x is changing at a rate of 1 m/s, y at -2 m/s and z at 1 m/s. Find the rate that the volume, surface area and diagonal length ($s = \sqrt{x^2+y^2+z^2}$) are changing at the instant when $x = 4$, $y = 3$ and $z = 2$.
I know I need to use to product rule. But my teacher didn't bother to teach it to us how to do it with three variables (ex. xyz for the volume). So I'm having trouble figuring out how to start this problem.
$\endgroup$2 Answers
$\begingroup$Write out your equations for volume, surface area, and diagonal length and take the derivative with respect to time. I assume you have gone over this in class, but for the surface area it should look like:
$\frac{d}{dt}s(x,y,z) = \frac{\partial s}{\partial x} \frac{dx}{dt} + \frac{\partial s}{\partial y}\frac{dy}{dt} + \frac{\partial s}{\partial z}\frac{dz}{dt}$. Then, you have equations for $\frac{dx}{dt}$ etc. From there it is just plug and chug.
$\endgroup$ 11 $\begingroup$If you know how to find the derivative of a product of $2$ terms, you can find the derivative of the product of any number of terms.
For example, let us find $(xyz)'$. So we want to find the derivative of the product $(xy)z$ of two terms.
By the product rule for two terms, the derivative is $(xy)z'+(xy)'z$. Applying the two variable product rule to $(xy)'$, we get that our derivative is $$(xy)z' +(xy'+x'y)z,$$ that is, $$xyz'+xy'z+x'yz.$$
In the same way, one finds that the derivative of $wxyz$ is $$wxyz'+wxy'z+wx'yz+w'xyz.$$ The obvious pattern continues for a product of $5$ terms, $6$, and so on.
$\endgroup$ 3