Writing *u* for *f*(*P*),
the equation for the level curve (or surface) is
*u* = *u*|_{P=P0}.
Writing Δ*u*
for *f*(*P* + Δ*P*) − *f*(*P*),
a quantity that depends on both a point *P* and a vector Δ*P*,
another equation for the level curve (or surface) is
Δ*u*|_{P=P0,ΔP=P−P0} =
0.
That is, you take the expression for Δ*u*,
which says how much *u* changes between two points,
put *P*_{0} in for the starting point *P*,
and then put *P* − *P*_{0} in
for the difference Δ*P* between the two points.
Since the value of *u* shouldn't change on the level curve (or surface),
this difference Δ*u* should be zero.
Notice that the meaning of *P*
changes over the course of this substitution;
originally it refers to the starting point,
which we set to *P*_{0},
but afterwards it refers to another point on the level curve (or surface),
so we set the displacement Δ*P* between the two points
to *P* − *P*_{0}.

The tangent line (or plane) is given by a very similar equation,
except that now we look at
how the curve (or surface) is changing infinitesimally at *P*_{0}
and extend this out to arbitrary distances.
Thus, the equation Δ*u* = 0 for the level curve (or surface)
becomes d*u* = 0 for the tangent line (or plane).
However, we're still looking for the values of *u* in the same place,
so the full equation is
d*u*|_{P=P0,dP=P−P0} =
0.
If you follow the definition of differential from my earlier handout,
then you'll see that this means precisely
∇*f*(*P*_{0}) ⋅
(*P* − *P*_{0}) =
0.

For example, if *u* = *x**y*
and *P*_{0} = (2, 3),
then the level curve is *x**y* = (2)(3),
or simply *x**y* = 6.
(Replace *x* with 2 and *y* with 3 on the right-hand side.)
Alternatively, Δ*u* =
(*x* + Δ*x*)(*y* + Δ*y*) −
*x**y* =
*y* Δ*x* + *x* Δ*y* +
Δ*x* Δ*y*,
so the level curve is
(3)(*x* − 2) + (2)(*y* − 3) +
(*x* − 2)(*y* − 3) =
0.
(Replace *x* with 2, *y* with 3,
Δ*x* with *x* − 2,
and Δ*y* with *y* − 3.)
This also simplifies to *x**y* = 6.

That was obviously more work than necessary for the level curve,
but now apply the same technique to the differential to get the tangent line:
d*u* = *y* d*x* + *x* d*y*,
so the tangent line is
(3)(*x* − 2) + (2)(*y* − 3) = 0.
(Replace *x* with 2, *y* with 3,
d*x* with *x* − 2,
and d*y* with *y* − 3.)
This simplifies to 3*x* + 2*y* = 12,
and now we learnt something that we didn't know before.

Because the normal line depends on the geometric notion of angle
(to tell you what's perpendicular to what),
this can't be done as slickly using only differentials.
Now we really do want to think of the gradient vector.
All the same, since this can be read off of the differential so easily,
you can still start with
d*u* = *y* d*x* + *x* d*y*.
First, replace only *x* with 2 and *y* with 3
to get 3 d*x* + 2 d*y*,
then read off the gradient vector ⟨3, 2⟩.
Since we started at the point (2, 3),
the parametric equation is
*P* = (2, 3) + *t*⟨3, 2⟩,
or (*x*, *y*) =
(3*t* + 2, 2*t* + 3)
in more detail.

None of this (beyond the level curve (or surface) itself) works right
if the gradient ∇*f*(*P*_{0}) is zero or undefined.
If the gradient is undefined, then of course we can't say anything using it;
but if the gradient is zero,
then these equations say
that every point belongs to the tangent line (or plane)
and only the point *P*_{0} belongs to the normal line.
Of course, that would mean
that they're not lines (or a plane and a line) at all!
When the gradient is zero,
the truth may be that there is no tangent
or that there is a tangent but it really does consist of everything,
or there may be an honest tangent line (or plane) after all;
but in any case, these formulas won't help you know that!

Go back to the course homepage.

This web page was written in 2016 by Toby Bartels, last edited on 2016 April 20. Toby reserves no legal rights to it.

The permanent URI of this web page
is
`http://tobybartels.name/MATH-2080/2016SP/tangents/`

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