On Partial Differentiation

This apparently isn’t on my a-level syllabus (or at least integration of this type isn’t) but, thanks to my pure teacher - who is ancient, a genius and a legend, I now know how to do it…

So differentiating with two variables

Lets take a simple equation like

z=x^2+y^2

To differentiate this we need to differentiate in respect to x and y separately. When we do this we are effectively ignoring x or y and taking a tangent of a line as if x or y was a constant so you just have a plane.

So if we want the rate of change of z in terms of x then we treat y as a constant.

dz/dx=2x

So as you can see we have differentiated the x term as normal but the y term has been treated like a constant and has thus disappeared. This is our first partial derivative. Our second partial derivative is,

dz/dy=2y

So therefore at x=4, y=2.

The gradient in the y-direction is 4 and in the x-direction is 8.

Now from this we can work out the tangent plane at x=4, y=2.

The formula for this will be,

z=mx+ny+c

Where m and n are the gradients in their respective directions.

So when x=4, y=2.

z=8x+4y+c

And

z=4^2+2^2=20

So

20=8(4)+4(2)+c

c=-20

So our tangent plane is

z=8x+4y-20

To verify this we can solve

z=8x+4y-20 And z=x^2+y^2

simultaneously and as the original formula is of degree 2 the should only be one solution, at (4,2,20).

Simultaneous equations solving

~ by Nathan Edwards on April 23, 2008.

One Response to “On Partial Differentiation”

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