Evaluating a First Partial at a Point

Printable View

June 8th 2010, 04:00 AM
Shananay

Evaluating a First Partial at a Point

Evaluate $z_x$ at $(1, 2, 3)$ if $x^3+y^3+z^3+6xyz=72$

Part of me says I must solve this in terms of z in order to find $z_x$ but I can't see how.

The other part thinks I can just find $z_x$ implicitly and plug in the point but I am getting an answer that doesn't make any sense.

What should I do?
June 8th 2010, 04:20 AM
p0oint

You need to find:
$\frac{\partial z}{\partial x}$

So differentiate implicitly with respect to x, assuming that y is constant and you will get $z_x$
June 8th 2010, 04:21 AM
p0oint

You need to find:
$\frac{\partial z}{\partial x}$

So differentiate implicitly with respect to x, assuming that y is constant and you will get $z_x$
June 8th 2010, 04:23 AM
Prove It

Quote:

Originally Posted by Shananay

Evaluate $z_x$ at $(1, 2, 3)$ if $x^3+y^3+z^3+6xyz=72$

Part of me says I must solve this in terms of z in order to find $z_x$ but I can't see how.

The other part thinks I can just find $z_x$ implicitly and plug in the point but I am getting an answer that doesn't make any sense.

What should I do?

You're not going to be able to define $z$ explicitly in terms of $x$ and $y$ .

So you will need to differentiate both sides implicitly, treating $y$ as constant.
June 8th 2010, 04:26 AM
Shananay

How do I treat z?

I have something like $3x^2+3z^2(z_x)+6y(z_x)=0$ then factor and solve for $z_x$ ?

I don't feel like that's right.
June 8th 2010, 04:43 AM
Prove It

Quote:

Originally Posted by Shananay

Evaluate $z_x$ at $(1, 2, 3)$ if $x^3+y^3+z^3+6xyz=72$

Part of me says I must solve this in terms of z in order to find $z_x$ but I can't see how.

The other part thinks I can just find $z_x$ implicitly and plug in the point but I am getting an answer that doesn't make any sense.

What should I do?

We assume that $z$ is a function of $x$ , so you need to use the Chain Rule...

$x^3 + y^3 + z^3 + 6xyz = 72$

$\frac{\partial}{\partial x}(x^3 + y^3 + z^3 + 6xyz) = \frac{\partial}{\partial x}(72)$

$\frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial x}(y^3) + \frac{\partial}{\partial x}(z^3) + \frac{\partial}{\partial x}(6xyz) = 0$

$3x^2 + 0 + \frac{\partial z}{\partial x}\,\frac{\partial}{\partial z}(z^3) + 6y\,\frac{\partial}{\partial x}(xz) = 0$ (Chain Rule)

$3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6y\left[x\,\frac{\partial}{\partial x}(z) + z\,\frac{\partial}{\partial x}(x)\right] = 0$ (Product Rule)

$3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6y\left[x\,\frac{\partial z}{\partial x} + z(1)\right] = 0$

$3x^2 + 3z^2\,\frac{\partial z}{\partial x} + 6xy\,\frac{\partial z}{\partial x} + 6yz = 0$

$3z^2\,\frac{\partial z}{\partial x} + 6xy\,\frac{\partial z}{\partial x} = -(3x^2 + 6yz)$

$(3z^2 + 6xy)\frac{\partial z}{\partial x} = -(3x^2 + 6yz)$

$\frac{\partial z}{\partial x} = -\frac{3x^2 + 6yz}{3z^2 + 6xy}$

$\frac{\partial z}{\partial x} = -\frac{x^2 + 2yz}{z^2 + 2xy}$ .

Now substitute $(x, y, z) = (1, 2, 3)$ .
June 8th 2010, 04:46 AM
p0oint

Quote:

Originally Posted by Shananay

How do I treat z?

I have something like $3x^2+3z^2(z_x)+6y(z_x)=0$ then factor and solve for $z_x$ ?

I don't feel like that's right.

Treat z as a composition which contains variables x,y. So at compositions you must apply the chain rule.

Now you did well except for one mistake:

$3x^2+3z^2z'+(6xyz)'=0$

$3x^2+3z^2z'+6y(xz)'=0$

$3x^2+3z^2z'+6y(x'z+xz')=0$

Now factor for z' and substitute (1,2,3) for x,y,z or first substitute and then factor z'. :D

Regards.
June 8th 2010, 04:56 AM
Shananay

Thank you both very much. I understand now.