Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Interpretations of
Partial Derivatives

This is a fairly short section and is here so we can acknowledge that the two main interpretations
of derivatives of functions of a single variable still hold for partial derivatives, with small
modifications of course to account of the fact that we now have more than one variable
...
As with
functions of single variables partial derivatives represent the rates of change of the functions as
the variables change
...


Example 1 Determine if f ( x, y )

x2
is increasing or decreasing at ( 2,5 ) ,
3
y

(a) if we allow x to vary and hold y fixed
...

Solution
(a) If we allow x to vary and hold y fixed
...


2x
f x ( x, y ) = 3
y

4
f x ( 2,5 ) =
>0
125

⇒

So, the partial derivative with respect to x is positive and so if we hold y fixed the function is
increasing at ( 2,5 ) as we vary x
...

For this part we will need f y ( x, y ) and its value at the point
...

Note that it is completely possible for a function to be increasing for a fixed y and decreasing for
a fixed x at a point as this example has shown
...


This is a graph of a hyperbolic paraboloid and we at the origin we can see that if we move in
along the y-axis the graph is increasing and if we move along the x-axis the graph is decreasing
...
We should never expect that the function will behave in exactly
the same way at a point as each variable changes
...
We know
from a Calculus I class that f ′ ( a ) represents the slope of the tangent line to y = f ( x ) at

x = a
...
The difference
here is the functions that they represent tangent lines to
...
The partial derivative f x ( a,b ) is the slope of the
trace of f ( x, y ) for the plane y = b at the point ( a,b )
...


Example 2 Find the slopes of the traces to z = 10 − 4x 2 − y 2 at the point (1, 2 )
...
For reference purposes here are the graphs of the traces
...


f x ( x, y ) = −8x

f y ( x, y ) = −2 y

To get the slopes all we need to do is evaluate the partial derivatives at the point in question
...
Also the tangent line at (1, 2 ) for the trace to z = 10 − 4x 2 − y 2 for the plane x = 1 has a
slope of -4
...
Recall that the equation
of a line in 3-D space is given by a vector equation
...

The point is easy
...
So, the point will be,

( a,b, f ( a,b ) )

The parallel (or tangent) vector is also just as easy
...
The same will hold true here
...
e
...


So, here is the tangent vector for traces with fixed y
...
Therefore the first component becomes a 1
and the second becomes a zero because we are treating y as a constant when we differentiate with
respect to x
...

For traces with fixed x the tangent vector is,

ry ( x, y ) = 0,1, f y ( x, y )

The equation for the tangent line to traces with fixed y is then,

r ( t ) = a,b, f ( a,b ) + t 1, 0, f x ( a,b )

and the tangent line to traces with fixed x is,

r ( t ) = a,b, f ( a,b ) + t 0,1, f y ( a,b )

Example 3 Write down the vector equations of the tangent lines to the traces to
z = 10 − 4x 2 − y 2 at the point (1, 2 )
...
We’ve already computed the derivatives and their values at (1, 2 ) in the
previous example and the point on each trace is,

(1, 2, f (1, 2 ) ) = (1, 2, 2 )

Here is the equation of the tangent line to the trace for the plane y = 2
...


r ( t ) = 1, 2, 2 + t 0,1, −4 = 1, 2 + t, 2 − 4t



---